ML Aggarwal Class 9 Maths Solutions Chapter 01 Rational and Irrational Numbers

Exercise 1.1

Question 1. Insert a rational number between \( \frac{2}{9} \) and \( \frac{3}{8} \) and arrange in descending order.
Answer: The L.C.M of 9 and 8 is 72.
\( \frac{2}{9} = \frac{2 \times 8}{9 \times 8} = \frac{16}{72} \)
\( \frac{3}{8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72} \)
Since 16 < 27, \( \frac{2}{9} < \frac{3}{8} \)
A rational number between \( \frac{2}{9} \) and \( \frac{3}{8} \)
\( = \frac{\frac{2}{9} + \frac{3}{8}}{2} = \frac{\frac{16 + 27}{72}}{2} = \frac{43}{144} \)
Numbers in descending order are: \( \frac{3}{8}, \frac{43}{144}, \frac{2}{9} \)
In simple words: Find a number exactly in the middle of two fractions. To do this, add them together and divide by 2. Then arrange all three from biggest to smallest.

Exam Tip: Always check which fraction is larger before finding the middle number. Use L.C.M to compare fractions easily.

Question 2. Insert two rational numbers between \( \frac{1}{4} \) and \( \frac{1}{3} \) and arrange in ascending order.
Answer: The L.C.M of 3 and 4 is 12.
\( \frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12} \)
\( \frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12} \)
Since 3 < 4, \( \frac{1}{4} < \frac{1}{3} \)
A rational number between \( \frac{1}{4} \) and \( \frac{1}{3} \)
\( = \frac{\frac{1}{4} + \frac{1}{3}}{2} = \frac{\frac{4 + 3}{12}}{2} = \frac{7}{24} \)
A rational number between \( \frac{1}{4} \) and \( \frac{7}{24} \)
\( = \frac{\frac{1}{4} + \frac{7}{24}}{2} = \frac{\frac{6 + 7}{24}}{2} = \frac{13}{48} \)
Numbers in ascending order are: \( \frac{1}{4}, \frac{13}{48}, \frac{7}{24}, \frac{1}{3} \)
In simple words: Insert two numbers between the given fractions by finding middle values step by step. Arrange them from smallest to largest.

Exam Tip: Each new rational number is found by adding two consecutive fractions and dividing by 2. This method always works.

Question 3. Insert two rational numbers between \( -\frac{1}{3} \) and \( -\frac{1}{2} \) and arrange in ascending order.
Answer: The L.C.M of 2 and 3 is 6.
\( -\frac{1}{3} = -\frac{1 \times 2}{3 \times 2} = -\frac{2}{6} \)
\( -\frac{1}{2} = -\frac{1 \times 3}{2 \times 3} = -\frac{3}{6} \)
Since - 3 < - 2, \( -\frac{1}{2} < -\frac{1}{3} \)
A rational number between \( -\frac{1}{2} \) and \( -\frac{1}{3} \)
\( = \frac{-\frac{1}{2} + (-\frac{1}{3})}{2} = \frac{\frac{-3 + (-2)}{6}}{2} = \frac{\frac{-3 - 2}{6}}{2} = -\frac{5}{12} \)
A rational number between \( -\frac{1}{2} \) and \( -\frac{5}{12} \)
\( = \frac{-\frac{1}{2} + (-\frac{5}{12})}{2} = \frac{\frac{-6 + (-5)}{12}}{2} = \frac{\frac{-6 - 5}{12}}{2} = -\frac{11}{24} \)
Numbers in ascending order are: \( -\frac{1}{2}, -\frac{11}{24}, -\frac{5}{12}, -\frac{1}{3} \)
In simple words: With negative fractions, remember that smaller numbers are more negative. So arrange them from most negative to least negative.

Exam Tip: Negative numbers work opposite to positive ones - the larger the negative value, the smaller the number itself. Keep this in mind when ordering.

Question 4. Insert three rational numbers between \( \frac{1}{3} \) and \( \frac{4}{5} \) and arrange in descending order.
Answer: The L.C.M of 3 and 5 is 15.
\( \frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15} \)
\( \frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15} \)
Since 5 < 12, \( \frac{1}{3} < \frac{4}{5} \)
A rational number between \( \frac{1}{3} \) and \( \frac{4}{5} \)
\( = \frac{\frac{1}{3} + \frac{4}{5}}{2} = \frac{\frac{5 + 12}{15}}{2} = \frac{17}{30} \)
A rational number between \( \frac{1}{3} \) and \( \frac{17}{30} \)
\( = \frac{\frac{1}{3} + \frac{17}{30}}{2} = \frac{\frac{10 + 17}{30}}{2} = \frac{27}{60} \)
A rational number between \( \frac{17}{30} \) and \( \frac{4}{5} \)
\( = \frac{\frac{17}{30} + \frac{4}{5}}{2} = \frac{\frac{17 + 24}{30}}{2} = \frac{41}{60} \)
Numbers in descending order are: \( \frac{4}{5}, \frac{41}{60}, \frac{17}{30}, \frac{27}{60}, \frac{1}{3} \)
In simple words: To insert three numbers, keep finding middle values between the given fractions and the numbers you already found. Then arrange from largest to smallest.

Exam Tip: The method of finding averages always produces a rational number between any two given rational numbers. This is a powerful tool for this type of question.

Question 5. Using the concept of decimals, insert (i) three rational numbers between 3 and 3.5 (ii) four rational numbers between \( \frac{1}{4} \) and \( \frac{2}{5} \) (iii) five rational numbers between \( 1\frac{1}{2} \) and \( 1\frac{3}{4} \)
Answer: (i) We need three rational numbers between 3 and 3.5. Since terminating decimals are rational numbers, we can pick: 3.1, 3.2, 3.3. We have \( 3 < 3.1 < 3.2 < 3.3 < 3.5 \). Thus, the three rational numbers between 3 and 3.5 are 3.1, 3.2, and 3.3.
(ii) Converting to decimal form: \( \frac{1}{4} = 0.25 \) and \( \frac{2}{5} = 0.40 \). We can pick: 0.28, 0.30, 0.32, 0.34. We have \( 0.25 < 0.28 < 0.30 < 0.32 < 0.34 < 0.40 \). Thus, four rational numbers between \( \frac{1}{4} \) and \( \frac{2}{5} \) are 0.28, 0.30, 0.32, and 0.34.
(iii) We need five rational numbers between \( 1\frac{1}{2} \) and \( 1\frac{3}{4} \), which equals 1.50 and 1.75. We can pick: 1.61, 1.62, 1.63, 1.64, 1.65. We have \( 1.50 < 1.61 < 1.62 < 1.63 < 1.64 < 1.65 < 1.75 \). Thus, five rational numbers between \( 1\frac{1}{2} \) and \( 1\frac{3}{4} \) are 1.61, 1.62, 1.63, 1.64, and 1.65.
In simple words: Decimals are easy to work with because you can simply choose any decimal between your two numbers. Just pick decimals that fall in the middle range.

Exam Tip: The decimal method is the quickest way to insert rational numbers. Always convert fractions to decimals first if asked to use the decimal method.

Question 6. Find six rational numbers between 3 and 4.
Answer: Write 3 and 4 as \( \frac{3}{1} \) and \( \frac{4}{1} \). Since we want six rational numbers between these, multiply the numerator and denominator by 6 + 1 = 7, giving \( \frac{21}{7} \) and \( \frac{28}{7} \), which are equivalent to the original numbers. As \( 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28 \), we have \( \frac{21}{7} < \frac{22}{7} < \frac{23}{7} < \frac{24}{7} < \frac{25}{7} < \frac{26}{7} < \frac{27}{7} < \frac{28}{7} \). Therefore, six rational numbers between 3 and 4 are: \( \frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7} \)
In simple words: Write whole numbers as fractions with 1 as the bottom. Multiply top and bottom by (n+1), where n is the count of numbers you need. This creates enough fractions between them.

Exam Tip: This method is efficient for finding many rational numbers at once. The key is multiplying by (n+1), not n - don't forget to add 1.

Question 7. Find five rational numbers between \( \frac{3}{5} \) and \( \frac{4}{5} \).
Answer: Since we need five rational numbers between these fractions, multiply the numerator and denominator by 5 + 1 = 6, giving \( \frac{18}{30} \) and \( \frac{24}{30} \), which are equivalent. As \( 18 < 19 < 20 < 21 < 22 < 23 < 24 \), we have \( \frac{18}{30} < \frac{19}{30} < \frac{20}{30} < \frac{21}{30} < \frac{22}{30} < \frac{23}{30} < \frac{24}{30} \). This simplifies to: \( \frac{3}{5} < \frac{19}{30} < \frac{2}{3} < \frac{7}{10} < \frac{11}{15} < \frac{23}{30} < \frac{4}{5} \). Therefore, five rational numbers between \( \frac{3}{5} \) and \( \frac{4}{5} \) are: \( \frac{19}{30}, \frac{2}{3}, \frac{7}{10}, \frac{11}{15}, \frac{23}{30} \)
In simple words: Multiply both the top and bottom of your fractions by 6. This gives you new fractions with the same value but with enough room between them to fit five more fractions.

Exam Tip: Always simplify the final fractions if possible, as examiners prefer simplified form. Check your work by converting to decimals and verifying the order.

Question 8. Find ten rational numbers between \( -\frac{2}{5} \) and \( \frac{1}{7} \).
Answer: Write the given numbers with the same denominator 35 (L.C.M. of 5 and 7): \( -\frac{2}{5} = -\frac{14}{35} \) and \( \frac{1}{7} = \frac{5}{35} \). As \( -14 < -13 < -12 < -11 < -10 < -9 < -8 < -7 < 0 < 1 < 2 < 5 \), we have \( -\frac{14}{35} < -\frac{13}{35} < -\frac{12}{35} < -\frac{11}{35} < -\frac{10}{35} < -\frac{9}{35} < -\frac{8}{35} < -\frac{7}{35} < 0 < \frac{1}{35} < \frac{2}{35} < \frac{5}{35} \). This simplifies to: \( -\frac{2}{5} < -\frac{13}{35} < -\frac{12}{35} < -\frac{11}{35} < -\frac{2}{7} < -\frac{9}{35} < -\frac{8}{35} < -\frac{1}{5} < 0 < \frac{1}{35} < \frac{2}{35} < \frac{1}{7} \). Therefore, ten rational numbers between \( -\frac{2}{5} \) and \( \frac{1}{7} \) are: \( -\frac{13}{35}, -\frac{12}{35}, -\frac{11}{35}, -\frac{2}{7}, -\frac{9}{35}, -\frac{8}{35}, -\frac{1}{5}, 0, \frac{1}{35}, \frac{2}{35} \)
In simple words: Find a common bottom number for the two fractions. Then count upward through the whole numbers between the new tops. Each step gives you a new rational number.

Exam Tip: When dealing with negative and positive numbers together, remember that negative fractions come before zero, which comes before positive fractions. Always simplify fractions at the end.

Question 9. Find six rational numbers between \( \frac{1}{2} \) and \( \frac{2}{3} \).
Answer: Write the given numbers with the same denominator 6 (L.C.M. of 2 and 3): \( \frac{1}{2} = \frac{3}{6} \) and \( \frac{2}{3} = \frac{4}{6} \). Since we need six rational numbers, multiply the numerator and denominator by 6 + 1 = 7, giving \( \frac{21}{42} \) and \( \frac{28}{42} \). As \( 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28 \), we have \( \frac{21}{42} < \frac{22}{42} < \frac{23}{42} < \frac{24}{42} < \frac{25}{42} < \frac{26}{42} < \frac{27}{42} < \frac{28}{42} \). This simplifies to: \( \frac{1}{2} < \frac{11}{21} < \frac{23}{42} < \frac{4}{7} < \frac{25}{42} < \frac{13}{21} < \frac{9}{14} < \frac{2}{3} \). Therefore, six rational numbers between \( \frac{1}{2} \) and \( \frac{2}{3} \) are: \( \frac{11}{21}, \frac{23}{42}, \frac{4}{7}, \frac{25}{42}, \frac{13}{21}, \frac{9}{14} \)
In simple words: Use the multiplication method with 7. This creates enough fractions packed between your two numbers that you can pick six of them.

Exam Tip: Always simplify the final answers when possible. Check that you have the correct count of rational numbers before finalizing your answer.

Exercise 1.2

Question 1. Prove that \( \sqrt{5} \) is an irrational number.
Answer: Assume \( \sqrt{5} \) is a rational number. Then \( \sqrt{5} = \frac{p}{q} \), where p and q are integers, q ≠ 0, and p, q have no common factors except 1. Squaring both sides: \( 5 = \frac{p^2}{q^2} \), which gives \( p^2 = 5q^2 \) ...(i). Since 5 divides \( 5q^2 \), 5 must divide \( p^2 \). As 5 is prime, 5 divides p. Let p = 5m, where m is an integer. Substituting into (i): \( (5m)^2 = 5q^2 \), so \( 25m^2 = 5q^2 \), giving \( 5m^2 = q^2 \). Since 5 divides \( 5m^2 \), 5 must divide \( q^2 \). As 5 is prime, 5 divides q. But now p and q share a common factor of 5, contradicting our assumption. Therefore, \( \sqrt{5} \) is not a rational number, so it must be irrational.
In simple words: Suppose the square root is a fraction. Show that both the top and bottom must be divisible by 5. This means they share a common factor, which is impossible. So square roots of non-perfect squares are irrational.

Exam Tip: This is a proof by contradiction. Always start by assuming the opposite of what you want to prove. Each step should follow logically from the previous one.

Question 2. Prove that \( \sqrt{7} \) is an irrational number.
Answer: Assume \( \sqrt{7} \) is a rational number. Then \( \sqrt{7} = \frac{p}{q} \), where p and q are integers, q ≠ 0, and p, q have no common factors except 1. Squaring both sides: \( 7 = \frac{p^2}{q^2} \), which gives \( p^2 = 7q^2 \) ...(i). Since 7 divides \( 7q^2 \), 7 must divide \( p^2 \). As 7 is prime, 7 divides p. Let p = 7m, where m is an integer. Substituting into (i): \( (7m)^2 = 7q^2 \), so \( 49m^2 = 7q^2 \), giving \( 7m^2 = q^2 \). Since 7 divides \( 7m^2 \), 7 must divide \( q^2 \). As 7 is prime, 7 divides q. But now p and q share a common factor of 7, which contradicts our assumption. Therefore, \( \sqrt{7} \) cannot be a rational number, so it is irrational.
In simple words: Use the same method as for square root of 5. Show that if square root of 7 were a fraction, both numerator and denominator would be divisible by 7, creating a shared factor that shouldn't exist.

Exam Tip: The pattern is identical for all primes. If n is prime, then the square root of n is irrational. Memorize the steps: assume rational, square both sides, show prime divides both numerator and denominator, reach contradiction.

Question 3. Prove that \( \sqrt{6} \) is an irrational number.
Answer: Assume \( \sqrt{6} \) is a rational number. Then \( \sqrt{6} = \frac{p}{q} \), where p and q are integers, q ≠ 0, and p, q have no common factors except 1. Squaring both sides: \( 6 = \frac{p^2}{q^2} \), which gives \( p^2 = 6q^2 \) ...(i). Since 2 divides \( 6q^2 \), 2 must divide \( p^2 \). As 2 is prime, 2 divides p. Let p = 2k, where k is an integer. Substituting into (i): \( (2k)^2 = 6q^2 \), so \( 4k^2 = 6q^2 \), giving \( 2k^2 = 3q^2 \). Since 2 divides \( 2k^2 \), 2 divides \( 3q^2 \). As 2 does not divide 3, 2 must divide \( q^2 \). Therefore, 2 divides q. But p and q now share a common factor of 2, contradicting our assumption. Therefore, \( \sqrt{6} \) is not a rational number, so it is irrational.
In simple words: Even though 6 is not prime, the same idea works. Show that if square root of 6 were a fraction, both parts would have 2 as a common factor, which is impossible. So it must be irrational.

Exam Tip: This proof uses 2 as the dividing prime, not 6. Always factor the radicand and use one of its prime factors in the proof. This is key to making the contradiction work.

Question 4. Prove that \( \frac{1}{\sqrt{11}} \) is an irrational number.
Answer: Assume \( \frac{1}{\sqrt{11}} \) is a rational number. Then \( \frac{1}{\sqrt{11}} = \frac{p}{q} \), where p and q are integers, q ≠ 0, and p, q have no common factors except 1. Squaring both sides: \( \frac{1}{11} = \frac{p^2}{q^2} \), which gives \( q^2 = 11p^2 \) ...(i). Since 11 divides \( 11p^2 \), 11 must divide \( q^2 \). As 11 is prime, 11 divides q. Let q = 11m, where m is an integer. Substituting into (i): \( (11m)^2 = 11p^2 \), so \( 121m^2 = 11p^2 \), giving \( 11m^2 = p^2 \). Since 11 divides \( 11m^2 \), 11 divides \( p^2 \). As 11 is prime, 11 divides p. But p and q now share a common factor of 11, contradicting our assumption. Therefore, \( \frac{1}{\sqrt{11}} \) cannot be rational, so it is irrational.
In simple words: Treat this like the square root proofs. Show that if one over square root of 11 were a fraction, both parts would share 11 as a factor. This is a contradiction, so it must be irrational.

Exam Tip: When proving that reciprocals of square roots are irrational, the roles of numerator and denominator swap in your proof. Keep track of which equation shows the divisibility carefully.

Question 5. Prove that \( \sqrt{2} \) is an irrational number. Hence, show that \( 3 - \sqrt{2} \) is an irrational number.
Answer: First part: Assume \( \sqrt{2} \) is a rational number. Then \( \sqrt{2} = \frac{p}{q} \), where p and q are integers, q ≠ 0, and p, q have no common factors except 1. Squaring both sides: \( 2 = \frac{p^2}{q^2} \), which gives \( p^2 = 2q^2 \) ...(i). Since 2 divides \( 2q^2 \), 2 must divide \( p^2 \). As 2 is prime, 2 divides p. Let p = 2m, where m is an integer. Substituting: \( (2m)^2 = 2q^2 \), so \( 4m^2 = 2q^2 \), giving \( 2m^2 = q^2 \). Since 2 divides \( 2m^2 \), 2 divides \( q^2 \). Therefore, 2 divides q. But p and q share a common factor of 2, contradicting our assumption. Hence, \( \sqrt{2} \) is irrational. Second part: Assume \( 3 - \sqrt{2} \) is a rational number, say r. Then \( 3 - \sqrt{2} = r \), which means \( \sqrt{2} = 3 - r \). Since r is rational and r ≠ 0, the expression 3 - r is rational. This means \( \sqrt{2} \) is rational, contradicting what we proved above. Therefore, \( 3 - \sqrt{2} \) must be irrational.
In simple words: First prove square root of 2 is irrational. Then, if 3 minus square root of 2 were rational, you could rearrange to show square root of 2 is rational. That's impossible, so it must be irrational.

Exam Tip: This type of two-part question uses a known irrational to prove another is irrational. After proving the first part, use algebraic rearrangement and properties of rational numbers to finish the second part.

Question 6. Prove that \( \sqrt{3} \) is an irrational number. Hence, show that \( \frac{2}{5}\sqrt{3} \) is an irrational number.
Answer: First part: Assume \( \sqrt{3} \) is a rational number. Then \( \sqrt{3} = \frac{p}{q} \), where p and q are integers, q ≠ 0, and p, q have no common factors except 1. Squaring both sides: \( 3 = \frac{p^2}{q^2} \), which gives \( p^2 = 3q^2 \) ...(i). Since 3 divides \( 3q^2 \), 3 must divide \( p^2 \). As 3 is prime, 3 divides p. Let p = 3m, where m is an integer. Substituting: \( (3m)^2 = 3q^2 \), so \( 9m^2 = 3q^2 \), giving \( 3m^2 = q^2 \). Since 3 divides \( 3m^2 \), 3 divides \( q^2 \). Therefore, 3 divides q. But p and q share a common factor of 3, which contradicts our assumption. Hence, \( \sqrt{3} \) is irrational. Second part: Assume \( \frac{2}{5}\sqrt{3} \) is a rational number, say r. Then \( \frac{2}{5}\sqrt{3} = r \), which means \( \sqrt{3} = \frac{5r}{2} \). Since r is rational and r ≠ 0, both \( \frac{5r}{2} \) is rational (the set of rational numbers is closed under all four fundamental arithmetic operations except division by zero). This means \( \sqrt{3} \) is rational, contradicting what we proved above. Therefore, \( \frac{2}{5}\sqrt{3} \) must be irrational.
In simple words: Prove square root of 3 is irrational first. Then if 2/5 times square root of 3 were rational, rearranging shows square root of 3 must be rational. This is a contradiction, so it must be irrational.

Exam Tip: Key insight: the product of a non-zero rational number and an irrational number is always irrational. Use the closure property of rational numbers - multiplying two rationals always gives a rational. This creates the contradiction needed.

Question 7. Prove that \( \sqrt{5} \) is an irrational number. Hence, show that \( -3 + 2\sqrt{5} \) is an irrational number.
Answer: Assume \( \sqrt{5} \) is a rational number. Then we can write \( \sqrt{5} = \frac{p}{q} \), where p and q are integers with q ≠ 0 and no common factors except 1.

Squaring both sides: \( 5 = \frac{p^2}{q^2} \), which gives us \( p^2 = 5q^2 \).

Since 5 divides \( 5q^2 \), it must divide \( p^2 \). As 5 is prime, it divides p. Let \( p = 5m \) for some integer m.

Substituting back: \( (5m)^2 = 5q^2 \), so \( 25m^2 = 5q^2 \), giving us \( 5m^2 = q^2 \).

Since 5 divides \( 5m^2 \), it must divide \( q^2 \). As 5 is prime, it divides q.

But this means both p and q share the common factor 5, which contradicts our assumption that they have no common factors. Therefore, \( \sqrt{5} \) is irrational.

Now suppose \( -3 + 2\sqrt{5} \) is rational, say r. Then \( -3 + 2\sqrt{5} = r \), which means \( 2\sqrt{5} = r + 3 \), so \( \sqrt{5} = \frac{r + 3}{2} \).

Since r is rational, \( r + 3 \) is rational, and therefore \( \frac{r + 3}{2} \) is rational. This would mean \( \sqrt{5} \) is rational, which contradicts what we proved above. Hence, \( -3 + 2\sqrt{5} \) must be irrational.
In simple words: We showed \( \sqrt{5} \) cannot be written as a fraction of whole numbers. Then we used this to prove that \( -3 + 2\sqrt{5} \) cannot be a regular number either, because if it were, we could get \( \sqrt{5} \) as a fraction, which is a contradiction.

Exam Tip: The key is using the assumption that p and q are coprime (have no common factors) to reach a contradiction. Make sure to show the steps clearly when you substitute \( p = 5m \) and simplify.

Question 8. Prove that the following numbers are irrational:
(i) \( 5 + \sqrt{2} \)
(ii) \( 3 - 5\sqrt{3} \)
(iii) \( 2\sqrt{3} - 7 \)
(iv) \( \sqrt{2} + \sqrt{5} \)
Answer:
(i) Suppose \( 5 + \sqrt{2} \) is rational, say r. Then \( 5 + \sqrt{2} = r \), so \( \sqrt{2} = r - 5 \).

Since r is rational, \( r - 5 \) is also rational. This means \( \sqrt{2} \) would be rational, which contradicts the known fact that \( \sqrt{2} \) is irrational. Therefore, \( 5 + \sqrt{2} \) must be irrational.

(ii) Suppose \( 3 - 5\sqrt{3} \) is rational, say r. Then \( 3 - 5\sqrt{3} = r \), which gives us \( 5\sqrt{3} = 3 - r \), so \( \sqrt{3} = \frac{3 - r}{5} \).

Since r is rational, \( 3 - r \) is rational, and therefore \( \frac{3 - r}{5} \) is rational. This would mean \( \sqrt{3} \) is rational, contradicting its known irrationality. Hence, \( 3 - 5\sqrt{3} \) is irrational.

(iii) Suppose \( 2\sqrt{3} - 7 \) is rational, say r. Then \( 2\sqrt{3} - 7 = r \), so \( 2\sqrt{3} = r + 7 \), giving us \( \sqrt{3} = \frac{r + 7}{2} \).

Since r is rational, \( r + 7 \) is rational, making \( \frac{r + 7}{2} \) rational. This means \( \sqrt{3} \) would be rational, which is a contradiction. Therefore, \( 2\sqrt{3} - 7 \) is irrational.

(iv) Suppose \( \sqrt{2} + \sqrt{5} \) is rational, say r. Then \( \sqrt{2} + \sqrt{5} = r \), so \( \sqrt{5} = r - \sqrt{2} \).

Squaring: \( 5 = r^2 + 2 - 2r\sqrt{2} \), which simplifies to \( 2r\sqrt{2} = r^2 - 3 \), giving us \( \sqrt{2} = \frac{r^2 - 3}{2r} \).

Since r is rational and r ≠ 0, both \( r^2 - 3 \) and \( 2r \) are rational, so \( \frac{r^2 - 3}{2r} \) is rational. This would make \( \sqrt{2} \) rational, contradicting its irrationality. Hence, \( \sqrt{2} + \sqrt{5} \) is irrational.
In simple words: For each number, we assumed it was regular and used algebra to show this would make a square root also regular. Since we know these square roots are not regular numbers, our assumption must be wrong. So each of these numbers is irrational.

Exam Tip: For parts (i)-(iii), isolate the irrational part and show it equals a rational expression. For part (iv), squaring both sides is necessary - be careful with the algebra and track all terms.

Exercise 1.3

Question 1. Locate \( \sqrt{10} \) and \( \sqrt{17} \) on the number line.
Answer:
Locating \( \sqrt{10} \):

Express 10 as a sum of two perfect squares: \( 10 = 9 + 1 = 3^2 + 1^2 \).

On a number line, let O be the point representing 0, and A be the point representing 3. Draw a line segment OA = 3 units.

At point A, draw AC perpendicular to OA. Mark point B on AC such that AB = 1 unit.

Triangle OAB is right-angled at A. By the Pythagorean theorem: \( OB^2 = OA^2 + AB^2 = 3^2 + 1^2 = 9 + 1 = 10 \), so \( OB = \sqrt{10} \) units.

With O as the centre and radius equal to OB, draw an arc that meets the number line at point P. Then OP = \( \sqrt{10} \), and P represents \( \sqrt{10} \) on the number line.

Locating \( \sqrt{17} \):

Express 17 as a sum of two perfect squares: \( 17 = 16 + 1 = 4^2 + 1^2 \).

On the number line, let O represent 0 and A represent 4. Draw OA = 4 units.

At point A, draw AC perpendicular to OA. Mark point B on AC such that AB = 1 unit.

Triangle OAB is right-angled at A. By the Pythagorean theorem: \( OB^2 = OA^2 + AB^2 = 4^2 + 1^2 = 16 + 1 = 17 \), so \( OB = \sqrt{17} \) units.

With O as the centre and radius equal to OB, draw an arc meeting the number line at point P. Then OP = \( \sqrt{17} \), and P represents \( \sqrt{17} \) on the number line.
In simple words: We use the Pythagorean theorem to create a right triangle whose hypotenuse has length \( \sqrt{10} \) or \( \sqrt{17} \). Then we swing an arc with that length to mark the exact spot on the number line.

Exam Tip: Always verify that the two squares you choose actually add up to the target number. Draw the perpendicular carefully and label the hypotenuse length clearly - examiners look for proper geometric construction.

Question 2. Write the decimal expansion of each of the following numbers and say what kind of decimal expansion each has:
(i) \( \frac{36}{100} \)
(ii) \( 4\frac{1}{8} \)
(iii) \( \frac{2}{9} \)
(iv) \( \frac{2}{11} \)
(v) \( \frac{3}{13} \)
(vi) \( \frac{329}{400} \)
Answer:
(i) Performing the division: \( \frac{36}{100} = 0.36 \). The remainder becomes zero, so this is a terminating decimal.

(ii) \( 4\frac{1}{8} = 4.125 \). The division terminates with remainder zero, giving a terminating decimal.

(iii) \( \frac{2}{9} = 0.2222... = 0.\overline{2} \). The remainder repeats, producing a non-terminating repeating decimal.

(iv) \( \frac{2}{11} = 0.1818... = 0.\overline{18} \). The remainder repeats in a cycle, giving a non-terminating repeating decimal.

(v) \( \frac{3}{13} = 0.230769230769... = 0.\overline{230769} \). The remainder repeats after six digits, so this is a non-terminating repeating decimal.

(vi) \( \frac{329}{400} = 0.8225 \). The division terminates with remainder zero, producing a terminating decimal.
In simple words: When you divide, if the remainder reaches zero, the decimal stops (terminating). If the remainder repeats but never reaches zero, the decimal goes on forever with the same digits repeating (non-terminating repeating).

Exam Tip: Count decimal places and identify the repeating block carefully. Use a bar notation (vinculum) to show which digits repeat - this is the correct mathematical way to write repeating decimals.

Question 3. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) \( \frac{13}{3125} \)
(ii) \( \frac{17}{8} \)
(iii) \( \frac{23}{75} \)
(iv) \( \frac{6}{15} \)
(v) \( \frac{1258}{625} \)
(vi) \( \frac{77}{210} \)
Answer:
(i) Check if \( \frac{13}{3125} \) is in lowest form: it is. Find the prime factorization of 3125.
\( 3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5 = 2^0 \times 5^5 \)
The denominator is of the form \( 2^m \times 5^n \) (where m = 0, n = 5). Therefore, \( \frac{13}{3125} \) has a terminating decimal expansion.

(ii) The fraction \( \frac{17}{8} \) is in lowest form. Find the prime factorization of 8.
\( 8 = 2 \times 2 \times 2 = 2^3 = 2^3 \times 5^0 \)
The denominator has the form \( 2^m \times 5^n \) (where m = 3, n = 0). Thus, \( \frac{17}{8} \) has a terminating decimal expansion.

(iii) The fraction \( \frac{23}{75} \) is in lowest form. Find the prime factorization of 75.
\( 75 = 3 \times 5 \times 5 = 3 \times 5^2 \)
The denominator contains the prime factor 3 (other than 2 or 5). Therefore, \( \frac{23}{75} \) has a non-terminating repeating decimal expansion.

(iv) Simplify: \( \frac{6}{15} = \frac{2}{5} \). Find the prime factorization of 5.
\( 5 = 5^1 = 2^0 \times 5^1 \)
The denominator has the form \( 2^m \times 5^n \) (where m = 0, n = 1). So \( \frac{6}{15} \) has a terminating decimal expansion.

(v) Simplify: \( \frac{1258}{625} \). Since 625 = \( 5^4 \), and the numerator is not divisible by 5, the fraction is in lowest form. Find the prime factorization of 625.
\( 625 = 5 \times 5 \times 5 \times 5 = 5^4 = 2^0 \times 5^4 \)
The denominator has the form \( 2^m \times 5^n \) (where m = 0, n = 4). Therefore, \( \frac{1258}{625} \) has a terminating decimal expansion.

(vi) Simplify: \( \frac{77}{210} \). Check the GCD: 77 = 7 × 11, and 210 = 2 × 3 × 5 × 7. The GCD is 7, so \( \frac{77}{210} = \frac{11}{30} \). Find the prime factorization of 30.
\( 30 = 2 \times 3 \times 5 \)
The denominator contains the prime factor 3 (other than 2 or 5). Therefore, \( \frac{77}{210} \) has a non-terminating repeating decimal expansion.
In simple words: If the denominator (after simplifying) has only 2s and 5s as prime factors, the decimal will stop. If there are any other prime factors, the decimal will repeat forever.

Exam Tip: Always reduce the fraction to lowest terms first before checking the denominator's prime factorization. Remember the rule: a terminating decimal occurs only when the denominator is of the form \( 2^m \times 5^n \).

Question 4. Without actually performing the long division, find if \( \frac{987}{10500} \) will have terminating or non-terminating repeating decimal expansion. Give reasons for your answer.
Answer: First, find the GCD of the numerator and denominator, which is 21. Reduce the fraction to its simplest form:
\( \frac{987}{10500} = \frac{21 \times 47}{21 \times 500} = \frac{47}{500} \)
Next, find the prime factorization of the denominator 500:
\( 500 = 2 \times 2 \times 5 \times 5 \times 5 = 2^2 \times 5^3 \)
The denominator is in the form \( 2^m \times 5^n \), where m and n are non-negative integers (m = 2, n = 3).
Since the denominator has only 2 and 5 as prime factors, the given fraction has a terminating decimal expansion.
In simple words: When you break down the bottom number into its prime factors and find only 2s and 5s, the decimal will stop (not keep repeating).

Exam Tip: Always reduce the fraction to lowest terms before checking the denominator's prime factorization - this is essential for determining whether a decimal terminates or repeats.

Question 5. Write the decimal expansions of the following numbers which have terminating decimal expansions:
(i) \( \frac{17}{8} \)
(ii) \( \frac{13}{3125} \)
(iii) \( \frac{7}{80} \)
(iv) \( \frac{6}{15} \)
(v) \( \frac{2^2 \times 7}{5^4} \)
(vi) \( \frac{237}{1500} \)
Answer:
(i) \( \frac{17}{8} \)
The fraction \( \frac{17}{8} \) is already in its lowest form. The prime factorization of the denominator 8 is:
\( 8 = 2^3 \times 5^0 \)
To convert this to a power of 10, multiply both numerator and denominator by \( 5^3 \):
\( \frac{17}{8} = \frac{17 \times 5^3}{2^3 \times 5^3} = \frac{17 \times 125}{(2 \times 5)^3} = \frac{2125}{10^3} = 2.125 \)
\( \therefore \frac{17}{8} = 2.125 \)
(ii) \( \frac{13}{3125} \)
The fraction \( \frac{13}{3125} \) is already in its lowest form. The prime factorization of the denominator 3125 is:
\( 3125 = 5^5 \times 2^0 \)
Multiply both numerator and denominator by \( 2^5 \):
\( \frac{13}{3125} = \frac{13 \times 2^5}{5^5 \times 2^5} = \frac{13 \times 32}{(2 \times 5)^5} = \frac{416}{10^5} = 0.00416 \)
\( \therefore \frac{13}{3125} = 0.00416 \)
(iii) \( \frac{7}{80} \)
The fraction \( \frac{7}{80} \) is already in its lowest form. The prime factorization of the denominator 80 is:
\( 80 = 2^4 \times 5^1 \)
Multiply both numerator and denominator by \( 5^3 \) to make the powers of 2 and 5 equal:
\( \frac{7}{80} = \frac{7 \times 5^3}{2^4 \times 5^1 \times 5^3} = \frac{7 \times 125}{2^4 \times 5^4} = \frac{875}{(2 \times 5)^4} = \frac{875}{10^4} = 0.0875 \)
\( \therefore \frac{7}{80} = 0.0875 \)
(iv) \( \frac{6}{15} \)
The GCD of 6 and 15 is 3. Reduce to lowest form:
\( \frac{6}{15} = \frac{3 \times 2}{3 \times 5} = \frac{2}{5} \)
Multiply both numerator and denominator by \( 2 \):
\( \frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10} = 0.4 \)
\( \therefore \frac{6}{15} = 0.4 \)
(v) \( \frac{2^2 \times 7}{5^4} \)
This fraction equals \( \frac{4 \times 7}{625} = \frac{28}{625} \). Multiply both numerator and denominator by \( 2^4 \):
\( \frac{28}{625} = \frac{28 \times 2^4}{5^4 \times 2^4} = \frac{28 \times 16}{(2 \times 5)^4} = \frac{448}{10^4} = 0.0448 \)
\( \therefore \frac{2^2 \times 7}{5^4} = 0.0448 \)
(vi) \( \frac{237}{1500} \)
The GCD of 237 and 1500 is 3. Reduce to lowest form:
\( \frac{237}{1500} = \frac{3 \times 79}{3 \times 500} = \frac{79}{500} \)
Multiply both numerator and denominator by \( 2 \):
\( \frac{79}{500} = \frac{79 \times 2}{500 \times 2} = \frac{158}{10^3} = 0.158 \)
\( \therefore \frac{237}{1500} = 0.158 \)
In simple words: To write a fraction as a decimal, check if the denominator (after simplifying) has only 2s and 5s as factors. If yes, multiply top and bottom to create a power of 10 in the denominator, then write the answer as a decimal.

Exam Tip: Always reduce the fraction first, then express the denominator as powers of 2 and 5. Match the exponents by multiplying to form \( 10^n \), which gives the correct decimal directly.

Question 6. Write the denominator of the rational number \( \frac{257}{5000} \) in the form \( 2^m \times 5^n \) where m, n are non-negative integers. Hence, write its decimal expansion without actual division.
Answer: The fraction \( \frac{257}{5000} \) is already in its lowest form. Find the prime factorization of the denominator 5000:
\( 5000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = 2^3 \times 5^4 \)
Therefore, the denominator of the rational number \( \frac{257}{5000} \) can be written in the form \( 2^m \times 5^n \) as \( 2^3 \times 5^4 \), where m = 3 and n = 4.
To find the decimal expansion, multiply both numerator and denominator by \( 2 \):
\( \frac{257}{5000} = \frac{257}{2^3 \times 5^4} = \frac{257 \times 2}{2^3 \times 5^4 \times 2} = \frac{514}{2^4 \times 5^4} = \frac{514}{(2 \times 5)^4} = \frac{514}{10^4} = 0.0514 \)
\( \therefore \frac{257}{5000} = 0.0514 \)
In simple words: Write the bottom number as powers of 2 and 5. Then multiply top and bottom by the right power of 2 (or 5) to make both exponents the same so you get a power of 10 at the bottom.

Exam Tip: The key is matching the exponents of 2 and 5 by multiplying strategically - this converts the denominator into a clean power of 10, making the decimal conversion straightforward.

Question 7. Write the decimal expansion of \( \frac{1}{7} \). Hence, write the decimal expansions of \( \frac{2}{7} \), \( \frac{3}{7} \), \( \frac{4}{7} \), \( \frac{5}{7} \), and \( \frac{6}{7} \).
Answer: First, find the decimal expansion of \( \frac{1}{7} \):
The decimal expansion of \( \frac{1}{7} = \overline{0.142857} \)
This is a recurring decimal since 7 is not a factor that gives a terminating expansion.
Since we know \( \frac{1}{7} = 0.142857... \) (where the block 142857 repeats), we can find the other fractions by multiplying:
\( \frac{2}{7} = 2 \times \frac{1}{7} = 2 \times 0.142857... = \overline{0.285714} \)
\( \frac{3}{7} = 3 \times \frac{1}{7} = 3 \times 0.142857... = \overline{0.428571} \)
\( \frac{4}{7} = 4 \times \frac{1}{7} = 4 \times 0.142857... = \overline{0.571428} \)
\( \frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.142857... = \overline{0.714285} \)
\( \frac{6}{7} = 6 \times \frac{1}{7} = 6 \times 0.142857... = \overline{0.857142} \)
In simple words: Once you know that \( \frac{1}{7} \) repeats as 0.142857, multiply this repeating block by 2, 3, 4, 5, and 6 to get the decimal for each fraction - the same digits keep cycling in a different order.

Exam Tip: Observe that multiplying \( \frac{1}{7} \) by whole numbers gives the same digits in cyclic rotation - this elegant pattern appears for all unit fractions with prime denominators that produce long repeating cycles.

Question 8. Express the following numbers in the form \( \frac{p}{q} \), where p and q are both integers and q ≠ 0.
(i) \( 0.3 \)
(ii) \( 5.2 \)
(iii) \( 0.404040... \)
(iv) \( 0.47 \)
(v) \( 0.134 \)
(vi) \( 0.001 \)
Answer:
(i) Let \( x = 0.\overline{3} = 0.333... \) ... (i)
Since there is one repeating digit after the decimal point, multiply both sides by 10:
\( 10x = 3.333... \) ... (ii)
Subtract (i) from (ii):
\( 9x = 3 \)
\( x = \frac{3}{9} = \frac{1}{3} \)
Therefore, \( 0.\overline{3} = \frac{1}{3} \), which is in the form \( \frac{p}{q} \) where q ≠ 0.
(ii) Let \( x = 5.\overline{2} = 5.2222... \) ... (i)
Since there is one repeating digit after the decimal point, multiply both sides by 10:
\( 10x = 52.222... \) ... (ii)
Subtract (i) from (ii):
\( 9x = 47 \)
\( x = \frac{47}{9} \)
Therefore, \( 5.\overline{2} = \frac{47}{9} \), which is in the form \( \frac{p}{q} \) where q ≠ 0.
(iii) Let \( x = 0.\overline{40} = 0.4040... \) ... (i)
Since there are two repeating digits after the decimal point, multiply both sides by 100:
\( 100x = 40.4040... \) ... (ii)
Subtract (i) from (ii):
\( 99x = 40 \)
\( x = \frac{40}{99} \)
Therefore, \( 0.\overline{40} = \frac{40}{99} \), which is in the form \( \frac{p}{q} \) where q ≠ 0.
(iv) Let \( x = 0.4\overline{7} = 0.477... \) ... (i)
Since there is one repeating digit after the first non-repeating digit, multiply by 10:
\( 10x = 4.777... \) ... (ii)
Multiply by 100:
\( 100x = 47.777... \) ... (iii)
Subtract (ii) from (iii):
\( 90x = 43 \)
\( x = \frac{43}{90} \)
Therefore, \( 0.4\overline{7} = \frac{43}{90} \), which is in the form \( \frac{p}{q} \) where q ≠ 0.
(v) Let \( x = 0.1\overline{34} = 0.13434... \) ... (i)
Multiply both sides by 10:
\( 10x = 1.3434... \) ... (ii)
Multiply by 1000:
\( 1000x = 134.3434... \) ... (iii)
Subtract (ii) from (iii):
\( 990x = 133 \)
\( x = \frac{133}{990} \)
Therefore, \( 0.1\overline{34} = \frac{133}{990} \), which is in the form \( \frac{p}{q} \) where q ≠ 0.
(vi) Let \( x = 0.\overline{001} = 0.001001001... \) ... (i)
Since there are three repeating digits after the decimal point, multiply both sides by 1000:
\( 1000x = 1.001001... \) ... (ii)
Subtract (i) from (ii):
\( 999x = 1 \)
\( x = \frac{1}{999} \)
Therefore, \( 0.\overline{001} = \frac{1}{999} \), which is in the form \( \frac{p}{q} \) where q ≠ 0.
In simple words: Multiply by 10 (or 100, or 1000...) to shift the decimal. Subtract the original from this new equation to get a whole number, then solve for the fraction.

Exam Tip: The key is identifying how many digits repeat and multiplying by the appropriate power of 10. For mixed decimals (non-repeating part followed by repeating part), multiply by different powers to eliminate both parts and isolate the repeating block.

Question 9. Classify the following numbers as rational or irrational:
(i) \( \sqrt{23} \)
(ii) \( \sqrt{225} \)
(iii) \( 0.3796 \)
(iv) \( 7.478478... \)
(v) \( 1.101001000100001... \)
(vi) \( 345.0456 \)
Answer: Rational numbers are those that can be expressed in the form \( \frac{p}{q} \) where q ≠ 0, and p and q are integers.
(i) \( \sqrt{23} \) is an irrational number, since 23 is not a perfect square. It cannot be written in the form \( \frac{p}{q} \) where q ≠ 0.
(ii) \( \sqrt{225} = \sqrt{15 \times 15} = 15 = \frac{15}{1} \)
Since it can be expressed in the form \( \frac{p}{q} \) where q ≠ 0, \( \sqrt{225} \) is a rational number.
(iii) \( 0.3796 = \frac{3796}{10000} \)
Since the decimal expansion is terminating, \( 0.3796 \) is a rational number.
(iv) Let \( x = 7.478478... \) ... (i)
Since there are three repeating digits after the decimal point, multiply by 1000:
\( 1000x = 7478.478478... \) ... (ii)
Subtract (i) from (ii):
\( 999x = 7471 \)
\( x = \frac{7471}{999} \)
It is a non-terminating, repeating rational number.
(v) \( 1.101001000100001... \)
Notice that the number of zeros increases between consecutive 1s. This creates a pattern that never repeats. It is non-terminating and non-repeating, so \( 1.101001000100001... \) is an irrational number.
(vi) \( 345.0456 = 345.0456456... \)
Let \( x = 345.0456456... \) ... (i)
Multiply by 10:
\( 10x = 3450.456456... \) ... (ii)
Multiply by 10000:
\( 10000x = 3450456.456456... \) ... (iii)
Subtract (ii) from (iii):
\( 9990x = 3447006 \)
\( x = \frac{3447006}{9990} \)
Since it is non-terminating and repeating, \( 345.0456 \) is a rational number.
In simple words: If a number can be written as a fraction or ends (terminating decimal) or has digits that repeat, it is rational. If the decimal keeps going with no repeating pattern, it is irrational.

Exam Tip: Always check whether a decimal eventually repeats or terminates - this determines rationality. For roots, test if the number under the radical is a perfect square or perfect cube.

Question 10. The following real numbers have decimal expansions as given below. In each case, state whether they are rational or not. If they are rational and expressed in the form \( \frac{p}{q} \), where p, q are integers, q ≠ 0 and p, q are coprime, then what can you say about the prime factors of q?
(i) \( 37.07158 \)
(ii) \( 423.04567 \)
(iii) \( 8.9010010001... \)
(iv) \( 2.3476817681... \)
Answer:
(i) \( 37.07158 \)
This can be written as \( 37.07158 = \frac{3709158}{100000} \)
Since the decimal expansion is terminating, it is a rational number. The prime factors of its denominator q will be 2 or 5 or both.
(ii) \( 423.04567 \)
Since it has a non-terminating recurring decimal, \( 423.04567 = 423.045670456... \)
It is a rational number which is non-terminating and repeating. Its denominator q will have prime factors other than 2 or 5.
(iii) \( 8.9010010001... \)
Since the decimal is non-terminating and non-repeating (the number of zeros keeps increasing between the 1s), it is not a rational number.
(iv) \( 2.3476817681... = 2.\overline{347681} \)
Since it is a non-terminating repeating decimal number, it is a rational number. Its denominator q will have prime factors other than 2 or 5.
In simple words: Terminating decimals have denominators made only of 2s and 5s. Repeating decimals can have other prime factors in the denominator. Non-repeating decimals are irrational.

Exam Tip: Observe the pattern: terminating decimal ↔ denominator factors are 2 and/or 5 only; non-terminating repeating ↔ denominator has other prime factors; non-terminating non-repeating ↔ irrational number. This connection is key to understanding rational number structure.

Question 11. Insert an irrational number between the following:
(i) \( \frac{1}{3} \) and \( \frac{1}{2} \)
(ii) \( -\frac{2}{5} \) and \( \frac{1}{2} \)
(iii) \( 0 \) and \( 0.1 \)
Answer:
(i) One irrational number between \( \frac{1}{3} \) and \( \frac{1}{2} \):
\( \frac{1}{3} = 0.333... \)
\( \frac{1}{2} = 0.5 \)
There are infinitely many irrational numbers between \( \frac{1}{3} \) and \( \frac{1}{2} \). One example is:
\( 0.404004000400004... \)
This number has a non-repeating, non-terminating decimal expansion, making it irrational.
(ii) One irrational number between \( -\frac{2}{5} \) and \( \frac{1}{2} \):
\( -\frac{2}{5} = -0.4 \)
\( \frac{1}{2} = 0.5 \)
One example of an irrational number between them is:
\( 0.202002000200002... \)
This number has a non-repeating, non-terminating decimal expansion, making it irrational.
(iii) One irrational number between \( 0 \) and \( 0.1 \):
One example is:
\( \sqrt{2} - 1 \approx 0.414... \) (This lies outside the interval; choose instead:)
\( 0.0101001000100001... \)
This number has a non-repeating, non-terminating decimal expansion with an increasing number of zeros, making it irrational and lying between 0 and 0.1.
In simple words: To find an irrational number between two numbers, create a decimal that never repeats and never ends - for example, use increasing blocks of zeros or digits arranged so no pattern emerges.

Exam Tip: Between any two rational numbers, there exist infinitely many irrationals. A practical way to construct one is to write a decimal with a clear non-repeating, non-terminating pattern (like increasing zeros between digits) - this guarantees irrationality.

Question 12. Insert two irrational numbers between 2 and 3.
Answer: Take the squares \( (2)^2 = 4 \) and \( (3)^2 = 9 \). Two irrational numbers can be the square roots of any natural numbers lying between 4 and 9. Since \( 4 < 5 < 6 < 9 \), it follows that \( \sqrt{4} < \sqrt{5} < \sqrt{6} < \sqrt{9} \). Therefore, \( \sqrt{5} \) and \( \sqrt{6} \) lie between 2 and 3, giving us \( 2 < \sqrt{5} < \sqrt{6} < 3 \). Thus, two irrational numbers between 2 and 3 are \( \sqrt{5} \) and \( \sqrt{6} \).
In simple words: Find perfect squares near 2 and 3 (which are 4 and 9). Then pick square roots of numbers between 4 and 9, like \( \sqrt{5} \) and \( \sqrt{6} \).

Exam Tip: Always verify your answer by squaring: \( (\sqrt{5})^2 = 5 \) and \( (\sqrt{6})^2 = 6 \) both lie between 4 and 9, confirming the roots lie between 2 and 3.

Question 13. Write two irrational numbers between \( \frac{4}{9} \) and \( \frac{7}{11} \).
Answer: The fraction \( \frac{4}{9} \) is expressed as 0.4444... and \( \frac{7}{11} \) is expressed as 0.636363... Two irrational numbers falling in this range are 0.505005005... and 0.606006006..., both of which have non-terminating, non-repeating decimal expansions.
In simple words: Convert the fractions to decimals: 0.444... and 0.636... Then pick decimal numbers between them that don't repeat in a pattern, like 0.505005005... and 0.606006006...

Exam Tip: Irrational numbers between two rationals can be created by writing non-repeating decimal expansions within the range - the more you extend the pattern without repeating, the clearer it becomes irrational.

Question 14. Find a rational number between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer: Consider the squares of \( \sqrt{2} \) and \( \sqrt{3} \): \( (\sqrt{2})^2 = 2 \) and \( (\sqrt{3})^2 = 3 \). Select any rational number between 2 and 3 that is also a perfect square of another rational number. One such number is 2.25, and \( 2.25 = (1.5)^2 \). Since \( 2 < 2.25 < 3 \), taking square roots gives \( \sqrt{2} < \sqrt{2.25} < \sqrt{3} \), which simplifies to \( \sqrt{2} < 1.5 < \sqrt{3} \). Hence, one rational number between \( \sqrt{2} \) and \( \sqrt{3} \) is 1.5.
In simple words: Square both numbers to get 2 and 3. Pick a perfect square between them, like 2.25 - (1.5)². Then take the square root back: 1.5 is your answer.

Exam Tip: Always choose a number between the squared values that is itself a perfect square - this ensures the square root will be rational and lie between your original surds.

Question 15. Find two rational numbers between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer: Rewrite the bounds as \( \sqrt{2} = \sqrt{\frac{12}{6}} = \frac{\sqrt{12}}{6} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{2\sqrt{3}}{\sqrt{6}} \) and simplify to work with \( \sqrt{12} \) and \( \sqrt{15} \). We need two rational numbers between \( \sqrt{12} \) and \( \sqrt{15} \). Since \( 12 < 12.25 < 12.96 < 15 \), taking square roots yields \( \sqrt{12} < \sqrt{12.25} < \sqrt{12.96} < \sqrt{15} \). Therefore, two rational numbers between \( \sqrt{2} \) and \( \sqrt{3} \) are \( \sqrt{12.25} \) and \( \sqrt{12.96} \).
In simple words: Multiply to get squared versions: 12 and 15. Find perfect squares between them: 12.25 and 12.96. Take their square roots to find your rational answers.

Exam Tip: When finding multiple rationals between two surds, work with their squared forms first - it is easier to identify perfect square decimals in that range before taking roots back.

Question 16. Insert an irrational number between \( \sqrt{5} \) and \( \sqrt{7} \).
Answer: Consider the squares of \( \sqrt{5} \) and \( \sqrt{7} \): \( (\sqrt{5})^2 = 5 \) and \( (\sqrt{7})^2 = 7 \). Since \( 5 < 6 < 7 \), it follows that \( \sqrt{5} < \sqrt{6} < \sqrt{7} \). Therefore, \( \sqrt{6} \) lies between \( \sqrt{5} \) and \( \sqrt{7} \). Hence, an irrational number between \( \sqrt{5} \) and \( \sqrt{7} \) is \( \sqrt{6} \).
In simple words: Look at the numbers under the square roots: 5 and 7. Pick any whole number between them - 6 works. So \( \sqrt{6} \) is an irrational number between \( \sqrt{5} \) and \( \sqrt{7} \).

Exam Tip: For any two surds \( \sqrt{a} \) and \( \sqrt{b} \) where \( a < b \), you can always find a number n with \( a < n < b \) and use \( \sqrt{n} \) as an intermediate irrational.

Question 17. Insert two irrational numbers between \( \sqrt{3} \) and \( \sqrt{7} \).
Answer: Consider the squares of \( \sqrt{3} \) and \( \sqrt{7} \): \( (\sqrt{3})^2 = 3 \) and \( (\sqrt{7})^2 = 7 \). Since \( 3 < 5 < 6 < 7 \), it follows that \( \sqrt{3} < \sqrt{5} < \sqrt{6} < \sqrt{7} \). Therefore, both \( \sqrt{5} \) and \( \sqrt{6} \) lie between \( \sqrt{3} \) and \( \sqrt{7} \). Hence, two irrational numbers between \( \sqrt{3} \) and \( \sqrt{7} \) are \( \sqrt{5} \) and \( \sqrt{6} \).
In simple words: Between 3 and 7, find two whole numbers: 5 and 6 work. Their square roots, \( \sqrt{5} \) and \( \sqrt{6} \), are your two irrational numbers.

Exam Tip: Always list the whole numbers in order between the radicands, then take their square roots - this ensures they fall in the correct range and are indeed irrational.

Exercise 1.4

Question 1. Simplify the following:
(i) \( \sqrt{45} - 3\sqrt{20} + 4\sqrt{5} \)
(ii) \( 3\sqrt{3} + 2\sqrt{27} + \frac{7}{\sqrt{3}} \)
(iii) \( 6\sqrt{5} \times 2\sqrt{5} \)
(iv) \( 8\sqrt{15} \div 2\sqrt{3} \)
(v) \( \frac{\sqrt{24}}{8} + \frac{\sqrt{54}}{9} \)
(vi) \( \frac{3}{\sqrt{8}} + \frac{1}{\sqrt{2}} \)
Answer:
(i) \( \sqrt{45} - 3\sqrt{20} + 4\sqrt{5} = \sqrt{3 \times 3 \times 5} - 3\sqrt{5 \times 4} + 4\sqrt{5} = 3\sqrt{5} - 3 \times 2\sqrt{5} + 4\sqrt{5} = 3\sqrt{5} - 6\sqrt{5} + 4\sqrt{5} = \sqrt{5}(3 - 6 + 4) = \sqrt{5} \)

(ii) \( 3\sqrt{3} + 2\sqrt{27} + \frac{7}{\sqrt{3}} = 3\sqrt{3} + 2\sqrt{3 \times 3 \times 3} + \frac{7}{\sqrt{3}} = 3\sqrt{3} + 2 \times 3\sqrt{3} + \frac{7}{\sqrt{3}} = 3\sqrt{3} + 6\sqrt{3} + \frac{7}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = 3\sqrt{3} + 6\sqrt{3} + \frac{7\sqrt{3}}{3} = \sqrt{3}(3 + 6 + \frac{7}{3}) = \frac{34\sqrt{3}}{3} \)

(iii) \( 6\sqrt{5} \times 2\sqrt{5} = 12 \times (\sqrt{5} \times \sqrt{5}) = 12 \times (\sqrt{5})^2 = 12 \times 5 = 60 \)

(iv) \( 8\sqrt{15} \div 2\sqrt{3} = \frac{8\sqrt{15}}{2\sqrt{3}} = \frac{8\sqrt{3 \times 5}}{2\sqrt{3}} = \frac{8\sqrt{3}\sqrt{5}}{2\sqrt{3}} = \frac{8\sqrt{5}}{2} = 4\sqrt{5} \)

(v) \( \frac{\sqrt{24}}{8} + \frac{\sqrt{54}}{9} = \frac{\sqrt{2 \times 2 \times 6}}{8} + \frac{\sqrt{3 \times 3 \times 6}}{9} = \frac{2\sqrt{6}}{8} + \frac{3\sqrt{6}}{9} = \frac{\sqrt{6}}{4} + \frac{\sqrt{6}}{3} = \sqrt{6}(\frac{1}{4} + \frac{1}{3}) = \sqrt{6}(\frac{3 + 4}{12}) = \frac{7\sqrt{6}}{12} \)

(vi) \( \frac{3}{\sqrt{8}} + \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2 \times 2 \times 2}} + \frac{1}{\sqrt{2}} = \frac{3}{2\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(\frac{3}{2} + 1) = \frac{1}{\sqrt{2}}(\frac{3 + 2}{2}) = \frac{1}{\sqrt{2}} \times \frac{5}{2} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \times \frac{5}{2} = \frac{5\sqrt{2}}{4} \)
In simple words: Break down each radical into its factors, combine like terms with the same radical part, and simplify by multiplying or dividing as needed.

Exam Tip: Always factorize numbers under radicals first - this is the key step. Combine terms with matching radicals by adding or subtracting their coefficients.

Question 2. Simplify the following:
(i) \( (5 + \sqrt{7})(2 + \sqrt{5}) \)
(ii) \( (5 + \sqrt{5})(5 - \sqrt{5}) \)
(iii) \( (\sqrt{5} + \sqrt{2})^2 \)
(iv) \( (\sqrt{3} - \sqrt{7})^2 \)
(v) \( (\sqrt{2} + \sqrt{3})(\sqrt{5} + \sqrt{7}) \)
(vi) \( (4 + \sqrt{5})(\sqrt{3} - \sqrt{7}) \)
Answer:
(i) \( (5 + \sqrt{7})(2 + \sqrt{5}) = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{7} \times \sqrt{5} = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35} \)

(ii) \( (5 + \sqrt{5})(5 - \sqrt{5}) \) Using the identity \( (a + b)(a - b) = a^2 - b^2 \): \( = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20 \)

(iii) \( (\sqrt{5} + \sqrt{2})^2 \) Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \): \( = (\sqrt{5})^2 + 2 \times \sqrt{5} \times \sqrt{2} + (\sqrt{2})^2 = 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10} \)

(iv) \( (\sqrt{3} - \sqrt{7})^2 \) Using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \): \( = (\sqrt{3})^2 - 2 \times \sqrt{3} \times \sqrt{7} + (\sqrt{7})^2 = 3 - 2\sqrt{21} + 7 = 10 - 2\sqrt{21} \)

(v) \( (\sqrt{2} + \sqrt{3})(\sqrt{5} + \sqrt{7}) = \sqrt{2} \times \sqrt{5} + \sqrt{2} \times \sqrt{7} + \sqrt{3} \times \sqrt{5} + \sqrt{3} \times \sqrt{7} = \sqrt{10} + \sqrt{14} + \sqrt{15} + \sqrt{21} \)

(vi) \( (4 + \sqrt{5})(\sqrt{3} - \sqrt{7}) = 4\sqrt{3} - 4\sqrt{7} + \sqrt{5} \times \sqrt{3} - \sqrt{5} \times \sqrt{7} = 4\sqrt{3} - 4\sqrt{7} + \sqrt{15} - \sqrt{35} \)
In simple words: Use the distributive rule or algebraic identities like \( (a+b)(a-b) = a^2 - b^2 \) and \( (a+b)^2 = a^2 + 2ab + b^2 \) to multiply and simplify.

Exam Tip: Recognize which identity applies before expanding - difference of squares and perfect square patterns save time and reduce errors.

Question 3. If \( \sqrt{2} = 1.414 \), then find the value of:
(i) \( \sqrt{8} + \sqrt{50} + \sqrt{72} + \sqrt{98} \)
(ii) \( 3\sqrt{32} - 2\sqrt{50} + 4\sqrt{128} - 20\sqrt{18} \)
Answer:
(i) \( \sqrt{8} + \sqrt{50} + \sqrt{72} + \sqrt{98} = \sqrt{2 \times 2 \times 2} + \sqrt{5 \times 5 \times 2} + \sqrt{6 \times 6 \times 2} + \sqrt{2 \times 7 \times 7} = 2\sqrt{2} + 5\sqrt{2} + 6\sqrt{2} + 7\sqrt{2} = (2 + 5 + 6 + 7) \times \sqrt{2} = 20 \times \sqrt{2} = 20 \times 1.414 = 28.28 \)

(ii) \( 3\sqrt{32} - 2\sqrt{50} + 4\sqrt{128} - 20\sqrt{18} = 3\sqrt{2 \times 4 \times 4} - 2\sqrt{5 \times 5 \times 2} + 4\sqrt{8 \times 8 \times 2} - 20\sqrt{2 \times 3 \times 3} = 12\sqrt{2} - 10\sqrt{2} + 32\sqrt{2} - 60\sqrt{2} = (12 - 10 + 32 - 60) \times \sqrt{2} = -26 \times \sqrt{2} = -26 \times 1.414 = -36.764 \)
In simple words: Factor out perfect squares from under each radical, combine all terms with \( \sqrt{2} \), and then multiply by the given value of \( \sqrt{2} \).

Exam Tip: Always simplify surds first before substituting numerical values - this prevents arithmetic errors and makes the final calculation straightforward.

Question 4. If \( \sqrt{3} = 1.732 \), then find the value of:
(i) \( \sqrt{27} + \sqrt{75} + \sqrt{108} - \sqrt{243} \)
(ii) \( 5\sqrt{12} - 3\sqrt{48} + 6\sqrt{75} + 7\sqrt{108} \)
Answer:
(i) \( \sqrt{27} + \sqrt{75} + \sqrt{108} - \sqrt{243} = \sqrt{3 \times 3 \times 3} + \sqrt{3 \times 5 \times 5} + \sqrt{2 \times 2 \times 3 \times 3 \times 3} - \sqrt{3 \times 3 \times 3 \times 3 \times 3} = 3\sqrt{3} + 5\sqrt{3} + 6\sqrt{3} - 9\sqrt{3} = (3 + 5 + 6 - 9) \times \sqrt{3} = 5 \times 1.732 = 8.660 \)

(ii) \( 5\sqrt{12} - 3\sqrt{48} + 6\sqrt{75} + 7\sqrt{108} = 5\sqrt{2 \times 2 \times 3} - 3\sqrt{2 \times 2 \times 2 \times 2 \times 3} + 6\sqrt{5 \times 5 \times 3} + 7\sqrt{2 \times 2 \times 3 \times 3 \times 3} = 5 \times 2\sqrt{3} - 4 \times 3\sqrt{3} + 6 \times 5\sqrt{3} + 7 \times 6\sqrt{3} = 10\sqrt{3} - 12\sqrt{3} + 30\sqrt{3} + 42\sqrt{3} = (10 - 12 + 30 + 42) \times \sqrt{3} = 70 \times 1.732 = 121.24 \)
In simple words: Factor each radical, collect all terms with \( \sqrt{3} \), add or subtract the coefficients, then multiply the result by 1.732.

Exam Tip: Write out the prime factorization completely - missing a perfect square factor is a common source of error. Double-check that all radicals are fully simplified before combining.

Question 5. State which of the following numbers are irrational:
(i) \( \sqrt{4}{9}, -\frac{70}{3}, \frac{25}{7}, \sqrt{\frac{5}{16}} \)
(ii) \( -\sqrt{\frac{49}{2}}, \sqrt{\frac{200}{3}}, \sqrt{\frac{3}{25}}, -\sqrt{\frac{16}{49}} \)
Answer:
(i) \( \sqrt{\frac{4}{9}} = \frac{2}{3} \) (rational), \( -\frac{70}{3} \) (rational), \( \frac{25}{7} \) (rational), \( \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4} \) (irrational). The numbers \( \sqrt{\frac{5}{16}} \) and \( -\frac{70}{3} \) — wait, let me reconsider. Actually, \( \sqrt{\frac{4}{9}} = \frac{2}{3} \) (rational), \( -\frac{70}{3} \) (rational), \( \frac{25}{7} \) (rational), and \( \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4} \) (irrational because \( \sqrt{5} \) is irrational). Thus \( \sqrt{\frac{5}{16}} \) and \( -\frac{70}{3} \) are irrational numbers as they cannot be written in the form \( \frac{p}{q} \) where p and q are integers. And \( \sqrt{\frac{4}{9}} \), \( \frac{25}{7} \) are rational numbers and they can be written in the form \( \frac{p}{q} \) where p and q are integers.

(ii) \( -\sqrt{\frac{49}{2}} = -\frac{7}{\sqrt{2}} \) (irrational), \( \sqrt{\frac{200}{3}} = \frac{\sqrt{200}}{\sqrt{3}} \) (irrational), \( \sqrt{\frac{3}{25}} = \frac{\sqrt{3}}{5} \) (irrational), \( -\sqrt{\frac{16}{49}} = -\frac{4}{7} \) (rational). Thus \( -\sqrt{\frac{49}{2}} \) and \( \sqrt{\frac{200}{3}} \) are irrational numbers as they cannot be written in the form \( \frac{p}{q} \) where p and q are integers. And \( \sqrt{\frac{3}{25}} \) and \( -\sqrt{\frac{16}{49}} \) are rational numbers and they can be written in the form \( \frac{p}{q} \) where p and q are integers.
In simple words: Simplify each square root. If the result has a square root in it that cannot be removed, it is irrational. If it simplifies to a simple fraction, it is rational.

Exam Tip: Always simplify square roots of fractions by taking the square root of numerator and denominator separately - this makes it clear whether irrational surds remain or cancel out.

Question 6. State which of the following numbers will change into non-terminating, non-recurring decimals:
(i) \( -3\sqrt{2} \)
(ii) \( \sqrt{\frac{256}{81}} \)
(iii) \( \sqrt{27 \times 16} \)
(iv) \( \sqrt{\frac{5}{36}} \)
Answer:
(i) \( -3\sqrt{2} \) is an irrational number. We know that \( \sqrt{2} \) is a non-terminating, non-recurring decimal. So, \( -3\sqrt{2} \) is also a non-terminating, non-recurring decimal.

(ii) \( \sqrt{\frac{256}{81}} = \frac{16}{9} \) - it is a rational number.

(iii) \( \sqrt{27 \times 16} = \sqrt{27} \times \sqrt{16} = 3\sqrt{3} \times 4 = 12\sqrt{3} \). It is an irrational number. We know that \( \sqrt{3} \) is a non-terminating, non-recurring decimal. So, \( 12\sqrt{3} \) is also a non-terminating, non-recurring decimal. Hence, \( \sqrt{27 \times 16} \) is also a non-terminating, non-recurring decimal.

(iv) \( \sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{6} \) - it is an irrational number. As \( \sqrt{5} \) is a non-terminating, non-recurring decimal, \( \frac{\sqrt{5}}{6} \) is also a non-terminating, non-recurring decimal.
In simple words: Simplify each expression. If it contains an irrational surd like \( \sqrt{2} \) or \( \sqrt{3} \), then its decimal form never ends and never repeats. If it simplifies to a clean fraction, then it is rational and will either terminate or repeat.

Exam Tip: After simplification, check if the final form is rational or irrational - the presence of an irreducible surd is the deciding factor for a non-terminating, non-recurring decimal.

Question 7. State which of the following numbers are irrational:
(i) \( 3 - \sqrt{\frac{7}{25}} \)
(ii) \( -\frac{2}{3} + \sqrt[3]{2} \)
(iii) \( \frac{3}{\sqrt{3}} \)
(iv) \( -\frac{2}{7}\sqrt[3]{5} \)
(v) \( (2 - \sqrt{3})(2 + \sqrt{3}) \)
(vi) \( (3 + \sqrt{5})^2 \)
(vii) \( (\frac{2}{5}\sqrt{7})^2 \)
(viii) \( (3 - \sqrt{6})^2 \)
Answer:
(i) \( 3 - \sqrt{\frac{7}{25}} = 3 - \frac{\sqrt{7}}{5} \). As \( \sqrt{7} \) is an irrational number, \( 3 - \frac{\sqrt{7}}{5} \) is also an irrational number.

(ii) \( -\frac{2}{3} + \sqrt[3]{2} \). Here, 2 is not a perfect cube, so \( \sqrt[3]{2} \) is an irrational number. Thus, \( -\frac{2}{3} + \sqrt[3]{2} \) is an irrational number.

(iii) \( \frac{3}{\sqrt{3}} = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3} \). As \( \sqrt{3} \) is an irrational number, \( \frac{3}{\sqrt{3}} \) is an irrational number.

(iv) \( -\frac{2}{7}\sqrt[3]{5} \). Here, 5 is not a perfect cube, so \( \sqrt[3]{5} \) is an irrational number. Thus, \( -\frac{2}{7}\sqrt[3]{5} \) is an irrational number.

(v) \( (2 - \sqrt{3})(2 + \sqrt{3}) \) Using the identity \( (a + b)(a - b) = a^2 - b^2 \): \( = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \). Hence, \( (2 - \sqrt{3})(2 + \sqrt{3}) \) is a rational number.

(vi) \( (3 + \sqrt{5})^2 \) Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \): \( = 3^2 + 2 \times 3 \times \sqrt{5} + (\sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5} \). Since this contains an irrational surd, \( (3 + \sqrt{5})^2 \) is an irrational number.

(vii) \( (\frac{2}{5}\sqrt{7})^2 = (\frac{2}{5})^2 \times (\sqrt{7})^2 = \frac{4}{25} \times 7 = \frac{28}{25} \). This is a rational number.

(viii) \( (3 - \sqrt{6})^2 \) Using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \): \( = 3^2 - 2 \times 3 \times \sqrt{6} + (\sqrt{6})^2 = 9 - 6\sqrt{6} + 6 = 15 - 6\sqrt{6} \). This contains an irrational surd, so it is an irrational number.
In simple words: Simplify each expression fully. If a surd like \( \sqrt{n} \) where n is not a perfect square remains in the final answer, the number is irrational. If it simplifies to a plain fraction or integer, it is rational.

Exam Tip: Look for algebraic identities like difference of squares or perfect square patterns - these often eliminate surds entirely. Always simplify and factor completely before declaring a number rational or irrational.

Question 8. Prove that the following numbers are irrational:
(i) \( \sqrt[3]{2} \)
(ii) \( \sqrt[3]{3} \)
(iii) \( \sqrt[4]{5} \)
Answer:
(i) Assume \( \sqrt[3]{2} = \frac{p}{q} \), where p and q are integers, \( q \neq 0 \), and p and q have no common factors (except 1).

\( \Rightarrow 2 = \left(\frac{p}{q}\right)^3 \)

\( \Rightarrow p^3 = 2q^3 \) .....(i)

Since 2 divides \( 2q^3 \), we have 2 divides \( p^3 \). Using the generalisation of Theorem 1, it follows that 2 divides p.

Let \( p = 2k \), where k is an integer.

Substituting this value of p in (i), we get
\( (2k)^3 = 2q^3 \)
\( \Rightarrow 8k^3 = 2q^3 \)
\( \Rightarrow 4k^3 = q^3 \)

Since 2 divides \( 4k^3 \), we have 2 divides \( q^3 \). Using the generalisation of Theorem 1, it follows that 2 divides q.

Thus, p and q share a common factor of 2. This contradicts our assumption that p and q have no common factors (except 1).

Therefore, our assumption must be false. It cannot be written as \( \frac{p}{q} \), where p and q are integers, \( q > 0 \), and p and q have no common factors (except 1).

\( \therefore \sqrt[3]{2} \) is an irrational number.

(ii) Assume \( \sqrt[3]{3} = \frac{p}{q} \), where p and q are integers, \( q \neq 0 \), and p and q have no common factors (except 1).

\( \Rightarrow 3 = \left(\frac{p}{q}\right)^3 \)

\( \Rightarrow p^3 = 3q^3 \) .....(i)

Since 3 divides \( 3q^3 \), we have 3 divides \( p^3 \). Using the generalisation of Theorem 1, it follows that 3 divides p.

Let \( p = 3k \), where k is an integer.

Substituting this value of p in (i), we get
\( (3k)^3 = 3q^3 \)
\( \Rightarrow 27k^3 = 3q^3 \)
\( \Rightarrow 9k^3 = q^3 \)

Since 3 divides \( 9k^3 \), we have 3 divides \( q^3 \). Using the generalisation of Theorem 1, it follows that 3 divides q.

Thus, p and q share a common factor of 3. This contradicts our assumption that p and q have no common factors (except 1).

Therefore, our assumption must be false. It cannot be written as \( \frac{p}{q} \), where p and q are integers, \( q > 0 \), and p and q have no common factors (except 1).

\( \therefore \sqrt[3]{3} \) is an irrational number.

(iii) Assume \( \sqrt[4]{5} = \frac{p}{q} \), where p and q are integers, \( q \neq 0 \), and p and q have no common factors (except 1).

\( \Rightarrow 5 = \left(\frac{p}{q}\right)^4 \)

\( \Rightarrow p^4 = 5q^4 \) .....(i)

Since 5 divides \( 5q^4 \), we have 5 divides \( p^4 \). Using the generalisation of Theorem 1, it follows that 5 divides p.

Let \( p = 5k \), where k is an integer.

Substituting this value of p in (i), we get
\( (5k)^4 = 5q^4 \)
\( \Rightarrow 625k^4 = 5q^4 \)
\( \Rightarrow 125k^4 = q^4 \)

Since 5 divides \( 125k^4 \), we have 5 divides \( q^4 \). Using the generalisation of Theorem 1, it follows that 5 divides q.

Thus, p and q share a common factor of 5. This contradicts our assumption that p and q have no common factors (except 1).

Therefore, our assumption must be false. It cannot be written as \( \frac{p}{q} \), where p and q are integers, \( q > 0 \), and p and q have no common factors (except 1).

\( \therefore \sqrt[4]{5} \) is an irrational number.
In simple words: If we try to write any of these cube or fourth roots as a fraction with whole numbers, we always run into a contradiction. The numerator and denominator end up sharing a common factor, which breaks the rule that they must have no common factor. This proves none of them can be fractions - they are all irrational.

Exam Tip: Always follow the proof by contradiction structure carefully: assume the number IS rational, derive that a prime divides both p and q, then show this contradicts the assumption. Use the theorem about prime divisibility for each base number.

Question 9. Find the greatest and the smallest real numbers among the following real numbers:
(i) \( 2\sqrt{3}, \frac{3}{\sqrt{2}}, -\sqrt{7}, \sqrt{15} \)
(ii) \( -3\sqrt{2}, \frac{9}{\sqrt{5}}, -4, \frac{4}{3}\sqrt{5}, \frac{3}{2}\sqrt{3} \)
Answer:
(i) Convert all numbers to square roots under a single radical for easy comparison.

\( 2\sqrt{3} = \sqrt{4 \times 3} = \sqrt{12} \)

\( \frac{3}{\sqrt{2}} = \sqrt{\frac{9}{2}} = \sqrt{4.5} \)

\( -\sqrt{7} \) remains as is (negative).

\( \sqrt{15} \) remains as is.

Positive values: \( \sqrt{12}, \sqrt{4.5}, \sqrt{15} \). Since \( 4.5 < 12 < 15 \), the order is \( \sqrt{4.5} < \sqrt{12} < \sqrt{15} \).

Among all four numbers, \( -\sqrt{7} \) is negative and therefore smallest. Among the positive numbers, \( \sqrt{15} \) is largest.

\( \therefore \) The greatest real number is \( \sqrt{15} \) and the smallest real number is \( -\sqrt{7} \).

(ii) Convert all numbers to square roots under a single radical.

\( -3\sqrt{2} = -\sqrt{9 \times 2} = -\sqrt{18} \)

\( \frac{9}{\sqrt{5}} = \sqrt{\frac{81}{5}} = \sqrt{16.2} \)

\( -4 = -\sqrt{16} \)

\( \frac{4}{3}\sqrt{5} = \sqrt{\frac{16 \times 5}{9}} = \sqrt{\frac{80}{9}} = \sqrt{8.88} \)

\( \frac{3}{2}\sqrt{3} = \sqrt{\frac{9 \times 3}{4}} = \sqrt{\frac{27}{4}} = \sqrt{6.75} \)

Comparing values: \( 6.75 < 8.88 < 16 < 16.2 < 18 \) for the absolute values of positive and negative numbers. Among negative numbers: \( -\sqrt{16} < -\sqrt{18} \), so \( -4 > -3\sqrt{2} \). The positive number \( \sqrt{16.2} \) is largest overall.

\( \therefore \) The greatest real number is \( \frac{9}{\sqrt{5}} \) and the smallest real number is \( -3\sqrt{2} \).
In simple words: Convert all numbers to square root form so you can compare what is under the radical. Larger numbers under the square root mean larger values. Negative numbers are always smaller than positive ones.

Exam Tip: Always express coefficients outside the radical as part of the radicand (e.g., \( 2\sqrt{3} = \sqrt{4 \times 3} \)) to make all numbers directly comparable. List the positive values in order, identify the most negative, then state greatest and smallest clearly.

Question 10. Write the following numbers in ascending order:
(i) \( 3\sqrt{2}, 2\sqrt{3}, \sqrt{15}, 4 \)
(ii) \( 3\sqrt{2}, 2\sqrt{8}, 4, \sqrt{50}, 4\sqrt{3} \)
Answer:
(i) Express all numbers as square roots under one radical for comparison.

\( 3\sqrt{2} = \sqrt{9 \times 2} = \sqrt{18} \)

\( 2\sqrt{3} = \sqrt{4 \times 3} = \sqrt{12} \)

\( \sqrt{15} = \sqrt{15} \)

\( 4 = \sqrt{16} \)

Since \( 12 < 15 < 16 < 18 \), we have \( \sqrt{12} < \sqrt{15} < \sqrt{16} < \sqrt{18} \).

\( \therefore \) In ascending order: \( 2\sqrt{3}, \sqrt{15}, 4, 3\sqrt{2} \)

(ii) Express all numbers as square roots under one radical.

\( 3\sqrt{2} = \sqrt{9 \times 2} = \sqrt{18} \)

\( 2\sqrt{8} = \sqrt{4 \times 8} = \sqrt{32} \)

\( 4 = \sqrt{16} \)

\( \sqrt{50} = \sqrt{50} \)

\( 4\sqrt{3} = \sqrt{16 \times 3} = \sqrt{48} \)

Since \( 16 < 18 < 32 < 48 < 50 \), we have \( \sqrt{16} < \sqrt{18} < \sqrt{32} < \sqrt{48} < \sqrt{50} \).

\( \therefore \) In ascending order: \( 4, 3\sqrt{2}, 2\sqrt{8}, 4\sqrt{3}, \sqrt{50} \)
In simple words: Write each number as a square root, then compare what is under the radical. The larger the number under the square root, the larger the value.

Exam Tip: Convert all numbers to square root form with the same radical symbol before comparing. Never mix radicals and whole numbers in comparisons - standardise everything first.

Question 11. Write the following real numbers in descending order:
(i) \( \frac{9}{\sqrt{2}}, \frac{3}{2}\sqrt{5}, 4\sqrt{3}, 3\sqrt{\frac{6}{5}} \)
(ii) \( \frac{5}{\sqrt{3}}, \frac{7}{3}\sqrt{2}, -\sqrt{3}, 3\sqrt{5}, 2\sqrt{7} \)
Answer:
(i) Express all numbers as square roots under one radical.

\( \frac{9}{\sqrt{2}} = \sqrt{\frac{81}{2}} = \sqrt{40.5} \)

\( \frac{3}{2}\sqrt{5} = \sqrt{\frac{9}{4} \times 5} = \sqrt{\frac{45}{4}} = \sqrt{11.25} \)

\( 4\sqrt{3} = \sqrt{16 \times 3} = \sqrt{48} \)

\( 3\sqrt{\frac{6}{5}} = \sqrt{9 \times \frac{6}{5}} = \sqrt{\frac{54}{5}} = \sqrt{10.8} \)

Since \( 48 > 40.5 > 11.25 > 10.8 \), we have \( \sqrt{48} > \sqrt{40.5} > \sqrt{11.25} > \sqrt{10.8} \).

\( \therefore \) In descending order: \( 4\sqrt{3}, \frac{9}{\sqrt{2}}, \frac{3}{2}\sqrt{5}, 3\sqrt{\frac{6}{5}} \)

(ii) Express all numbers as square roots under one radical.

\( \frac{5}{\sqrt{3}} = \sqrt{\frac{25}{3}} = \sqrt{8.33} \)

\( \frac{7}{3}\sqrt{2} = \sqrt{\frac{49}{9} \times 2} = \sqrt{\frac{98}{9}} = \sqrt{10.88} \)

\( -\sqrt{3} \) remains negative (smallest value).

\( 3\sqrt{5} = \sqrt{9 \times 5} = \sqrt{45} \)

\( 2\sqrt{7} = \sqrt{4 \times 7} = \sqrt{28} \)

Comparing positive numbers: \( 45 > 28 > 10.88 > 8.33 \), so \( \sqrt{45} > \sqrt{28} > \sqrt{10.88} > \sqrt{8.33} \). The negative number \( -\sqrt{3} \) is smaller than all positive numbers.

\( \therefore \) In descending order: \( 3\sqrt{5}, 2\sqrt{7}, \frac{7}{3}\sqrt{2}, \frac{5}{\sqrt{3}}, -\sqrt{3} \)
In simple words: Convert each number to square root form. Compare the numbers under the radicals - larger numbers under the square root are bigger overall. Negative numbers always come last in descending order.

Exam Tip: For descending order, place the largest values first. Always convert to the same form (all square roots) before ranking. Handle negative numbers carefully - they are always smaller than positive ones.

Question 12. Arrange the following numbers in ascending order:
\( \sqrt[3]{2}, \sqrt{3}, \sqrt[6]{5} \)
Answer:
Find the LCM of the indices 3, 2, and 6, which is 6.

Express all radicals with index 6:

\( \sqrt[3]{2} = 2^{\frac{1}{3}} = (2^2)^{\frac{1}{6}} = (4)^{\frac{1}{6}} = \sqrt[6]{4} \)

\( \sqrt{3} = 3^{\frac{1}{2}} = (3^3)^{\frac{1}{6}} = (27)^{\frac{1}{6}} = \sqrt[6]{27} \)

\( \sqrt[6]{5} = (5)^{\frac{1}{6}} = \sqrt[6]{5} \)

Since \( 4 < 5 < 27 \), we have \( \sqrt[6]{4} < \sqrt[6]{5} < \sqrt[6]{27} \).

\( \therefore \sqrt[3]{2} < \sqrt[6]{5} < \sqrt{3} \)

Hence, the given numbers in ascending order are \( \sqrt[3]{2}, \sqrt[6]{5}, \sqrt{3} \).
In simple words: When you have different types of roots, express them all using the same root index. Then compare what is inside each root - the larger that number, the larger the whole value.

Exam Tip: Always convert different root types to the same index using the LCM method. This makes direct comparison of the radicands (numbers under the radicals) possible.

Exercise 1.5

Question 1. Rationalise the denominator of the following:
(i) \( \frac{3}{4\sqrt{5}} \)
(ii) \( \frac{5\sqrt{7}}{\sqrt{3}} \)
(iii) \( \frac{3}{4 - \sqrt{7}} \)
(iv) \( \frac{17}{3\sqrt{2} + 1} \)
(v) \( \frac{16}{\sqrt{41} - 5} \)
(vi) \( \frac{1}{\sqrt{7} - \sqrt{6}} \)
(vii) \( \frac{1}{\sqrt{5} + \sqrt{2}} \)
(viii) \( \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \)
Answer:
(i) To rationalise, multiply both numerator and denominator by \( \sqrt{5} \):

\( \frac{3}{4\sqrt{5}} = \frac{3 \times \sqrt{5}}{4\sqrt{5} \times \sqrt{5}} = \frac{3\sqrt{5}}{4 \times 5} = \frac{3\sqrt{5}}{20} \)

(ii) Multiply both numerator and denominator by \( \sqrt{3} \):

\( \frac{5\sqrt{7}}{\sqrt{3}} = \frac{5\sqrt{7} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{5\sqrt{21}}{3} \)

(iii) Multiply both numerator and denominator by the conjugate \( 4 + \sqrt{7} \):

\( \frac{3}{4 - \sqrt{7}} = \frac{3}{4 - \sqrt{7}} \times \frac{4 + \sqrt{7}}{4 + \sqrt{7}} = \frac{3(4 + \sqrt{7})}{(4)^2 - (\sqrt{7})^2} = \frac{3(4 + \sqrt{7})}{16 - 7} = \frac{3(4 + \sqrt{7})}{9} = \frac{4 + \sqrt{7}}{3} \)

(iv) Multiply both numerator and denominator by the conjugate \( 3\sqrt{2} - 1 \):

\( \frac{17}{3\sqrt{2} + 1} = \frac{17}{3\sqrt{2} + 1} \times \frac{3\sqrt{2} - 1}{3\sqrt{2} - 1} = \frac{17(3\sqrt{2} - 1)}{(3\sqrt{2})^2 - 1} = \frac{17(3\sqrt{2} - 1)}{18 - 1} = \frac{17(3\sqrt{2} - 1)}{17} = 3\sqrt{2} - 1 \)

(v) Multiply both numerator and denominator by the conjugate \( \sqrt{41} + 5 \):

\( \frac{16}{\sqrt{41} - 5} = \frac{16}{(\sqrt{41} - 5)} \times \frac{\sqrt{41} + 5}{\sqrt{41} + 5} = \frac{16(\sqrt{41} + 5)}{(\sqrt{41})^2 - 5^2} = \frac{16(\sqrt{41} + 5)}{41 - 25} = \frac{16(\sqrt{41} + 5)}{16} = \sqrt{41} + 5 \)

(vi) Multiply both numerator and denominator by the conjugate \( \sqrt{7} + \sqrt{6} \):

\( \frac{1}{\sqrt{7} - \sqrt{6}} = \frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6} \)

(vii) Multiply both numerator and denominator by the conjugate \( \sqrt{5} - \sqrt{2} \):

\( \frac{1}{\sqrt{5} + \sqrt{2}} = \frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3} \)

(viii) Multiply both numerator and denominator by the conjugate \( \sqrt{2} + \sqrt{3} \):

\( \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \times \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} = \frac{(\sqrt{2} + \sqrt{3})^2}{(\sqrt{2})^2 - (\sqrt{3})^2} \)

\( = \frac{(\sqrt{2})^2 + (\sqrt{3})^2 + 2 \times \sqrt{2} \times \sqrt{3}}{2 - 3} = \frac{2 + 3 + 2\sqrt{6}}{-1} = -(5 + 2\sqrt{6}) \)
In simple words: To remove a square root from a denominator, multiply the fraction by a form of 1 - use the conjugate (change the sign in the middle) or simply the same root. This makes the denominator a whole number.

Exam Tip: Always identify whether the denominator has one radical (multiply by that radical) or two radicals (use the conjugate formula with opposite sign). Apply \( (a+b)(a-b) = a^2 - b^2 \) correctly to eliminate the radical from the denominator.

Question 2. Simplify each of the following by rationalising the denominator:
(i) \( \frac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} \)
(ii) \( \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} \)
(iii) \( \frac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} \)
Answer:
(i) Multiply both numerator and denominator by the conjugate of the denominator, \( 7 + 3\sqrt{5} \):

\( \frac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} = \frac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} \times \frac{7 + 3\sqrt{5}}{7 + 3\sqrt{5}} = \frac{(7 + 3\sqrt{5})^2}{7^2 - (3\sqrt{5})^2} \)

\( = \frac{7^2 + (3\sqrt{5})^2 + 2 \times 7 \times 3\sqrt{5}}{49 - 45} = \frac{49 + 45 + 42\sqrt{5}}{4} = \frac{94 + 42\sqrt{5}}{4} = \frac{2(47 + 21\sqrt{5})}{4} = \frac{47 + 21\sqrt{5}}{2} \)

(ii) Multiply both numerator and denominator by the conjugate of the denominator, \( 3 - 2\sqrt{2} \):

\( \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{(3 - 2\sqrt{2})^2}{3^2 - (2\sqrt{2})^2} \)

\( = \frac{3^2 + (2\sqrt{2})^2 - 2 \times 3 \times 2\sqrt{2}}{9 - 8} = \frac{9 + 8 - 12\sqrt{2}}{1} = 17 - 12\sqrt{2} \)

(iii) Multiply both numerator and denominator by the conjugate of the denominator, \( 7 - 2\sqrt{14} \):

\( \frac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} = \frac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} \times \frac{7 - 2\sqrt{14}}{7 - 2\sqrt{14}} = \frac{(5 - 3\sqrt{14})(7 - 2\sqrt{14})}{7^2 - (2\sqrt{14})^2} \)

\( = \frac{35 - 10\sqrt{14} - 21\sqrt{14} + 6 \times 14}{49 - 56} = \frac{35 - 31\sqrt{14} + 84}{-7} = \frac{119 - 31\sqrt{14}}{-7} \)
In simple words: Multiply top and bottom by the conjugate (change the sign between the two terms in the denominator). Use the squared difference formula to eliminate radicals from the denominator, then simplify.

Exam Tip: Always use the conjugate: if the denominator is \( a + b\sqrt{c} \), multiply by \( a - b\sqrt{c} \). Expand the numerator carefully using FOIL or the perfect square formula. Simplify the final answer completely.

Question 3. Simplify: \( \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} - \frac{7 + 2\sqrt{14}}{5 - 3\sqrt{14}} \)
Answer: To simplify the first fraction, multiply numerator and denominator by the conjugate \( 3 - 2\sqrt{2} \):
\( \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{(3 - 2\sqrt{2})^2}{3^2 - (2\sqrt{2})^2} \)
\( \implies \frac{9 + 8 - 12\sqrt{2}}{9 - 8} = \frac{17 - 12\sqrt{2}}{1} = 17 - 12\sqrt{2} \)
For the second fraction, multiply numerator and denominator by the conjugate \( 7 - 2\sqrt{14} \):
\( \frac{7 + 2\sqrt{14}}{5 - 3\sqrt{14}} \times \frac{7 - 2\sqrt{14}}{7 - 2\sqrt{14}} = \frac{(7 + 2\sqrt{14})(7 - 2\sqrt{14})}{(5 - 3\sqrt{14})(7 - 2\sqrt{14})} \)
\( \implies \frac{49 - 56}{35 - 10\sqrt{14} - 21\sqrt{14} + 84} = \frac{-7}{119 - 31\sqrt{14}} \)
Multiply by -1 to get: \( \frac{7}{-119 + 31\sqrt{14}} = \frac{-119 + 31\sqrt{14}}{7} \)
Therefore: \( (17 - 12\sqrt{2}) - \frac{-119 + 31\sqrt{14}}{7} = 17 - 12\sqrt{2} + \frac{119 - 31\sqrt{14}}{7} \)
In simple words: Use the conjugate to remove square roots from denominators. The first term becomes \( 17 - 12\sqrt{2} \) and the second becomes a fraction with \( \sqrt{14} \).

Exam Tip: Always multiply by the conjugate of the denominator. Watch the signs carefully when subtracting negative fractions.

Question 3. Simplify: \( \frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} \)
Answer: Simplifying each term individually by rationalizing:
For \( \frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} \), multiply by \( \frac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} \):
\( \frac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{10 - 3} = \frac{7\sqrt{30} - 21}{7} = \sqrt{30} - 3 \)
For \( \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} \), multiply by \( \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} \):
\( \frac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{6 - 5} = 2\sqrt{30} - 10 \)
For \( \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} \), multiply by \( \frac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}} \):
\( \frac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{15 - 18} = \frac{3\sqrt{30} - 18}{-3} = 6 - \sqrt{30} \)
Combining all terms:
\( (\sqrt{30} - 3) - (2\sqrt{30} - 10) - (6 - \sqrt{30}) = \sqrt{30} - 3 - 2\sqrt{30} + 10 - 6 + \sqrt{30} = 1 \)
In simple words: Rationalize each fraction separately, then add and subtract them. Notice that many \( \sqrt{30} \) terms cancel out, leaving just 1.

Exam Tip: Work through each rationalization carefully. The final answer being a whole number suggests checking your arithmetic - often these problems are designed to have simple answers.

Question 4. Simplify: \( \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} \)
Answer: Rationalize each term individually:
\( \frac{1}{\sqrt{4} + \sqrt{5}} \times \frac{\sqrt{4} - \sqrt{5}}{\sqrt{4} - \sqrt{5}} = \frac{\sqrt{4} - \sqrt{5}}{4 - 5} = \sqrt{5} - \sqrt{4} \)
\( \frac{1}{\sqrt{5} + \sqrt{6}} \times \frac{\sqrt{5} - \sqrt{6}}{\sqrt{5} - \sqrt{6}} = \frac{\sqrt{5} - \sqrt{6}}{5 - 6} = \sqrt{6} - \sqrt{5} \)
\( \frac{1}{\sqrt{6} + \sqrt{7}} \times \frac{\sqrt{6} - \sqrt{7}}{\sqrt{6} - \sqrt{7}} = \frac{\sqrt{6} - \sqrt{7}}{6 - 7} = \sqrt{7} - \sqrt{6} \)
\( \frac{1}{\sqrt{7} + \sqrt{8}} \times \frac{\sqrt{7} - \sqrt{8}}{\sqrt{7} - \sqrt{8}} = \frac{\sqrt{7} - \sqrt{8}}{7 - 8} = \sqrt{8} - \sqrt{7} \)
\( \frac{1}{\sqrt{8} + \sqrt{9}} \times \frac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}} = \frac{\sqrt{8} - \sqrt{9}}{8 - 9} = \sqrt{9} - \sqrt{8} \)
Adding all these expressions (notice the telescoping pattern):
\( (\sqrt{5} - \sqrt{4}) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (\sqrt{8} - \sqrt{7}) + (\sqrt{9} - \sqrt{8}) = \sqrt{9} - \sqrt{4} = 3 - 2 = 1 \)
In simple words: Each fraction simplifies to a difference of two roots. When you add them all together, the middle terms cancel (telescoping), leaving only the first and last.

Exam Tip: Look for the telescoping pattern - this is a key feature of sums of conjugates. Always write out at least the first two and last two terms to spot the cancellation.

Question 5. Given a and b are rational numbers. Find a and b if:
(i) \( \frac{3 - \sqrt{5}}{3 + 2\sqrt{5}} = -\frac{19}{11} + a\sqrt{5} \)
Answer: Rationalizing the denominator by multiplying by \( \frac{3 - 2\sqrt{5}}{3 - 2\sqrt{5}} \):
\( \frac{(3 - \sqrt{5})(3 - 2\sqrt{5})}{(3 + 2\sqrt{5})(3 - 2\sqrt{5})} = \frac{9 - 6\sqrt{5} - 3\sqrt{5} + 10}{9 - 20} = \frac{19 - 9\sqrt{5}}{-11} = -\frac{19}{11} + \frac{9\sqrt{5}}{11} \)
Comparing with \( -\frac{19}{11} + a\sqrt{5} \), we get \( a = \frac{9}{11} \)
In simple words: Clear the square root from the denominator by multiplying by the conjugate. Then compare the rational and irrational parts separately to find a.

Exam Tip: Always separate rational from irrational parts when comparing. The coefficient of \( \sqrt{5} \) on the left must equal the coefficient of \( \sqrt{5} \) on the right.

Question 5. (ii) \( \frac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a - b\sqrt{6} \)
Answer: Rationalizing by multiplying by \( \frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} \):
\( \frac{(\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2})^2 - (2\sqrt{3})^2} = \frac{6 + 2\sqrt{6} + 3\sqrt{6} + 6}{18 - 12} = \frac{12 + 5\sqrt{6}}{6} = 2 + \frac{5\sqrt{6}}{6} \)
Writing as \( a - b\sqrt{6} \), we get \( 2 - (-\frac{5}{6})\sqrt{6} \), so \( a = 2 \) and \( b = -\frac{5}{6} \)
In simple words: Multiply by the conjugate to remove all roots from the denominator. Split the answer into a whole number part and a square root part to find both a and b.

Exam Tip: Pay attention to signs when matching against the given form. If the coefficient is negative, b itself might be negative.

Question 5. (iii) \( \frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11}b\sqrt{5} \)
Answer: Finding a common denominator:
\( \frac{(7 + \sqrt{5})^2 - (7 - \sqrt{5})^2}{(7 - \sqrt{5})(7 + \sqrt{5})} = \frac{(49 + 14\sqrt{5} + 5) - (49 - 14\sqrt{5} + 5)}{49 - 5} = \frac{28\sqrt{5}}{44} = \frac{7\sqrt{5}}{11} \)
Comparing with \( a + \frac{7}{11}b\sqrt{5} \), we get \( 0 + \frac{7}{11}(1)\sqrt{5} = \frac{7\sqrt{5}}{11} \)
Therefore, \( a = 0 \) and \( b = 1 \)
In simple words: Expand each squared term, subtract them, and simplify the denominator using difference of squares. The result is purely a multiple of \( \sqrt{5} \), so the rational part is zero.

Exam Tip: Use the identity \( (x + y)^2 - (x - y)^2 = 4xy \) to simplify the numerator quickly. This saves time and reduces errors.

Question 5. (iv) \( \frac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} = a + b\sqrt{24} \)
Answer: Rationalizing by multiplying by \( \frac{2\sqrt{2} - 2\sqrt{3}}{2\sqrt{2} - 2\sqrt{3}} \):
\( \frac{(2\sqrt{2} - \sqrt{3})(2\sqrt{2} - 2\sqrt{3})}{(2\sqrt{2})^2 - (2\sqrt{3})^2} = \frac{8 - 4\sqrt{6} - 2\sqrt{6} + 6}{8 - 12} = \frac{14 - 6\sqrt{6}}{-4} = -\frac{7}{2} + \frac{3\sqrt{6}}{2} \)
Since \( \sqrt{24} = 2\sqrt{6} \), we write \( -\frac{7}{2} + \frac{3}{4}\sqrt{24} \)
Therefore, \( a = -\frac{7}{2} \) and \( b = \frac{3}{4} \)
In simple words: After rationalizing, simplify the resulting fraction. Express \( \sqrt{6} \) as \( \frac{\sqrt{24}}{2} \) to match the required form.

Exam Tip: Remember that \( \sqrt{24} = 2\sqrt{6} \), so converting between forms requires careful algebra. Always double-check your conversion before comparing coefficients.

Question 6. If \( \frac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = p + q\sqrt{5} \), find the values of p and q where p and q are rational numbers.
Answer: Finding a common denominator:
\( \frac{(7 + 3\sqrt{5})(3 - \sqrt{5}) - (7 - 3\sqrt{5})(3 + \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} \)
Expanding the numerator:
\( (7 + 3\sqrt{5})(3 - \sqrt{5}) = 21 - 7\sqrt{5} + 9\sqrt{5} - 15 = 6 + 2\sqrt{5} \)
\( (7 - 3\sqrt{5})(3 + \sqrt{5}) = 21 + 7\sqrt{5} - 9\sqrt{5} - 15 = 6 - 2\sqrt{5} \)
Subtracting:
\( (6 + 2\sqrt{5}) - (6 - 2\sqrt{5}) = 4\sqrt{5} \)
The denominator is \( 9 - 5 = 4 \)
Therefore: \( \frac{4\sqrt{5}}{4} = \sqrt{5} = 0 + 1 \cdot \sqrt{5} \)
So \( p = 0 \) and \( q = 1 \)
In simple words: Expand both products carefully, subtract them, and simplify. The result contains only the square root term, so p is zero.

Exam Tip: When products contain many terms, organize the work by grouping rational and irrational parts. This prevents careless sign errors.

Question 7. Rationalise the denominator of the following and hence evaluate by taking \( \sqrt{2} = 1.414 \) and \( \sqrt{3} = 1.732 \), upto three places of decimal:
(i) \( \frac{\sqrt{2}}{2 + \sqrt{2}} \)
Answer: Rationalizing by multiplying by \( \frac{2 - \sqrt{2}}{2 - \sqrt{2}} \):
\( \frac{\sqrt{2}(2 - \sqrt{2})}{(2)^2 - (\sqrt{2})^2} = \frac{2\sqrt{2} - 2}{4 - 2} = \frac{2(\sqrt{2} - 1)}{2} = \sqrt{2} - 1 \)
Substituting \( \sqrt{2} = 1.414 \):
\( 1.414 - 1 = 0.414 \)
In simple words: Use the conjugate to eliminate the root from the denominator. Then substitute the given value to get the decimal answer.

Exam Tip: Always rationalize first before substituting numerical values - this gives a cleaner form to evaluate.

Question 7. (ii) \( \frac{1}{\sqrt{3} + \sqrt{2}} \)
Answer: Rationalizing by multiplying by \( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \):
\( \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2} \)
Substituting \( \sqrt{3} = 1.732 \) and \( \sqrt{2} = 1.414 \):
\( 1.732 - 1.414 = 0.318 \)
In simple words: The conjugate gives a simple difference of the two square root values. Subtract the given decimal approximations to find the answer.

Exam Tip: After rationalization, the denominator often becomes 1, making evaluation straightforward. Check this before moving to decimal approximation.

Question 8. If \( a = 2 + \sqrt{3} \), then find the value \( a - \frac{1}{a} \).
Answer: Finding \( \frac{1}{a} \):
\( \frac{1}{a} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \)
Therefore:
\( a - \frac{1}{a} = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3} \)
In simple words: First rationalize to find the reciprocal, then subtract it from the original value. The rational parts cancel, leaving only the square root part.

Exam Tip: Notice that \( a \) and \( \frac{1}{a} \) are conjugates - this makes their difference particularly simple.

Question 9. If \( x = 1 - \sqrt{2} \), find the value of \( \left( x - \frac{1}{x} \right)^4 \).
Answer: Finding \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{1 - \sqrt{2}} \times \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1 + \sqrt{2}}{1 - 2} = \frac{1 + \sqrt{2}}{-1} = -(1 + \sqrt{2}) \)
Finding \( x - \frac{1}{x} \):
\( x - \frac{1}{x} = (1 - \sqrt{2}) - (-(1 + \sqrt{2})) = 1 - \sqrt{2} + 1 + \sqrt{2} = 2 \)
Therefore:
\( \left( x - \frac{1}{x} \right)^4 = 2^4 = 16 \)
In simple words: Rationalize the reciprocal, then subtract it from x. The result is a simple number that can be raised to the fourth power easily.

Exam Tip: When a difference simplifies to a whole number, raising it to a power becomes much easier. Always check if intermediate simplifications occur.

Question 10. If \( x = 5 - 2\sqrt{6} \), find the value of \( x^2 + \frac{1}{x^2} \).
Answer: Finding \( \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{5 - 2\sqrt{6}} \times \frac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} = \frac{5 + 2\sqrt{6}}{25 - 24} = 5 + 2\sqrt{6} \)
Finding \( x + \frac{1}{x} \):
\( x + \frac{1}{x} = (5 - 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10 \)
Squaring both sides:
\( \left( x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2} = 100 \)
Therefore:
\( x^2 + \frac{1}{x^2} = 100 - 2 = 98 \)
In simple words: Find the reciprocal and add it to x to get 10. Square this sum to find \( x^2 + \frac{1}{x^2} \) by using the expansion formula.

Exam Tip: Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \) with \( ab = 1 \) is the key shortcut here - it avoids computing \( x^2 \) and \( \frac{1}{x^2} \) separately.

Question 11. If p = \( \frac{2 - \sqrt{5}}{2 + \sqrt{5}} \) and q = \( \frac{2 + \sqrt{5}}{2 - \sqrt{5}} \) find the values of:
(i) p + q
(ii) p - q
(iii) p² + q²
(iv) p² - q²
Answer:
(i) Add p and q. To find p + q, express both terms with a common denominator:
\( p + q = \frac{2 - \sqrt{5}}{2 + \sqrt{5}} + \frac{2 + \sqrt{5}}{2 - \sqrt{5}} = \frac{(2 - \sqrt{5})^2 + (2 + \sqrt{5})^2}{(2 - \sqrt{5})(2 + \sqrt{5})} \)
\( \implies \frac{4 + 5 - 4\sqrt{5} + 4 + 5 + 4\sqrt{5}}{4 - 5} = \frac{18}{-1} \)
\( \therefore p + q = -18 \) .....(i)

(ii) Find p - q using a similar method with the common denominator:
\( p - q = \frac{2 - \sqrt{5}}{2 + \sqrt{5}} - \frac{2 + \sqrt{5}}{2 - \sqrt{5}} = \frac{(2 - \sqrt{5})^2 - (2 + \sqrt{5})^2}{(2 - \sqrt{5})(2 + \sqrt{5})} \)
\( \implies \frac{4 + 5 - 4\sqrt{5} - 4 - 5 - 4\sqrt{5}}{-1} = \frac{-8\sqrt{5}}{-1} \)
\( \therefore p - q = 8\sqrt{5} \) .....(ii)

(iii) To get p² + q², use the identity (p + q)² = p² + q² + 2pq, which gives p² + q² = (p + q)² - 2pq:
First find pq: \( pq = \frac{2 - \sqrt{5}}{2 + \sqrt{5}} \times \frac{2 + \sqrt{5}}{2 - \sqrt{5}} = 1 \) .....(iv)
Then: \( p^2 + q^2 = (-18)^2 - 2(1) = 324 - 2 = 322 \)

(iv) For p² - q², use the factorisation (p² - q²) = (p + q)(p - q):
\( p^2 - q^2 = (-18) \times 8\sqrt{5} = -144\sqrt{5} \)
In simple words: When you add two square root fractions that are reciprocals, you get a negative whole number. When you subtract them, you get a multiple of the square root. Multiplying the sum and difference gives you another square root expression.

Exam Tip: Recognise that p and q are reciprocals (pq = 1). Use the identities for (p + q)² and p² - q² = (p + q)(p - q) to avoid lengthy algebraic expansion.

Question 1. Choose the correct statement:
(1) Reciprocal of every rational number is a rational number.
(2) The square roots of all positive integers are irrational numbers.
(3) The product of a rational and an irrational number is an irrational number.
(4) The difference of a rational number and an irrational number is an irrational number.
Answer: (4) The difference of a rational number and an irrational number is an irrational number.
In simple words: When you take any whole number (like 2) and subtract an irrational number (like \( \sqrt{3} \)), the answer will always be irrational. The remaining options fail because 0 has no reciprocal, not all roots are irrational, and a rational times irrational can be rational in some cases.

Exam Tip: Test option (1) by considering reciprocal of 0 - it doesn't exist, making the statement false. Option (4) is safest because subtraction from a rational always leaves an irrational unchanged in type.

Question 2. Every rational number is
(1) a natural number
(2) an integer
(3) a real number
(4) a whole number
Answer: (3) a real number
In simple words: Rational numbers are fractions where both the top and bottom parts are whole numbers. All such fractions belong to the set of real numbers, which is the largest group here.

Exam Tip: Remember the hierarchy: natural numbers ⊂ whole numbers ⊂ integers ⊂ rational numbers ⊂ real numbers. Every rational is automatically a real number.

Question 3. Between two rational numbers
(1) there is no rational number
(2) there is exactly one rational number
(3) there are infinitely many rational numbers
(4) there are only rational numbers and no irrational numbers
Answer: (3) there are infinitely many rational numbers
In simple words: No matter how close two rational numbers are, you can always find another one between them. In fact, you can keep doing this forever - there is no limit.

Exam Tip: The rational numbers are dense - between any two rationals, you can always construct another by taking their average.

Question 4. Decimal representation of a rational number cannot be
(1) terminating
(2) non-terminating
(3) non-terminating repeating
(4) non-terminating non-repeating
Answer: (4) non-terminating non-repeating
In simple words: Every rational number's decimal either stops or repeats at some point. It can never go on forever without any pattern - that would make it irrational.

Exam Tip: A key property: rational numbers have terminating or eventually repeating decimal expansions. Non-repeating, non-terminating decimals are exclusively irrational.

Question 5. The product of any two irrational numbers is
(1) always an irrational number
(2) always a rational number
(3) always an integer
(4) sometimes rational, sometimes irrational
Answer: (4) sometimes rational, sometimes irrational
In simple words: Multiply \( 2\sqrt{3} \) by \( 3\sqrt{3} \) and you get 18 - a regular number. But multiply \( 2\sqrt{2} \) by \( 3\sqrt{3} \) and you get \( 6\sqrt{6} \) - still irrational. It depends on what you multiply.

Exam Tip: Look for examples where the square roots "cancel out" (like \( \sqrt{3} \times \sqrt{3} = 3 \)) versus cases where they don't - this shows the product can be either type.

Question 6. The division of two irrational numbers is
(1) a rational number
(2) an irrational number
(3) either a rational number or an irrational number
(4) neither rational number nor irrational number
Answer: (3) either a rational number or an irrational number
In simple words: When you divide \( 2\sqrt{3} \) by \( 3\sqrt{3} \), the square roots cancel to give \( \frac{2}{3} \) - a rational number. But dividing \( 2\sqrt{3} \) by \( 3\sqrt{5} \) gives \( \frac{2\sqrt{3}}{3\sqrt{5}} \) - still irrational.

Exam Tip: The outcome hinges on whether the irrational parts simplify away or remain. Always test with specific examples before concluding.

Question 7. Which of the following is an irrational number
(1) \( \sqrt{\frac{4}{9}} \)
(2) \( \frac{\sqrt{12}}{\sqrt{3}} \)
(3) \( \sqrt{7} \)
(4) \( \sqrt{81} \)
Answer: (3) \( \sqrt{7} \)
In simple words: Option (1) simplifies to \( \frac{2}{3} \) - a fraction. Option (2) simplifies to 2 - a whole number. Option (4) is 9 - another whole number. But 7 is not a perfect square, so \( \sqrt{7} \) stays as an irrational number.

Exam Tip: Always simplify radicals by factorising - if the radicand is a perfect square (or the fraction reduces to one), the result is rational.

Question 8. Which of the following numbers has terminating decimal representation?
(1) \( \frac{3}{7} \)
(2) \( \frac{3}{5} \)
(3) \( \frac{1}{3} \)
(4) \( \frac{3}{11} \)
Answer: (2) \( \frac{3}{5} = 0.6 \) is terminating decimal representation.
In simple words: A fraction has a terminating decimal only if the denominator (in simplest form) has only the prime factors 2 and 5. Since 5 = 5, we get a terminating decimal. The others have prime factors like 3, 7, or 11, which cause endless repetition.

Exam Tip: Check the prime factorisation of the denominator - if it contains ONLY 2s and 5s (or is empty after cancelling), the decimal terminates.

Question 9. Which of the following is an irrational number?
(1) 0.14
(2) \( 0.\overline{1416} \)
(3) \( 0.1\overline{416} \)
(4) 0.4014001400014...
Answer: (4) 0.4014001400014...
In simple words: Option (1) stops after two decimal places - it's rational. Options (2) and (3) repeat at some point - they are also rational because they repeat. But option (4) never repeats and never stops - it has no pattern, making it irrational.

Exam Tip: Irrational decimals are non-terminating AND non-repeating. If you see a bar over digits or any repeating cycle, it's rational.

Question 10. Which of the following numbers has non-terminating repeating decimal expansion?
(1) \( \frac{11}{80} \)
(2) \( \frac{17}{160} \)
(3) \( \frac{63}{240} \)
(4) \( \frac{93}{420} \)
Answer: (4) \( \frac{93}{420} = 0.22142857... \) has non-terminating repeating decimal expansion.
In simple words: Fractions (1), (2), and (3) have denominators with only 2s and 5s as prime factors after simplifying, so they terminate. Fraction (4) reduces to a form containing the prime 3, so it repeats forever without ever ending.

Exam Tip: Simplify each fraction first, then check the prime factors of the denominator. Only 2 and 5 mean terminating; any other prime means repeating.

Question 11. A rational number between \( \sqrt{2} \) and \( \sqrt{3} \) is
(1) \( \frac{\sqrt{2} + \sqrt{3}}{2} \)
(2) \( \frac{\sqrt{2} \times \sqrt{3}}{2} \)
(3) 1.5
(4) 1.8
Answer: (3) 1.5
In simple words: Square \( \sqrt{2} \) and \( \sqrt{3} \) to get 2 and 3. Now find a rational perfect square between 2 and 3 - for example, 2.25. Then \( \sqrt{2.25} = 1.5 \). Since 2 < 2.25 < 3, we have \( \sqrt{2} < 1.5 < \sqrt{3} \).

Exam Tip: To find a rational between two irrationals, square them first, find a rational perfect square between the results, then take the square root.

Question 12. The decimal expansion of 2 - \( \sqrt{3} \) is
(1) terminating and non-repeating
(2) terminating and repeating
(3) non-terminating and non-repeating
(4) non-terminating and repeating
Answer: (3) non-terminating and non-repeating
In simple words: \( \sqrt{3} \) is irrational, meaning its decimal never stops and never repeats. When you subtract it from 2 (a rational number), the result stays irrational - still non-terminating and non-repeating.

Exam Tip: Combining a rational and irrational through arithmetic always yields an irrational result with the same decimal properties as the irrational component.

Question 13. The decimal expansion of the rational number \( \frac{33}{2^2 \times 5} \) will terminate after
(1) one decimal place
(2) two decimal places
(3) three decimal places
(4) four decimal places
Answer: (2) two decimal places
In simple words: Rewrite the denominator: \( \frac{33}{2^2 \times 5} = \frac{33}{4 \times 5} = \frac{33}{20} \). Convert to a power of 10 by multiplying: \( \frac{33}{20} = \frac{33 \times 5}{20 \times 5} = \frac{165}{100} = 1.65 \). The result has exactly 2 decimal places.

Exam Tip: To find the number of decimal places, express the denominator as \( 2^a \times 5^b \) and take the maximum of a and b - that is the number of decimal places.

Question 14. \( \sqrt{10} \times \sqrt{15} \) is equal to
(1) \( 6\sqrt{5} \)
(2) \( 5\sqrt{6} \)
(3) \( \sqrt{25} \)
(4) \( 10\sqrt{5} \)
Answer: (2) \( 5\sqrt{6} \)
In simple words: Multiply the numbers under the radicals: \( \sqrt{10} \times \sqrt{15} = \sqrt{10 \times 15} = \sqrt{150} \). Break 150 into prime factors: \( 150 = 2 \times 75 = 2 \times 25 \times 3 \). So \( \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6} \).

Exam Tip: Always multiply inside the radicals first, then factorise and extract perfect squares to simplify.

Question 15. \( 2\sqrt{3} + \sqrt{3} \) is equal to
(1) \( 2\sqrt{6} \)
(2) 6
(3) \( 3\sqrt{3} \)
(4) \( 4\sqrt{6} \)
Answer: (3) \( 3\sqrt{3} \)
In simple words: Both terms have the same irrational part \( \sqrt{3} \), so you simply add the numbers in front: 2 + 1 = 3. The result is \( 3\sqrt{3} \).

Exam Tip: Combine like radical terms the same way you combine like algebraic terms - add/subtract the coefficients and keep the radical unchanged.

Question 16. The value of \( \sqrt{8} + \sqrt{18} \) is
(1) \( \sqrt{26} \)
(2) \( 2(\sqrt{2} + \sqrt{3}) \)
(3) \( 5\sqrt{2} \)
(4) \( 6\sqrt{2} \)
Answer: (3) \( 5\sqrt{2} \)
In simple words: Simplify each radical separately. \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \) and \( \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \). Add them: \( 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2} \).

Exam Tip: Always reduce radicals to simplest form before combining - extract perfect square factors from under the radical sign.

Question 17. The number \( (2 - \sqrt{3})^2 \) is
(1) a natural number
(2) an integer
(3) a rational number
(4) an irrational number
Answer: (4) an irrational number
In simple words: Expand using the formula: \( (2 - \sqrt{3})^2 = 4 + 3 - 4\sqrt{3} = 7 - 4\sqrt{3} \). Because this expression contains \( 4\sqrt{3} \) (an irrational term), the entire result is irrational.

Exam Tip: When expanding \( (a - b)^2 \), if b contains an irrational part, the result will include an irrational cross-term unless it cancels (which it doesn't here).

Question 18. If x is a positive rational number which is not a perfect square, then \( -5\sqrt{x} \) is
(1) a negative integer
(2) an integer
(3) a rational number
(4) an irrational number
Answer: (4) an irrational number
In simple words: Since x is not a perfect square, \( \sqrt{x} \) is irrational. Multiplying an irrational by -5 (a rational) always gives an irrational result.

Exam Tip: The product of a non-zero rational and an irrational is always irrational - this is a key theorem.

Question 19. If x, y are both positive rational numbers, then \( (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) \) is
(1) a rational number
(2) an irrational number
(3) neither rational nor irrational number
(4) both rational as well as irrational number
Answer: (1) a rational number
In simple words: Use the difference-of-squares formula: \( (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = (\sqrt{x})^2 - (\sqrt{y})^2 = x - y \). Since x and y are both rational, their difference is also rational.

Exam Tip: Recognise the pattern (a + b)(a - b) = a² - b² - it instantly eliminates the irrational middle terms and leaves only the difference of two rationals.

Question 20. After rationalising the denominator of \( \frac{7}{3\sqrt{3} - 2\sqrt{2}} \), we get the denominator as
(1) 13
(2) 19
(3) 5
(4) 35
Answer: (2) 19
In simple words: Multiply the fraction by the conjugate \( \frac{3\sqrt{3} + 2\sqrt{2}}{3\sqrt{3} + 2\sqrt{2}} \). The new denominator becomes \( (3\sqrt{3})^2 - (2\sqrt{2})^2 = 27 - 8 = 19 \).

Exam Tip: For denominators of the form \( a\sqrt{m} - b\sqrt{n} \), multiply by the conjugate with a plus sign and apply the difference-of-squares identity to eliminate the radicals.

Question 21. The number obtained on rationalising the denominator of \( \frac{1}{\sqrt{7} - 2} \) is
(1) \( \frac{\sqrt{7} + 2}{3} \)
(2) \( \frac{\sqrt{7} - 2}{5} \)
(3) \( \frac{\sqrt{7} + 2}{5} \)
(4) \( \frac{\sqrt{7} + 2}{45} \)
Answer: (1) \( \frac{\sqrt{7} + 2}{3} \)
In simple words: Multiply by the conjugate: \( \frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3} \).

Exam Tip: Always use the conjugate (change the sign in the middle) to remove radicals from the denominator - the denominator becomes a simple integer.

Question 22. The number 0.25 is equal to
(1) \( \frac{65}{99} \)
(2) \( \frac{37}{99} \)
(3) \( \frac{5}{9} \)
(4) \( \frac{25}{99} \)
Answer: (4) \( \frac{25}{99} \)
In simple words: Let x = 0.25 = 0.252525... Multiply by 100 to shift the decimal two places: 100x = 25.252525... Subtract the original: 99x = 25, so x = \( \frac{25}{99} \).

Exam Tip: For a repeating decimal with period 2, multiply by 100; if period is 1, multiply by 10; if period is 3, multiply by 1000. Then subtract and solve for x.

Question 23. The value of \( 1.\overline{999} \) in the form of \( \frac{p}{q} \), where p and q are integers and q ≠ 0, is
(1) \( \frac{19}{20} \)
(2) \( \frac{1999}{1000} \)
(3) 2
(4) \( \frac{1}{9} \)
Answer: (3) 2
In simple words: Let x = 1.999... Multiply by 10: 10x = 19.999... Subtract the original: 9x = 18, so x = 2. This shows that 1.999... is actually equal to 2 exactly.

Exam Tip: The repeating decimal 0.999... equals 1. Similarly, 1.999... equals 2. Whenever the repeating digit is 9, the result is the next whole number.