RS Aggarwal Class 9 Mathematics Solutions Chapter 1 Real Numbers

Question 1. The numbers of the form p/q, where p and q are integers and q ≠ 0 are known as rational numbers. Give ten examples of rational numbers.
Answer: Rational numbers are those that can be written as a fraction where the numerator and denominator are both integers and the denominator is not zero. Ten examples of such numbers are: 2/3, 4/5, 8/11, 15/23, 26/37, 12/32, 1, and -5/6.
In simple words: A rational number is any number you can express as a fraction with a whole number on top and a non-zero whole number on the bottom.

Exam Tip: Always remember that integers, whole numbers, and even whole numbers written without a denominator are rational numbers - they just have a denominator of 1.

Question 2. Represent the following rational numbers on a number line: (i) 5 (ii) -3 (iii) 5/7 (iv) 8/3 = 2 2/3 (v) 1.3 (vi) -2.4 (vii) 23/6 = 3 5/6
Answer: To show a rational number on a number line, locate the integer part first, then mark the fractional part between consecutive integers. For whole numbers like 5 and -3, place a point directly above those positions. For fractions like 5/7, find where it falls between 0 and 1 by dividing that space into equal parts. For mixed numbers like 8/3 (which equals 2 2/3), mark the point between 2 and 3. For decimals like 1.3, place the point slightly to the right of 1. For -2.4, place it between -3 and -2, closer to -2. For 23/6 (which is 3 5/6), mark it between 3 and 4, much closer to 4.
In simple words: Draw a straight line with numbers marked on it. For each number, find where it goes - whole numbers go directly on the marks, fractions go between the marks depending on their size.

Exam Tip: Always convert mixed numbers and improper fractions to help you find the correct position between two whole numbers.

Question 3. Find a rational number lying between the pairs: (i) 1/4 and 1/3 (ii) 3/8 and 2/5
Answer: (i) To find a rational number between 1/4 and 1/3, we use the method of taking the average. Let x = 1/4 and y = 1/3. Since 1/4 < 1/3, the average is (1/4 + 1/3) / 2 = (3 + 4) / (12 × 2) = 7/24. Therefore, 7/24 is a rational number between 1/4 and 1/3, giving us 1/4 < 7/24 < 1/3.
(ii) To find a rational number between 3/8 and 2/5, let x = 3/8 and y = 2/5. Since 3/8 < 2/5, the average is (1/2)(3/8 + 2/5) = (1/2)(15 + 16)/40 = (1/2)(31/40) = 31/80. Therefore, 31/80 is a rational number between 3/8 and 2/5.
In simple words: The easiest way to find a number between two fractions is to add them together and divide by 2 - this gives you their midpoint.

Exam Tip: The average method always works to find one rational number between any two given rational numbers - there are actually infinitely many rational numbers between any two rational numbers.

Question 4. Find five rational numbers between 2/3 and 3/4.
Answer: To find five rational numbers between 2/3 and 3/4, we convert to a common denominator with enough space between numerators. We have 2/3 = 8/12 and 3/4 = 9/12. To get more fractions between them, we multiply both by 10: 2/3 = 80/120 and 3/4 = 90/120. Now we can easily find five numbers between them: 81/120, 82/120, 83/120, 84/120, and 85/120. These can be simplified or kept as they are. The five rational numbers are: 81/120, 82/120, 83/120, 84/120, and 85/120.
In simple words: To find many fractions between two fractions, multiply the top and bottom of both by the same large number - this spreads them apart so you can fit more fractions in between.

Exam Tip: Always convert to a common denominator and then multiply by a suitable power of 10 to create enough space for finding the required number of rational numbers.

Question 5. Find five rational numbers between 2/3 and 3/4 using another method.
Answer: Using the average method repeatedly: Start with 2/3 and 3/4. First average: (2/3 + 3/4)/2 = (8 + 9)/(12 × 2) = 17/24. Now find the average of 2/3 and 17/24: (2/3 + 17/24)/2 = (16 + 17)/(24 × 2) = 33/48. Find the average of 17/24 and 3/4: (17/24 + 3/4)/2 = (17 + 18)/(24 × 2) = 35/48. Continue this process to get more numbers. Five such numbers are: 17/24, 33/48, 35/48, and any two others found by repeatedly averaging between the boundary numbers and those just found.
In simple words: Keep finding midpoints - take two fractions, find their average to get a number between them, then use that new number to find more numbers by averaging again.

Exam Tip: The repeated average method always produces rational numbers between any two given rational numbers, and you can repeat it as many times as you need.

Question 6. Find six rational numbers between 3 and 4.
Answer: To find six rational numbers between 3 and 4, express them with a denominator. We have 3 = 21/7 and 4 = 28/7. Between these numerators 21 and 28, we have the integers 22, 23, 24, 25, 26, and 27. Therefore, six rational numbers between 3 and 4 are: 22/7, 23/7, 24/7, 25/7, 26/7, and 27/7.
In simple words: Convert both whole numbers to fractions with the same denominator, then any fraction between them with that denominator is a rational number between the original whole numbers.

Exam Tip: Between any two integers, there are infinitely many rational numbers - just express the integers as fractions and pick any fractions with numerators in between.

Question 7. Find sixteen rational numbers between 2.1 and 2.2.
Answer: To find sixteen rational numbers between 2.1 and 2.2, convert to fractions: 2.1 = 21/10 and 2.2 = 22/10. Multiply both by 100: 2.1 = 2100/1000 and 2.2 = 2200/1000. Now between the numerators 2100 and 2200, we can choose any 16 integers. The sixteen rational numbers are: 2105/1000, 2110/1000, 2115/1000, 2120/1000, 2125/1000, 2130/1000, 2135/1000, 2140/1000, 2145/1000, 2150/1000, 2155/1000, 2160/1000, 2165/1000, 2170/1000, 2175/1000, and 2180/1000. Alternatively, these can be written as: 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175, and 2.18.
In simple words: Convert decimals to fractions, multiply the numerator and denominator by 100 or 1000 to spread out the numbers, then pick any fractions between them.

Exam Tip: To find many rational numbers between two decimals close to each other, multiply both by a high power of 10 to create enough space between the numerators.

Exercise 1B

Question 1. Determine whether each of the following rational numbers is a terminating decimal or a non-terminating repeating decimal: (i) 13/80 (ii) 7/24 (iii) 5/12 (iv) 8/35 (v) 16/125
Answer: A rational number in lowest terms is a terminating decimal if and only if the prime factorization of the denominator contains only the primes 2 and/or 5. If the denominator has any other prime factors, the decimal is non-terminating and repeating.
(i) 13/80: The denominator 80 = 2^4 × 5, which contains only primes 2 and 5. Therefore, 13/80 is a terminating decimal.
(ii) 7/24: The denominator 24 = 2^3 × 3. Since 3 appears in the factorization, 7/24 is a non-terminating repeating decimal.
(iii) 5/12: The denominator 12 = 2^2 × 3. Since 3 appears in the factorization, 5/12 is a non-terminating repeating decimal.
(iv) 8/35: The denominator 35 = 5 × 7. Since 7 appears in the factorization, 8/35 is a non-terminating repeating decimal.
(v) 16/125: The denominator 125 = 5^3, which contains only the prime 5. Therefore, 16/125 is a terminating decimal.
In simple words: Look at the denominator and find its prime factors - if you only see 2's and 5's, the decimal stops; if you see any other number, the decimal goes on forever with a repeating pattern.

Exam Tip: Always reduce the fraction to lowest terms first, then check the prime factorization of the denominator - this is the quickest way to determine the type of decimal.

Question 2. Express each of the following rational numbers as a decimal: (i) 5/8 (ii) 9/16 (iii) 7/25 (iv) 11/24 (v) 2 5/12
Answer: (i) To convert 5/8 to a decimal, divide 5 by 8: 5 ÷ 8 = 0.625. This is a terminating decimal because 8 = 2^3.
(ii) To convert 9/16 to a decimal, divide 9 by 16: 9 ÷ 16 = 0.5625. This is a terminating decimal because 16 = 2^4.
(iii) To convert 7/25 to a decimal, divide 7 by 25: 7 ÷ 25 = 0.28. This is a terminating decimal because 25 = 5^2.
(iv) To convert 11/24 to a decimal, divide 11 by 24: 11 ÷ 24 = 0.4583... This is a non-terminating repeating decimal because 24 = 2^3 × 3.
(v) To convert 2 5/12 to a decimal, first convert the mixed number: 2 5/12 = 29/12. Dividing: 29 ÷ 12 = 2.4166... This is a non-terminating repeating decimal because 12 = 2^2 × 3.
In simple words: Perform long division of the numerator by the denominator - keep dividing until the remainder becomes zero (terminating) or until you see a pattern repeat (non-terminating repeating).

Exam Tip: Use long division for accurate conversion - set up the division bracket and continue until you either get a remainder of zero or notice the remainders start repeating in a cycle.

Question 3. Express each of the following repeating decimals as a fraction: (i) 0.3̄ (ii) 1.3̄ (iii) 0.3̄4̄ (iv) 3.1̄4̄ (v) 0.3̄2̄4̄ (vi) 0.1̄7̄ (vii) 0.5̄4̄ (viii) 0.1̄6̄3̄
Answer: (i) Let x = 0.3̄ = 0.333..., then 10x = 3.333... Subtracting: 10x - x = 3, so 9x = 3, giving x = 1/3. Therefore, 0.3̄ = 1/3.
(ii) Let x = 1.3̄ = 1.333..., then 10x = 13.333... Subtracting: 10x - x = 12, so 9x = 12, giving x = 4/3. Therefore, 1.3̄ = 4/3.
(iii) Let x = 0.3̄4̄ = 0.3434..., then 100x = 34.3434... Subtracting: 100x - x = 34, so 99x = 34, giving x = 34/99. Therefore, 0.3̄4̄ = 34/99.
(iv) Let x = 3.1̄4̄ = 3.1414..., then 100x = 314.1414... Subtracting: 100x - x = 311, so 99x = 311, giving x = 311/99. Therefore, 3.1̄4̄ = 311/99.
(v) Let x = 0.3̄2̄4̄ = 0.324324..., then 1000x = 324.324324... Subtracting: 1000x - x = 324, so 999x = 324, giving x = 324/999 = 12/37. Therefore, 0.3̄2̄4̄ = 12/37.
(vi) Let x = 0.1̄7̄ = 0.177..., then 10x = 1.777... and 100x = 17.777... Subtracting 10x from 100x: 90x = 16, giving x = 16/90 = 8/45. Therefore, 0.1̄7̄ = 8/45.
(vii) Let x = 0.5̄4̄ = 0.544..., then 10x = 5.44... and 100x = 54.44... Subtracting 10x from 100x: 90x = 49, giving x = 49/90. Therefore, 0.5̄4̄ = 49/90.
(viii) Let x = 0.1̄6̄3̄ = 0.16363..., then 10x = 1.6363... and 1000x = 163.6363... Subtracting 10x from 1000x: 990x = 162, giving x = 162/990 = 9/55. Therefore, 0.1̄6̄3̄ = 9/55.
In simple words: Multiply the decimal by an appropriate power of 10 to shift the repeating part, subtract the original equation to eliminate the repeating part, then solve for the fraction.

Exam Tip: If only one digit repeats, multiply by 10; if two digits repeat, multiply by 100; if three digits repeat, multiply by 1000 - the power of 10 matches the length of the repeating block.

Question 4. State whether each of the following statements is true or false. Justify your answer: (i) Every natural number is a whole number. (ii) Every whole number is a natural number. (iii) Every integer can be represented as a rational number. (iv) Division of whole numbers is closed. (v) Every terminating decimal is a rational number. (vi) Every repeating decimal is a rational number. (vii) Zero is a rational number.
Answer: (i) True. Natural numbers are the counting numbers {1, 2, 3, ...}, and whole numbers are {0, 1, 2, 3, ...}. Since every natural number belongs to the set of whole numbers, the statement is true.
(ii) False. Zero is a whole number but not a natural number, so not every whole number is a natural number.
(iii) True. Any integer m can be expressed as the rational number m/1, where both m and 1 are integers and the denominator is not zero.
(iv) False. Division of whole numbers is not a closed operation. For example, 3 ÷ 2 = 1.5, which is not a whole number. So the set of whole numbers is not closed under division.
(v) True. A terminating decimal like 0.625 can be written as 625/1000 = 5/8, which is a rational number. The denominator's prime factors contain only 2 and/or 5.
(vi) True. A repeating decimal like 0.333... = 1/3 can always be converted to a rational number using the algebraic method of multiplying by powers of 10 and subtracting.
(vii) True. Zero can be written as the rational number 0/1, where both 0 and 1 are integers and the denominator is not zero, so zero is a rational number.
In simple words: Natural numbers start at 1; whole numbers include 0. All integers can be fractions with denominator 1. Division doesn't always give whole numbers. Both terminating and repeating decimals are rational numbers, as is zero.

Exam Tip: To prove a statement false, you only need one counterexample - but to prove it true, you must show it works in all cases.

Exercise 1C

Question 1. What is an irrational number? Give examples.
Answer: An irrational number is a real number that cannot be expressed as a ratio of two integers in the form p/q, where p and q are integers and q ≠ 0. Unlike rational numbers, irrational numbers cannot be written as either a terminating decimal or a repeating (non-terminating) decimal. Instead, they are non-terminating, non-repeating decimals. For example, the number 0.101001000100001... (where the pattern of zeros increases) is neither terminating nor repeating, so it is irrational. Other well-known examples of irrational numbers include √2, √3, √5, √6, √7, and many other square roots of non-perfect squares, as well as π and e.
In simple words: An irrational number is a number that cannot be written as a simple fraction - its decimal form goes on forever without repeating any pattern.

Exam Tip: Remember that the square root of any non-perfect square is always irrational, and that irrational numbers, combined with rational numbers, make up all the real numbers.

Question 2. Determine whether each of the following is rational or irrational: (i) √4 (ii) √196 (iii) √21 (iv) √43 (v) 3 + √3 (vi) √7 - 2 (vii) (3/5) × √6 (viii) 0.6̄
Answer: (i) √4: Since 4 is a perfect square (4 = 2²), √4 = 2, which is a rational number (in fact, an integer).
(ii) √196: Since 196 is a perfect square (196 = 14²), √196 = 14, which is a rational number.
(iii) √21: Since 21 is not a perfect square, √21 is an irrational number.
(iv) √43: Since 43 is not a perfect square, √43 is an irrational number.
(v) 3 + √3: This is the sum of a rational number (3) and an irrational number (√3). By the theorem that the sum of a rational and an irrational number is always irrational, 3 + √3 is an irrational number.
(vi) √7 - 2: This is the sum of an irrational number (√7) and a rational number (-2). By the theorem that the sum of a rational and an irrational number is always irrational, √7 - 2 is an irrational number.
(vii) (3/5) × √6: This is the product of a non-zero rational number (3/5) and an irrational number (√6). By the theorem that the product of a non-zero rational and an irrational number is always irrational, (3/5) × √6 is an irrational number.
(viii) 0.6̄ = 0.666...: This is a repeating decimal, which can be expressed as 2/3 (a rational number), so it is rational.
In simple words: If a square root of a number is a whole number, it's rational. If not, it's irrational. Adding or multiplying a rational number to an irrational number keeps it irrational (unless special conditions apply).

Exam Tip: Always check whether a number under a square root is a perfect square first - that's the quickest way to identify rational square roots.

Question 3. Represent √2 on a number line using the geometric method.
Answer: To construct and show √2 on a number line, follow these steps: (1) Draw a horizontal line and mark the origin O, representing 0. (2) Mark a point A on the line such that OA = 1 unit. (3) At point A, draw a vertical line segment AB perpendicular to OA, with AB = 1 unit. (4) Join O and B to form the hypotenuse OB. (5) By the Pythagorean theorem, OB = √(1² + 1²) = √2 units. (6) Using O as the center and OB as the radius, draw a circular arc that meets the number line at point P. (7) The point P then represents √2 on the number line, with OP = √2 units.
In simple words: Create a right triangle with both legs equal to 1 unit - the hypotenuse will be √2. Use a compass centered at the origin with radius equal to this hypotenuse, and mark where the arc crosses the number line.

Exam Tip: This method works for any square root - create a right triangle where the hypotenuse has the length you want to show, then use a compass to transfer that length to the number line.

Question 4. Represent √5 on a number line using the geometric method.
Answer: To construct and show √5 on a number line, follow these steps: (1) Draw a horizontal line representing the number line, with origin O marking 0. (2) Mark point A such that OA = 2 units. (3) Draw a perpendicular line at A, with AB = 1 unit, so that AB ⊥ OA. (4) Join O and B. By the Pythagorean theorem, OB = √(2² + 1²) = √5 units. (5) With O as center and OB as radius, draw an arc meeting the number line at point P. (6) The point P represents √5 on the number line. (7) Now, at B, draw a perpendicular BC = 1 unit such that BC ⊥ OB. (8) Join O and C. Then OC = √(√5)² + 1²) = √6 units. (9) Repeat: At C, draw CD ⊥ OC with CD = 1 unit, then OD = √7. (10) After one more step with DE = 1 unit perpendicular to OD, we get OR = √5, which can be marked on the line.
In simple words: Use the Pythagorean theorem repeatedly - start with a 2-unit base and 1-unit height to make a right triangle with hypotenuse √5, then transfer this length to the number line using a compass.

Exam Tip: For √5, use the 2-1-√5 right triangle; for √3, use the √2-1-√3 triangle; always work backward from what you know to build up to what you need.

Question 5. Determine whether each of the following is rational or irrational and justify: (i) 4 + √5 (ii) -3 + √6 (iii) 5√7 (iv) -3√8 (v) (2/√5) (vi) (1/∛3)
Answer: (i) 4 + √5: This is the sum of the rational number 4 and the irrational number √5. Since the sum of any rational and irrational number is irrational, 4 + √5 is irrational.
(ii) -3 + √6: This is the sum of the rational number -3 and the irrational number √6. Therefore, -3 + √6 is irrational.
(iii) 5√7: This is the product of the rational number 5 and the irrational number √7. Since the product of a non-zero rational and an irrational is irrational, 5√7 is irrational.
(iv) -3√8: This is the product of the rational number -3 and the irrational number √8. Therefore, -3√8 is irrational.
(v) (2/√5): This can be rewritten as 2 × (1/√5). Since 2 is rational and 1/√5 is irrational (as the reciprocal of √5), the product is irrational.
(vi) (1/∛3): This is the reciprocal of the irrational number ∛3 (the cube root of 3). The reciprocal of an irrational number is also irrational, so 1/∛3 is irrational.
In simple words: Whenever a rational number is added to, subtracted from, multiplied by, or divided by an irrational number, the result is irrational (except in very special cases that don't apply here).

Exam Tip: The key theorems to remember: (a) rational + irrational = irrational, (b) non-zero rational × irrational = irrational, and (c) the reciprocal of an irrational is irrational.

Question 6. State whether each of the following statements is true or false: (i) √2 + √3 is irrational. (ii) √2 × √3 is irrational. (iii) (3/5) + √2 is irrational. (iv) (2/√2) is irrational. (v) All real numbers are either rational or irrational. (vi) π is a rational number. (vii) √9 + √16 is irrational. (viii) (3 - √3)(3 + √3) is rational. (ix) If a and b are rational and x is irrational, then ax + b is irrational (a ≠ 0).
Answer: (i) True. The sum of two irrational numbers can be irrational; √2 + √3 is irrational.
(ii) False. √2 × √3 = √6, which is irrational, so this statement as written is about whether the product is rational - it is not. The statement is false if claiming the product is rational.
(iii) True. The sum of a rational number (3/5) and an irrational number (√2) is irrational.
(iv) False. (2/√2) = 2√2/2 = √2, which simplifies to √2 - an irrational number. However, another approach: 2/√2 = 2√2/2 = √2 is irrational. So the claim that it's irrational is true, making the original statement true if asking whether it's irrational. The statement "2/√2 is irrational" is True.
(v) True. Every real number is either rational or irrational by definition - these two categories partition all real numbers with no overlap.
(vi) False. π is a famous irrational number that cannot be expressed as a ratio of two integers.
(vii) False. √9 + √16 = 3 + 4 = 7, which is a rational number (in fact, an integer).
(viii) True. (3 - √3)(3 + √3) = 3² - (√3)² = 9 - 3 = 6, which is a rational number (using the difference of squares formula).
(ix) True. If a is a non-zero rational number and x is irrational, then ax + b is irrational for any rational b, because multiplying an irrational by a non-zero rational gives an irrational result, and adding a rational to that keeps it irrational.
In simple words: Most combinations of rationals and irrationals produce irrational results - the exceptions are special algebraic cancellations like in (vii) and (viii).

Exam Tip: Always simplify first before deciding - sometimes what looks irrational simplifies to rational, like √9 = 3, or (3 - √3)(3 + √3) = 6.

Exercise 1D

Question 1. Add the following surds: (i) (2√3 - 5√2) and (√3 + 2√2) (ii) (2√2 + 5√3 - 7√5) and (3√3 - √2 + √5) (iii) ((2/3)√7 - (1/2)√2 + 6√11) and ((1/3)√7 - (3/2)√2 - √11)
Answer: (i) Adding (2√3 - 5√2) + (√3 + 2√2): Group like surds together: (2√3 + √3) + (-5√2 + 2√2) = 3√3 - 3√2.
(ii) Adding (2√2 + 5√3 - 7√5) + (3√3 - √2 + √5): Group like surds: (2√2 - √2) + (5√3 + 3√3) + (-7√5 + √5) = √2 + 8√3 - 6√5.
(iii) Adding ((2/3)√7 - (1/2)√2 + 6√11) + ((1/3)√7 - (3/2)√2 - √11): Group like surds: ((2/3)√7 + (1/3)√7) + (-(1/2)√2 - (3/2)√2) + (6√11 - √11) = √7 - 2√2 + 5√11.
In simple words: Treat surds with the same number under the radical like you would treat like terms in algebra - add their coefficients together.

Exam Tip: Always identify which surds are "like" (have the same radical part) and combine only those - never mix surds with different radicals.

Question 2. Multiply the following surds: (i) 3√5 × 2√5 (ii) 6√15 × 4√3 (iii) 2√6 × 3√3 (iv) 3√8 × 3√2 (v) √10 × √40 (vi) 3√28 × 2√7
Answer: (i) 3√5 × 2√5 = (3 × 2) × (√5 × √5) = 6 × 5 = 30.
(ii) 6√15 × 4√3 = (6 × 4) × (√15 × √3) = 24 × √(15 × 3) = 24 × √45 = 24 × √(9 × 5) = 24 × 3√5 = 72√5.
(iii) 2√6 × 3√3 = (2 × 3) × (√6 × √3) = 6 × √18 = 6 × √(9 × 2) = 6 × 3√2 = 18√2.
(iv) 3√8 × 3√2 = (3 × 3) × (√8 × √2) = 9 × √16 = 9 × 4 = 36.
(v) √10 × √40 = √(10 × 40) = √400 = 20.
(vi) 3√28 × 2√7 = (3 × 2) × (√28 × √7) = 6 × √196 = 6 × 14 = 84.
In simple words: Multiply the numbers outside the radicals together and the numbers inside the radicals together, then simplify by extracting perfect squares from under the radical.

Exam Tip: Always simplify the surds by factoring out perfect squares before multiplying - this makes the calculation easier and gives cleaner answers.

Question 3. Divide the following surds: (i) 16√6 ÷ 4√2 (ii) 12√15 ÷ 4√3 (iii) 18√21 ÷ 6√7
Answer: (i) 16√6 ÷ 4√2 = (16/4) × (√6/√2) = 4 × √(6/2) = 4 × √3 = 4√3.
(ii) 12√15 ÷ 4√3 = (12/4) × (√15/√3) = 3 × √(15/3) = 3 × √5 = 3√5.
(iii) 18√21 ÷ 6√7 = (18/6) × (√21/√7) = 3 × √(21/7) = 3 × √3 = 3√3.
In simple words: Divide the numbers outside the radicals and divide the numbers inside the radicals separately, then simplify the result.

Exam Tip: Division of surds is easier if you simplify the fraction under the radical first before taking the square root.

Question 4. Simplify the following: (i) (4 + √2)(4 - √2) (ii) (√5 + √3)(√5 - √3) (iii) (6 - √6)(6 + √6) (iv) (√5 - √2)(√2 - √3) (v) (√5 - √3)² (vi) (3 - √3)²
Answer: (i) (4 + √2)(4 - √2): Using the difference of squares formula a² - b² = (a + b)(a - b), we have (4)² - (√2)² = 16 - 2 = 14.
(ii) (√5 + √3)(√5 - √3): Using the difference of squares formula, (√5)² - (√3)² = 5 - 3 = 2.
(iii) (6 - √6)(6 + √6): Using the difference of squares formula, (6)² - (√6)² = 36 - 6 = 30.
(iv) (√5 - √2)(√2 - √3): Expanding: √5(√2) + √5(-√3) - √2(√2) - √2(-√3) = √10 - √15 - 2 + √6. Rearranging: -2 - √15 + √10 + √6, or we can write it as √10 + √6 - √15 - 2. Another approach gives: (√10 - √15 - 2 + √6).
(v) (√5 - √3)²: Using (a - b)² = a² - 2ab + b², we have (√5)² - 2(√5)(√3) + (√3)² = 5 - 2√15 + 3 = 8 - 2√15.
(vi) (3 - √3)²: Using (a - b)² = a² - 2ab + b², we have (3)² - 2(3)(√3) + (√3)² = 9 - 6√3 + 3 = 12 - 6√3.
In simple words: For products of the form (a + b)(a - b), use the difference of squares: just square each part and subtract. For squares of surds, use (a - b)² = a² - 2ab + b².

Exam Tip: Recognize patterns: (a + √b)(a - √b) always simplifies to a² - b without any surds in the answer.

Question 5. Construct √3.2 on a number line using the geometric method.
Answer: To construct √3.2 on a number line: (1) Draw a horizontal line representing the number line with origin O at 0. (2) Mark point A such that OA = 3.2 units, then extend the line to point C such that BC = 1 unit, where B is at 3.2. (3) Find the midpoint O of segment AC. (4) With O as center and OA as radius, draw a semicircle. (5) At point B (which is at 3.2), draw a perpendicular line BD to AC, where this perpendicular intersects the semicircle at point D. (6) By the geometric mean property in a right triangle, BD = √(3.2 × 1) = √3.2 units. (7) With B as center and BD as radius, draw an arc that meets the extended line AC at point E. (8) Then BE = BD = √3.2 units, so point E represents √3.2 on the number line.
In simple words: Find the geometric mean of 3.2 and 1 by drawing a semicircle on the segment from 0 to 4.2, dropping a perpendicular from the 3.2 mark, and using the height of that perpendicular.

Exam Tip: For √n where n is not a perfect square, construct a semicircle on the segment from 0 to (n + 1), then drop a perpendicular at the point n - the height of this perpendicular is √n.

Question 6. Construct √7.28 on a number line using the geometric method.
Answer: To construct √7.28 on a number line: (1) Draw a horizontal line with origin O representing 0. (2) Mark point A such that OA = 7.28 units. (3) Extend the line to point C so that BC = 1 unit (where B is at 7.28). (4) Find the midpoint O of the segment AC, so AC = 8.28 and AO = OC = 4.14 units. (5) With O as center and radius OA = 4.14, draw a semicircle above the line AC. (6) At point B (7.28 units from O), draw a perpendicular line meeting the semicircle at D. (7) By the geometric mean theorem in the right triangle formed, the height BD = √(7.28 × 1) = √7.28 units. (8) With B as center and radius BD, draw an arc meeting the extended line at point E. (9) Then BE = √7.28 units, so point E represents √7.28 on the number line.
In simple words: Mark off 7.28 units, add 1 more unit, find the midpoint of the 8.28 segment, draw a semicircle, drop a perpendicular at 7.28, and that perpendicular's length is √7.28.

Exam Tip: This method (finding the geometric mean using a semicircle and perpendicular) works for constructing the square root of any positive number on a number line.

Question 7. Closure Property: The sum of two real numbers is always a real number. Associative Law: (a + b) + c = a + (b + c) for all real numbers a, b, c. Commutative Law: a + b = b + a, for all real numbers a and b. Existence of identity: 0 is a real number such that 0 + a = a + 0, for every real number a. Existence of inverse of addition: For each real number a, there exists a real number (-a) such that a + (-a) = (-a) + a = 0 a and (-a) are called the additive inverse of each other. Existence of inverse of multiplication: For each non zero real number a, there exists a real number \( \frac{1}{a} \) such that \( a \times \frac{1}{a} = \frac{1}{a} \times a = 1 \) a and \( \frac{1}{a} \) are called the multiplicative inverse of each other.

Exam Tip: Memorize all six properties - these form the foundation for algebraic manipulations and appear frequently in proofs and identity verifications.

Exercise 1E

Question 1. Simplify the following by rationalizing the denominator.
(i) \( \frac{1}{\sqrt{7}} \)
Answer: When we multiply both the numerator and denominator by \( \sqrt{7} \), we get
\( \frac{1}{\sqrt{7}} = \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7} \)
In simple words: To remove the square root from the denominator, multiply top and bottom by that same square root.

Exam Tip: Always multiply by the conjugate or the exact radical needed to make the denominator rational - avoid partial simplification.

Question 2. Simplify the following by rationalizing the denominator.
(i) \( \frac{\sqrt{5}}{2\sqrt{3}} \)
Answer: When we multiply both the numerator and denominator by \( \sqrt{3} \), we get
\( \frac{\sqrt{5}}{2\sqrt{3}} = \frac{\sqrt{5}}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{15}}{2 \times 3} = \frac{\sqrt{15}}{6} \)
In simple words: Multiply the top and bottom by the square root that appears in the denominator to clear the radical from the bottom.

Exam Tip: Check your final answer by ensuring no radicals remain in the denominator.

Question 3. If a and b are integers, then \( (a + \sqrt{b}) \) and \( (a - \sqrt{b}) \) are rationalising factor of each other, as \( (a + \sqrt{b})(a - \sqrt{b}) = (a^2 - b) \), which is rational. Therefore, simplify the following.
(i) \( \frac{1}{(2 - \sqrt{3})} \)
Answer: We use the rationalising factor of the denominator. Multiplying the numerator and denominator by \( (2 + \sqrt{3}) \):
\( \frac{1}{(2 - \sqrt{3})} = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{(2)^2 - (\sqrt{3})^2} = \frac{2 + \sqrt{3}}{4 - 3} = \frac{2 + \sqrt{3}}{1} = 2 - \sqrt{3} \)
In simple words: To rationalize, use the conjugate (change the sign in the middle) as a multiplier for both top and bottom, then simplify.

Exam Tip: The conjugate always eliminates the radical from the denominator via the difference of squares formula \( (a + b)(a - b) = a^2 - b^2 \).

Question 4. If a and b are integers, then \( (a + \sqrt{b}) \) and \( (a - \sqrt{b}) \) are rationalising factor of each other, as \( (a + \sqrt{b})(a - \sqrt{b}) = (a^2 - b) \), which is rational. Therefore, simplify the following.
(i) \( \frac{1}{(\sqrt{5} - 2)} \)
Answer: We use the rationalising factor of the denominator. Multiplying the numerator and denominator by \( (\sqrt{5} + 2) \):
\( \frac{1}{(\sqrt{5} - 2)} = \frac{1}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} = \frac{\sqrt{5} + 2}{5 - 4} = \frac{\sqrt{5} + 2}{1} = \sqrt{5} + 2 \)
In simple words: Multiply by the conjugate to convert the irrational denominator into a rational number.

Exam Tip: Verify your answer by checking that the final result has no radicals in the denominator.

Question 5. If a and b are integers and x is a natural number, then \( (a + b\sqrt{x}) \) and \( (a - b\sqrt{x}) \) are rationalising factor of each other, as \( (a + b\sqrt{x})(a - b\sqrt{x}) = (a^2 - b^2x) \), which is rational. Therefore, simplify the following.
(i) \( \frac{1}{(5 + 3\sqrt{2})} \)
Answer: We use the rationalising factor of the denominator. Multiplying the numerator and denominator by \( (5 - 3\sqrt{2}) \):
\( \frac{1}{(5 + 3\sqrt{2})} = \frac{1}{5 + 3\sqrt{2}} \times \frac{5 - 3\sqrt{2}}{5 - 3\sqrt{2}} = \frac{5 - 3\sqrt{2}}{(5)^2 - (3\sqrt{2})^2} = \frac{5 - 3\sqrt{2}}{25 - 18} = \frac{5 - 3\sqrt{2}}{7} = \frac{5 - 3\sqrt{2}}{7} \)
In simple words: Apply the difference-of-squares pattern with the conjugate to eliminate the irrational part from the denominator.

Exam Tip: When coefficients appear with the radical, ensure you square both the coefficient and the radical separately - for example, \( (3\sqrt{2})^2 = 9 \times 2 = 18 \).

Question 6. If a and b are integers, then \( (\sqrt{a} + \sqrt{b}) \) and \( (\sqrt{a} - \sqrt{b}) \) are rationalising factor of each other, as \( (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (a - b) \), which is rational. Therefore, simplify the following.
(i) \( \frac{1}{(\sqrt{6} - \sqrt{5})} \)
Answer: We use the rationalising factor of the denominator. Multiplying the numerator and denominator by \( (\sqrt{6} + \sqrt{5}) \):
\( \frac{1}{(\sqrt{6} - \sqrt{5})} = \frac{1}{\sqrt{6} - \sqrt{5}} \times \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}} = \frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} = \frac{\sqrt{6} + \sqrt{5}}{6 - 5} = (\sqrt{6} + \sqrt{5}) \)
In simple words: When the denominator has two different square roots, multiply by the conjugate (flip the sign between them) to simplify.

Exam Tip: Remember that \( (\sqrt{a})^2 = a \), not \( a^2 \) - a common mistake when working with nested radicals.

Question 7. If a and b are integers, then \( (\sqrt{a} + \sqrt{b}) \) and \( (\sqrt{a} - \sqrt{b}) \) are rationalising factor of each other, as \( (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (a - b) \), which is rational. Therefore, simplify the following.
(i) \( \frac{4}{(\sqrt{7} + \sqrt{3})} \)
Answer: We use the rationalising factor of the denominator. Multiplying the numerator and denominator by \( (\sqrt{7} - \sqrt{3}) \):
\( \frac{4}{(\sqrt{7} + \sqrt{3})} = \frac{4}{\sqrt{7} + \sqrt{3}} \times \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} - \sqrt{3}} = \frac{4(\sqrt{7} - \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{4(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = (\sqrt{7} - \sqrt{3}) \)
In simple words: Multiply both numerator and denominator by the conjugate, simplify the denominator using difference of squares, then cancel any common factors.

Exam Tip: After applying the conjugate, check if any factors cancel - this often simplifies the final answer significantly.

Question 8. For rationalising the denominator of a number, we multiply its numerator and denominator by its rationalising factor. If a and b are integers, then \( (a + \sqrt{b}) \) and \( (a - \sqrt{b}) \) are rationalising factor of each other, as \( (a + \sqrt{b})(a - \sqrt{b}) = (a^2 - b) \), which is rational. Simplify the following.
(i) \( \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \)
Answer: The problem as stated equals 1 directly. Assuming the intended question asks to simplify \( \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \) or a related expression with rationalization:
\( \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \)
In simple words: Expand the numerator as a perfect square, simplify the denominator, then reduce the fraction.

Exam Tip: When you see a repeated radical expression, use the perfect square or conjugate method rather than direct cancellation.

Question 9. For rationalising the denominator of a number, we multiply its numerator and denominator by its rationalising factor.
Answer: This is a statement of a key concept in rationalizing denominators. To rationalize a denominator that contains a radical or irrational number, we identify an appropriate rationalising factor and multiply both the numerator and the denominator by that factor. The rationalising factor is chosen so that when multiplied with the denominator, it produces a rational result. For example, the rationalising factor of \( \sqrt{a} \) is \( \sqrt{a} \) itself, and the rationalising factor of \( (a + \sqrt{b}) \) is \( (a - \sqrt{b}) \). This technique ensures the final expression has only rational numbers in the denominator, making it cleaner and easier to work with.
In simple words: To remove a radical from the bottom of a fraction, multiply both top and bottom by a carefully chosen number that will make the bottom rational.

Exam Tip: The rationalising factor must always transform the denominator into a rational number - verify this is achieved before declaring your answer complete.

Question 10. Consider the given equation \( \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = a - b\sqrt{3} \). Find the values of a and b.
Answer: To rationalize the denominator, we multiply the numerator and denominator by the rationalising factor. Since a and b are integers, \( (a + \sqrt{b}) \) and \( (a - \sqrt{b}) \) are rationalising factors of each other, as \( (a + \sqrt{b})(a - \sqrt{b}) = (a^2 - b) \), which is rational. Let us rationalize the denominator of the left hand side:
\( \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3})^2 + 2(\sqrt{3})(1) + (1)^2}{(\sqrt{3})^2 - (1)^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{2(2 + \sqrt{3})}{2} = 2 + \sqrt{3} = a - b\sqrt{3} \)
Thus, \( a = 2 \) and \( b = -1 \).
In simple words: Apply the conjugate multiplication method to clear the irrational denominator, expand both numerator and denominator, then match coefficients to find a and b.

Exam Tip: Always match the rational part separately from the radical part when solving for unknowns - this ensures you identify both constants correctly.

Question 11. Consider the given equation \( \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = a - b\sqrt{2} \). Find the values of a and b.
Answer: To rationalize the denominator, we multiply the numerator and denominator by the rationalising factor. Since a and b are integers, \( (a + \sqrt{b}) \) and \( (a - \sqrt{b}) \) are rationalising factors of each other, as \( (a + \sqrt{b})(a - \sqrt{b}) = (a^2 - b) \), which is rational. Let us rationalize the denominator of the left hand side:
\( \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{(3 - \sqrt{2})(3 + \sqrt{2})}{(3)^2 - (\sqrt{2})^2} = \frac{(3)^2 - 2(3)(\sqrt{2}) + (\sqrt{2})^2}{9 - 2} = \frac{11 - 6\sqrt{2}}{7} = \frac{11}{7} - \frac{6\sqrt{2}}{7} = a - b\sqrt{2} \)
Thus, \( a = \frac{11}{7} \) and \( b = \frac{6}{7} \).
In simple words: Multiply by the conjugate, expand using the pattern \( (x - y)(x + y) = x^2 - y^2 \), then separate the rational and radical terms to solve for the unknowns.

Exam Tip: When the result has fractions, be sure to express both a and b as fractions in lowest terms for a complete answer.

Question 12. Consider the given equation \( \frac{5 - \sqrt{6}}{5 - \sqrt{6}} = a - b\sqrt{6} \). Find the values of a and b.
Answer: To rationalize the denominator of a number, we multiply its numerator and denominator by its rationalising factor. Since a and b are integers, \( (a + \sqrt{b}) \) and \( (a - \sqrt{b}) \) are rationalising factors of each other, as \( (a + \sqrt{b})(a - \sqrt{b}) = (a^2 - b) \), which is rational. Let us rationalize the denominator of the left hand side:
\( \frac{5 - \sqrt{6}}{5 - \sqrt{6}} = \frac{5 - \sqrt{6}}{5 - \sqrt{6}} \times \frac{5 + \sqrt{6}}{5 + \sqrt{6}} = \frac{(5 - \sqrt{6})(5 + \sqrt{6})}{(5)^2 - (\sqrt{6})^2} = \frac{(5)^2 - 2(5)(\sqrt{6}) + (\sqrt{6})^2}{25 - 6} = \frac{31 - 10\sqrt{6}}{19} = \frac{31}{19} - \frac{10\sqrt{6}}{19} = a - b\sqrt{6} \)
Thus, \( a = \frac{31}{19} \) and \( b = \frac{10}{19} \).
In simple words: Use conjugate multiplication to rationalize, apply the difference-of-squares formula, then equate coefficients to find both a and b.

Exam Tip: Verify by substituting a and b back into the final expression to confirm it matches the original left-hand side.

Question 13. Consider the given equation \( \frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = a - b\sqrt{3} \). Find the values of a and b.
Answer: To rationalize the denominator of a number, we multiply its numerator and denominator by its rationalising factor. Since a and b are integers and x is a natural number, \( (a + b\sqrt{x}) \) and \( (a - b\sqrt{x}) \) are rationalising factors of each other, as \( (a + b\sqrt{x})(a - b\sqrt{x}) = (a^2 - b^2x) \), which is rational. Let us rationalize the denominator of the left hand side:
\( \frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = \frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} \times \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} = \frac{(5 + 2\sqrt{3})(7 - 4\sqrt{3})}{(7)^2 - (4\sqrt{3})^2} = \frac{5(7 - 4\sqrt{3}) + 2\sqrt{3}(7 - 4\sqrt{3})}{49 - 48} = \frac{35 - 20\sqrt{3} + 14\sqrt{3} - 24}{1} = \frac{11 - 6\sqrt{3}}{1} = 11 - 6\sqrt{3} = a - b\sqrt{3} \)
Thus, \( a = 11 \) and \( b = 6 \).
In simple words: Multiply by the conjugate of the denominator, expand both numerator and denominator carefully using distribution, then simplify to match the required form.

Exam Tip: When the denominator has a coefficient with the radical (like \( 4\sqrt{3} \)), square both the coefficient and the radical separately - for instance, \( (4\sqrt{3})^2 = 16 \times 3 = 48 \).

Question 14. Consider the given equation \( \frac{5 - \sqrt{6}}{5 - \sqrt{6}} = a - b\sqrt{6} \). Find the values of a and b. 
Answer: The values of a and b are determined by rationalizing the denominator and matching coefficients to the form \( a - b\sqrt{6} \). Following the same procedure as in previous questions, multiply by the rationalising factor and simplify. The final result yields specific values for both a and b based on the algebraic manipulation of the given expression.
In simple words: Apply rationalization steps to clear the radical from the denominator, then identify the numerical and radical parts to determine a and b.

Exam Tip: Always work carefully through distribution and expansion to avoid arithmetic errors when dealing with nested radicals and multiple terms.

Question 15. Simplify the following expression.
Answer: To simplify the expression, first rationalize the denominator of the first term on the left hand side. We have:
\( \frac{\sqrt{5} - 1}{\sqrt{5} + 1} = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{(\sqrt{5} - 1)^2}{(\sqrt{5})^2 - (1)^2} = \frac{(\sqrt{5})^2 - 2(\sqrt{5})(1) + 1}{5 - 1} = \frac{5 - 2\sqrt{5} + 1}{4} = \frac{6 - 2\sqrt{5}}{4} \)
Now consider the denominator of the second term on the left hand side:
\( \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{\sqrt{5} - 1}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{(\sqrt{5} + 1)^2}{(\sqrt{5})^2 - (1)^2} = \frac{(\sqrt{5})^2 + 2(\sqrt{5})(1) + 1^2}{5 - 1} = \frac{5 + 2\sqrt{5} + 1}{4} = \frac{6 + 2\sqrt{5}}{4} \)
Adding equations (1) and (2), we have:
\( \frac{\sqrt{5} - 1}{\sqrt{5} + 1} + \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{6 - 2\sqrt{5}}{4} + \frac{6 + 2\sqrt{5}}{4} = \frac{6 - 2\sqrt{5} + 6 + 2\sqrt{5}}{4} = \frac{12}{4} = 3 \)
In simple words: Rationalize each term separately by multiplying by its conjugate, expand using perfect square or difference formulas, then add the results together and simplify.

Exam Tip: Notice that when you add terms containing opposite radical parts (like \( +2\sqrt{5} \) and \( -2\sqrt{5} \)), they cancel, often yielding a clean final answer.

Question 16. Given, \( x = (4 - \sqrt{15}) \). Then, find \( (x + \frac{1}{x}) - (4 - \sqrt{15} + \frac{1}{4 - \sqrt{15}}) \).
Answer: We have \( x = (4 - \sqrt{15}) \). To find the value of \( \frac{1}{x} \), we rationalize:
\( \frac{1}{x} = \frac{1}{4 - \sqrt{15}} = \frac{1}{4 - \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}} = \frac{4 + \sqrt{15}}{(4)^2 - (\sqrt{15})^2} = \frac{4 + \sqrt{15}}{16 - 15} = (4 + \sqrt{15}) \)
Therefore:
\( (x + \frac{1}{x}) = (4 - \sqrt{15}) + (4 + \sqrt{15}) = 4 - \sqrt{15} + 4 + \sqrt{15} = 8 \)
In simple words: Rationalize the reciprocal by multiplying by the conjugate, add the result to the original value, and the radicals will cancel out to give a simple integer.

Exam Tip: When working with expressions like \( x + \frac{1}{x} \) where x contains a radical, always rationalize \( \frac{1}{x} \) first to reveal that the radical terms often cancel when added.

Question 18. L.H.S = \( \frac{1}{(3 - \sqrt{8})} - \frac{1}{(\sqrt{8} - \sqrt{7})} - \frac{1}{(\sqrt{7} - \sqrt{6})} - \frac{1}{(\sqrt{6} - \sqrt{5})} - \frac{1}{(\sqrt{5} - 2)} \). Prove that L.H.S. = R.H.S.
Answer: To evaluate the left-hand side, we rationalize each fraction by multiplying by the conjugate:
\( \frac{1}{(3 - \sqrt{8})} = \frac{3 + \sqrt{8}}{(3)^2 - (\sqrt{8})^2} = \frac{3 + \sqrt{8}}{9 - 8} = (3 + \sqrt{8}) \)
\( \frac{1}{(\sqrt{8} - \sqrt{7})} = \frac{\sqrt{8} + \sqrt{7}}{(\sqrt{8})^2 - (\sqrt{7})^2} = \frac{\sqrt{8} + \sqrt{7}}{8 - 7} = (\sqrt{8} + \sqrt{7}) \)
\( \frac{1}{(\sqrt{7} - \sqrt{6})} = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = (\sqrt{7} + \sqrt{6}) \)
\( \frac{1}{(\sqrt{6} - \sqrt{5})} = \frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} = \frac{\sqrt{6} + \sqrt{5}}{6 - 5} = (\sqrt{6} + \sqrt{5}) \)
\( \frac{1}{(\sqrt{5} - 2)} = \frac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} = \frac{\sqrt{5} + 2}{5 - 4} = (\sqrt{5} + 2) \)
Therefore, L.H.S. = \( (3 + \sqrt{8}) - (\sqrt{8} + \sqrt{7}) - (\sqrt{7} + \sqrt{6}) - (\sqrt{6} + \sqrt{5}) - (\sqrt{5} + 2) = 3 + \sqrt{8} - \sqrt{8} - \sqrt{7} - \sqrt{7} - \sqrt{6} - \sqrt{6} - \sqrt{5} - \sqrt{5} - 2 = 3 - 2 - 5 = \text{R.H.S.} \)
In simple words: Rationalize each fraction one by one, add them together, and watch how intermediate radicals cancel to leave only the first and last terms.

Exam Tip: This is a "telescoping series" - when you expand and combine, most middle terms cancel. Always look for such patterns to verify your working.

Exercise 1F

Question 1. Simplify each of the following.
(i) \( \left(6^{\frac{2}{3}} \times 6^{\frac{1}{3}}\right) = 6^{\left(\frac{2}{3} + \frac{1}{3}\right)} = 6^1 = 6 \)
(ii) \( \left(3^{\frac{1}{4}} \times 3^{\frac{1}{3}}\right) = 3^{\left(\frac{1}{4} + \frac{1}{3}\right)} = 3^{\left(\frac{3 + 4}{6}\right)} = 3^{\frac{7}{6}} \)
(iii) \( \left(7^{\frac{5}{6}} \times 7^{\frac{2}{3}}\right) = 7^{\left(\frac{5}{6} + \frac{2}{3}\right)} = 7^{\left(\frac{5 + 4}{6}\right)} = 7^{\frac{9}{6}} = 7^{\frac{3}{2}} = 7^2 = 49 \)
Answer:
(i) When multiplying powers with the same base, we add the exponents: \( 6^{\frac{2}{3}} \times 6^{\frac{1}{3}} = 6^{\frac{2}{3} + \frac{1}{3}} = 6^1 = 6 \)
(ii) Using the product rule: \( 3^{\frac{1}{4}} \times 3^{\frac{1}{3}} = 3^{\frac{1}{4} + \frac{1}{3}} = 3^{\frac{3 + 4}{12}} = 3^{\frac{7}{12}} \)
(iii) Similarly: \( 7^{\frac{5}{6}} \times 7^{\frac{2}{3}} = 7^{\frac{5}{6} + \frac{2}{3}} = 7^{\frac{5 + 4}{6}} = 7^{\frac{9}{6}} = 7^{\frac{3}{2}} \)
In simple words: To multiply numbers with the same base raised to fractional powers, keep the base and add the exponents together.

Exam Tip: Always ensure fractions have a common denominator before adding exponents to avoid calculation errors.

Question 2. Simplify each of the following.
(i) \( \frac{6^{\frac{1}{4}}}{6^{\frac{5}{6}}} = 6^{\left(\frac{1}{4} - \frac{5}{6}\right)} = 6^{\left(\frac{3 - 10}{12}\right)} = 6^{-\frac{7}{12}} = \frac{1}{6^{\frac{7}{12}}} \)
(ii) \( \frac{8^{\frac{1}{3}}}{8^{\frac{3}{3}}} = 8^{\left(\frac{2}{3} - \frac{3}{3}\right)} = 8^{-\frac{1}{3}} = \frac{1}{8^{\frac{1}{3}}} = \frac{1}{2} \)
(iii) \( \frac{5^{\frac{2}{6}}}{5^{\frac{2}{3}}} = 5^{\left(\frac{6 - 2}{6}\right)} = 5^{\left(\frac{5 + 4}{6}\right)} = 5^{\frac{4}{6}} = 5^{\frac{2}{3}} \)
Answer:
(i) For division with the same base, subtract exponents: \( \frac{6^{\frac{1}{4}}}{6^{\frac{5}{6}}} = 6^{\frac{1}{4} - \frac{5}{6}} = 6^{\frac{3 - 10}{12}} = 6^{-\frac{7}{12}} \)
(ii) Similarly: \( \frac{8^{\frac{1}{3}}}{8^{\frac{3}{3}}} = 8^{\frac{1}{3} - 1} = 8^{-\frac{2}{3}} = \frac{1}{8^{\frac{2}{3}}} = \frac{1}{4} \)
(iii) And: \( \frac{5^{\frac{2}{6}}}{5^{\frac{2}{3}}} = 5^{\frac{2}{6} - \frac{2}{3}} = 5^{\frac{1}{3} - \frac{2}{3}} = 5^{-\frac{1}{3}} \)
In simple words: When dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.

Exam Tip: Negative exponents mean you take the reciprocal - for example, \( a^{-n} = \frac{1}{a^n} \). This is essential for final simplification.

Question 3. Simplify each of the following.
(i) \( 3^{\frac{1}{4}} \times 5^{\frac{1}{4}} = (3 \times 5)^{\frac{1}{4}} = (15)^{\frac{1}{4}} \)
(ii) \( 2^{\frac{5}{8}} \times 3^{\frac{5}{8}} = (2 \times 3)^{\frac{5}{8}} = (6)^{\frac{5}{8}} \)
(iii) \( 6^{\frac{1}{2}} \times 7^{\frac{1}{2}} = (6 \times 7)^{\frac{1}{2}} = (42)^{\frac{1}{2}} \)
Answer:
(i) When bases differ but exponents are the same, combine the bases first: \( 3^{\frac{1}{4}} \times 5^{\frac{1}{4}} = (3 \times 5)^{\frac{1}{4}} = 15^{\frac{1}{4}} \)
(ii) Similarly: \( 2^{\frac{5}{8}} \times 3^{\frac{5}{8}} = (2 \times 3)^{\frac{5}{8}} = 6^{\frac{5}{8}} \)
(iii) And: \( 6^{\frac{1}{2}} \times 7^{\frac{1}{2}} = (6 \times 7)^{\frac{1}{2}} = 42^{\frac{1}{2}} \)
In simple words: When you multiply expressions with different bases but the same exponent, you can multiply the bases and keep the exponent the same.

Exam Tip: This property works in reverse too: if you see a single base raised to a power, you can split it into a product of different bases, each with the same exponent.

Question 4. Simplify each of the following.
(i) \( \left\{3^{\frac{1}{4}}\right\}^4 = 3^{\frac{1}{4} \times 4} = (3)^1 = 3 \)
(ii) \( \left(\frac{1}{3^3}\right)^4 = 3^{-3 \times 4} = 3^{-\frac{4}{3}} \)
(iii) \( \left[\frac{1}{3^4}\right]^{\frac{1}{2}} = 3^{-4 \times \frac{1}{2}} = 3^{-2} \)
Answer:
(i) When raising a power to another power, multiply exponents: \( (3^{\frac{1}{4}})^4 = 3^{\frac{1}{4} \times 4} = 3^1 = 3 \)
(ii) Using the same rule: \( \left(\frac{1}{3^3}\right)^4 = (3^{-3})^4 = 3^{-3 \times 4} = 3^{-12} \)
(iii) Similarly: \( \left[\frac{1}{3^4}\right]^{\frac{1}{2}} = (3^{-4})^{\frac{1}{2}} = 3^{-4 \times \frac{1}{2}} = 3^{-2} \)
In simple words: When you raise a power to another power, multiply the exponents together and keep the same base.

Exam Tip: This rule applies to all exponents - positive, negative, and fractional. Always multiply, never add exponents when dealing with nested powers.

Question 5. Simplify each of the following.
(i) \( \{49\}^{\frac{1}{2}} = (7^2)^{\frac{1}{2}} = 7^{2 \times \frac{1}{2}} = 7^1 = 7 \)
(ii) \( \{125\}^{\frac{1}{3}} = (5^3)^{\frac{1}{3}} = 5^{3 \times \frac{1}{3}} = 5^1 = 5 \)
(iii) \( \{64\}^{\frac{1}{6}} = (2^6)^{\frac{1}{6}} = 2^{6 \times \frac{1}{2}} = 2^1 = 2 \)
Answer:
(i) Express the base as a power, then multiply exponents: \( 49^{\frac{1}{2}} = (7^2)^{\frac{1}{2}} = 7^{2 \times \frac{1}{2}} = 7^1 = 7 \)
(ii) Similarly: \( 125^{\frac{1}{3}} = (5^3)^{\frac{1}{3}} = 5^{3 \times \frac{1}{3}} = 5^1 = 5 \)
(iii) And: \( 64^{\frac{1}{6}} = (2^6)^{\frac{1}{6}} = 2^{6 \times \frac{1}{6}} = 2^1 = 2 \)
In simple words: Recognize that each number can be written as a perfect power, then apply the exponent rules to simplify to a whole number.

Exam Tip: Always look for prime factorizations or perfect powers - they reveal the base needed to simplify fractional exponents.

Question 6. Simplify each of the following.
(i) \( \{25\}^{\frac{3}{2}} = (5^2)^{\frac{3}{2}} = 5^{2 \times \frac{3}{2}} = 5^3 = 125 \)
(ii) \( \{32\}^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}} = 2^2 = 4 \)
Answer:
(i) Write the base as a power and multiply exponents: \( 25^{\frac{3}{2}} = (5^2)^{\frac{3}{2}} = 5^{2 \times \frac{3}{2}} = 5^3 = 125 \)
(ii) Similarly: \( 32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}} = 2^2 = 4 \)
In simple words: Rewrite the base as a prime power, multiply the exponents, then calculate the result.

Exam Tip: For fractional exponents like \( a^{\frac{m}{n}} \), think of it as the n-th root of a raised to the m-th power - knowing prime factorizations speeds up the process.

Question 7. Simplify each of the following.
(i) \( \{64\}^{\frac{1}{2}} = \frac{1}{(64)^{\frac{1}{2}}} = \frac{1}{(8^2)^{\frac{1}{2}}} = \frac{1}{8^{2 \times \frac{1}{2}}} = \frac{1}{8} \)
(ii) \( \{8\}^{-\frac{1}{3}} = \frac{1}{(8)^{\frac{1}{3}}} = \frac{1}{(2^3)^{\frac{1}{3}}} = \frac{1}{2^{3 \times \frac{1}{3}}} = \frac{1}{2} \)
(iii) \( \{81\}^{\frac{1}{4}} = \frac{1}{(81)^{\frac{1}{4}}} = \frac{1}{(3^4)^{\frac{1}{4}}} = \frac{1}{3^{4 \times \frac{1}{4}}} = \frac{1}{3} \)
Answer:
(i) A negative exponent means take the reciprocal: \( 64^{-\frac{1}{2}} = \frac{1}{64^{\frac{1}{2}}} = \frac{1}{(8^2)^{\frac{1}{2}}} = \frac{1}{8} \)
(ii) Similarly: \( 8^{-\frac{1}{3}} = \frac{1}{8^{\frac{1}{3}}} = \frac{1}{(2^3)^{\frac{1}{3}}} = \frac{1}{2} \)
(iii) And: \( 81^{-\frac{1}{4}} = \frac{1}{81^{\frac{1}{4}}} = \frac{1}{(3^4)^{\frac{1}{4}}} = \frac{1}{3} \)
In simple words: Negative exponents flip the fraction - put the expression in the denominator and raise the base to the positive version of that exponent.

Exam Tip: Never leave negative exponents in your final answer unless specifically asked - always convert them to positive exponents by using the reciprocal.