CBSE Class 10 Maths HOTs Real Numbers Set C

TOPIC 2 : Irrational Numbers, Terminating And Non-Terminating, Recurring Decimals

Very Short Answer Type Questions

Question. What is the condition for the decimal expansion of a rational number to terminate? Explain with the help of an example.
Answer: The decimal expansion of a rational number terminates, if the denominator of rational number can be expressed as \(2^n 5^m\) where \(m\) and \(n\) are non negative integers and \(p\) and \(q\) both co-primes.
e.g. \(\frac{3}{10} = \frac{3}{2^1 \times 5^1} = 0.3\)

Question. Find the smallest positive rational number by which \(\frac{1}{7}\) should be multiplied so that its decimal expansion terminates after 2 places of decimal.
Answer: Since \(\frac{1}{7} \times \frac{7}{100} = \frac{1}{100} = 0.01\).
Thus smallest rational number is \(\frac{7}{100}\)

Question. What type of decimal expansion does a rational number has? How can you distinguish it from decimal expansion of irrational numbers?
Answer: A rational number has its decimal expansion either terminating or non-terminating, repeating. An irrational numbers has its decimal expansion non-repeating and non-terminating.

Question. Calculate \(\frac{3}{8}\) in the decimal form.
Answer: We have \(\frac{3}{8} = \frac{3}{2^3} = \frac{3 \times 5^3}{2^3 \times 5^3} = \frac{375}{10^3} = \frac{375}{1000} = 0.375\)

Question. The decimal representation of \(\frac{6}{1250}\) will terminate after how many places of decimal?
Answer: We have \(\frac{6}{1250} = \frac{6}{2 \times 5^4} = \frac{6 \times 2^3}{2 \times 2^3 \times 5^4} = \frac{6 \times 2^3}{2^4 \times 5^4} = \frac{48}{(10)^4} = \frac{48}{10000} = 0.0048\)
Thus \(\frac{6}{1250}\) will terminate after 4 decimal places.

Question. Write whether rational number \(\frac{7}{75}\) will have terminating decimal expansion or a non-terminating decimal.
Answer: We have \(\frac{7}{75} = \frac{7}{3 \times 5^2}\).
Since denominator of given rational number is not of form \(2^n \times 5^m\), Hence, It is non-terminating decimal expansion.

Short Answer Type Questions - I

Question. Show that \(5\sqrt{6}\) is an irrational number.
Answer: Let \(5\sqrt{6}\) be a rational number, which can be expressed as \(\frac{a}{b}\), where \(b \neq 0\); \(a\) and \(b\) are co-primes.
Now \(5\sqrt{6} = \frac{a}{b}\)
\(\sqrt{6} = \frac{a}{5b}\)
or, \(\sqrt{6} = \text{rational}\)
But, \(\sqrt{6}\) is an irrational number. Thus, our assumption is wrong. Hence, \(5\sqrt{6}\) is an irrational number.

Question. Write the denominator of the rational number \(\frac{257}{500}\) in the form \(2^m \times 5^n\), where \(m\) and \(n\) are non-negative integers. Hence write its decimal expansion without actual division.
Answer: We have \(500 = 25 \times 20 = 5^2 \times 5 \times 4 = 5^3 \times 2^2\)
Here denominator is 500 which can be written as \(2^2 \times 5^3\).
Now decimal expansion,
\(\frac{257}{500} = \frac{257 \times 2}{2 \times 2^2 \times 5^3} = \frac{514}{10^3} = 0.514\)

Question. Write a rational number between \(\sqrt{2}\) and \(\sqrt{3}\).
Answer: We have \(\sqrt{2} = \sqrt{\frac{200}{100}}\) and \(\sqrt{3} = \sqrt{\frac{300}{100}}\).
We need to find a rational number \(x\) such that \(\frac{1}{10}\sqrt{200} < x < \frac{1}{10}\sqrt{300}\).
Choosing any perfect square such as 225 or 256 in between 200 and 300, we have
\(x = \frac{\sqrt{225}}{10} = \frac{15}{10} = 1.5\) or \(x = \frac{\sqrt{256}}{10} = \frac{16}{10} = 1.6\)

Question. Write the rational number \( \frac{7}{75} \) will have a terminating decimal expansion. or a non-terminating repeating decimal.
Answer: We have \( \frac{7}{75} = \frac{7}{3 \times 5^2} \). The denominator of rational number \( \frac{7}{75} \) can not be written in form \( 2^n 5^m \). So it is non-terminating repeating decimal expansion.

Short Answer Type Questions - II

Question. Express \( (\frac{15}{4} + \frac{5}{40}) \) as a decimal fraction without actual division.
Answer: We have \( \frac{15}{4} + \frac{5}{40} = \frac{15 \times 25}{4 \times 25} + \frac{5 \times 25}{40 \times 25} \)
\( = \frac{375}{100} + \frac{125}{1000} \)
\( = 3.75 + 0.125 = 3.875 \)

Question. Express the number \( 0.3\overline{178} \) in the form of rational number \( \frac{a}{b} \).
Answer: Let \( x = 0.3\overline{178} \)
\( x = 0.3178178178... \)
\( 10,000x = 3178.178178... \)
\( 10x = 3.178178... \)
Subtracting, \( 9990x = 3175 \)
or, \( x = \frac{3175}{9990} = \frac{635}{1998} \)

Question. Prove that \( \sqrt{2} \) is an irrational number.
Answer: Let \( \sqrt{2} \) be a rational number.
Then \( \sqrt{2} = \frac{p}{q} \), where \( p \) and \( q \) are co-prime integers and \( q \neq 0 \). On squaring both the sides we have,
\( 2 = \frac{p^2}{q^2} \)
or, \( p^2 = 2q^2 \)
Since \( p^2 \) is divisible by 2, thus \( p \) is also divisible by 2.
Let \( p = 2r \) for some positive integer \( r \), then we have
\( (2r)^2 = 2q^2 \)
\( 4r^2 = 2q^2 \)
or, \( q^2 = 2r^2 \)
Since \( q^2 \) is divisible by 2, thus \( q \) is also divisible by 2.
We have seen that \( p \) and \( q \) are divisible by 2, which contradicts the fact that \( p \) and \( q \) are co-primes. Hence, our assumption is false and \( \sqrt{2} \) is irrational.

Question. If \( p \) is prime number, then prove that \( \sqrt{p} \) is an irrational.
Answer: Let \( p \) be a prime number and if possible, let \( \sqrt{p} \) be rational.
Thus \( \sqrt{p} = \frac{m}{n} \), where \( m \) and \( n \) are co-primes and \( n \neq 0 \).
Squaring on both sides, we get
\( p = \frac{m^2}{n^2} \)
or, \( pn^2 = m^2 \) ...(1)
Here \( p \) divides \( pn^2 \). Thus \( p \) divides \( m^2 \) and in result \( p \) also divides \( m \).
Let \( m = pq \) for some integer \( q \) and putting \( m = pq \) in eq. (1), we have
\( pn^2 = (pq)^2 = p^2 q^2 \)
or, \( n^2 = pq^2 \)
Here \( p \) divides \( pq^2 \). Thus \( p \) divides \( n^2 \) and in result \( p \) also divides \( n \).
[\( \because p \) is prime and \( p \) divides \( n^2 \implies p \) divides \( n \)]
Thus \( p \) is a common factor of \( m \) and \( n \) but this contradicts the fact that \( m \) and \( n \) are primes. The contradiction arises by assuming that \( \sqrt{p} \) is rational. Hence, \( \sqrt{p} \) is irrational.

Question. Prove that \( 3 + \sqrt{5} \) is an irrational number.
Answer: Assume that \( 3 + \sqrt{5} \) is a rational number, then we have
\( 3 + \sqrt{5} = \frac{p}{q}, q \neq 0 \)
\( \sqrt{5} = \frac{p}{q} - 3 \)
\( \sqrt{5} = \frac{p - 3q}{q} \)
Here \( \sqrt{5} \) is irrational and \( \frac{p - 3q}{q} \) is rational. But rational number cannot be equal to an irrational number. Hence \( 3 + \sqrt{5} \) is an irrational number.

Question. Prove that \( \sqrt{5} \) is an irrational number and hence show that \( 2 - \sqrt{5} \) is also an irrational number.
Answer: Assume that \( \sqrt{5} \) be a rational number then we have
\( \sqrt{5} = \frac{a}{b} \), (\( a, b \) are co-primes and \( b \neq 0 \))
\( a = b\sqrt{5} \)
Squaring both the sides, we have
\( a^2 = 5b^2 \)
Thus 5 is a factor of \( a^2 \) and in result 5 is also a factor of \( a \).
Let \( a = 5c \) where \( c \) is some integer, then we have
\( (5c)^2 = 5b^2 \)
\( 25c^2 = 5b^2 \)
\( 5c^2 = b^2 \)
Thus 5 is a factor of \( b^2 \) and in result 5 is also a factor of \( b \).
Thus 5 is a common factor of \( a \) and \( b \). But this contradicts the fact that \( a \) and \( b \) are co-primes. Thus, our assumption that \( \sqrt{5} \) is rational number is wrong. Hence \( \sqrt{5} \) is irrational.
Let us assume that \( 2 - \sqrt{5} \) be rational equal to \( a \), then we have
\( 2 - \sqrt{5} = a \)
\( 2 - a = \sqrt{5} \)
Since we have assume \( 2 - a \) is rational, but \( \sqrt{5} \) is not rational. Rational number cannot be equal to an irrational number. Thus \( 2 - \sqrt{5} \) is irrational.

Question. If two positive integers \( p \) and \( q \) are written as \( p = a^2b^3 \) and \( q = a^3b \), \( a \) and \( b \) are prime number then. Verify. \( \text{LCM}(p, q) \times \text{HCF}(p, q) = pq \).
Answer: We have \( p = a^2b^3 = a \times a \times b \times b \times b \)
and \( q = a^3b = a \times a \times a \times b \)
Now \( \text{LCM}(p, q) = a \times a \times a \times b \times b \times b = a^3b^3 \)
and \( \text{HCF}(p, q) = a \times a \times b = a^2b \)
\( \text{LCM}(p, q) \times \text{HCF}(p, q) = a^3b^3 \times a^2b = a^5b^4 \)
\( = a^2b^3 \times a^3b = pq \)

Long Answer Type Questions

Question. Prove that \( \sqrt{3} \) is an irrational number. Hence, show that \( 7 + 2\sqrt{3} \) is also an irrational number.
Answer: Assume that \( \sqrt{3} \) be a rational number then we have
\( \sqrt{3} = \frac{a}{b} \), (\( a, b \) are co-primes and \( b \neq 0 \))
\( a = b\sqrt{3} \)
Squaring both the sides, we have
\( a^2 = 3b^2 \)
Thus 3 is a factor of \( a^2 \) and in result 3 is also a factor of \( a \).
Let \( a = 3c \) where \( c \) is some integer, then we have
\( (3c)^2 = 3b^2 \)
\( 9c^2 = 3b^2 \)
\( 3c^2 = b^2 \)
Thus 3 is a factor of \( b^2 \) and in result 3 is also a factor of \( b \).
Thus 3 is a common factor of \( a \) and \( b \). But this contradicts the fact that \( a \) and \( b \) are co-primes. Thus, our assumption that \( \sqrt{3} \) is rational number is wrong. Hence \( \sqrt{3} \) is irrational.
Let us assume that \( 7 + 2\sqrt{3} \) be rational equal to \( \frac{p}{q} \), \( q \neq 0 \) and \( p \) and \( q \) are co-primes.
\( 2\sqrt{3} = \frac{p}{q} - 7 = \frac{p - 7q}{q} \)
or \( \sqrt{3} = \frac{p - 7q}{2q} \)
Here \( p - 7q \) and \( 2q \) both are integers, hence \( \sqrt{3} \) should be a rational number. But this contradicts the fact that \( \sqrt{3} \) is an irrational number. Hence our assumption is not correct and \( 7 + 2\sqrt{3} \) is irrational.

Question. Show that there is no positive integer \( n \), for which \( \sqrt{n+1} + \sqrt{n-1} \) is rational.
Answer: Let us assume that there is a positive integer \( n \) for which \( \sqrt{n+1} + \sqrt{n-1} \) is rational and equal to \( \frac{p}{q} \), where \( p \) and \( q \) are positive integers and \( q \neq 0 \).
\( \sqrt{n+1} + \sqrt{n-1} = \frac{p}{q} \) ...(1)
or, \( \frac{1}{\sqrt{n+1} + \sqrt{n-1}} = \frac{q}{p} \)
\( \frac{\sqrt{n+1} - \sqrt{n-1}}{(\sqrt{n+1} + \sqrt{n-1})(\sqrt{n+1} - \sqrt{n-1})} = \frac{q}{p} \)
\( \frac{\sqrt{n+1} - \sqrt{n-1}}{(n+1) - (n-1)} = \frac{q}{p} \)
or \( \frac{\sqrt{n+1} - \sqrt{n-1}}{2} = \frac{q}{p} \)
\( \sqrt{n+1} - \sqrt{n-1} = \frac{2q}{p} \) ...(2)
Adding (1) and (2), we get
\( 2\sqrt{n+1} = \frac{p}{q} + \frac{2q}{p} = \frac{p^2 + 2q^2}{pq} \) ...(3)
Subtracting (2) from (1) we have
\( 2\sqrt{n-1} = \frac{p}{q} - \frac{2q}{p} = \frac{p^2 - 2q^2}{pq} \) ...(4)
From (3) and (4), we observe that \( \sqrt{n+1} \) and \( \sqrt{n-1} \) both are rational because \( p \) and \( q \) both are rational. But it possible only when \( (n+1) \) and \( (n-1) \) both are perfect squares. But they differ by 2 and two perfect squares never differ by 2. So both \( (n+1) \) and \( (n-1) \) cannot be perfect squares, hence there is no positive integer \( n \) for which \( \sqrt{n+1} + \sqrt{n-1} \) is rational.

HOTS QUESTIONS

Question. Show that 571 is a prime number.
Answer: Let \( x = 571 \)
\( \sqrt{x} = \sqrt{571} \)
Now 571 lies between the perfect squares of \( (23)^2 = 529 \) and \( (24)^2 = 576 \). Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23. Here 571 is not divisible by any of the above numbers, thus 571 is a prime number.

Question. Find the least number that is divisible by all numbers between 1 and 10 (both inclusive).
Answer: The required number is the LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
\( \text{LCM} = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 \)
\( = 2520 \)

Question. An army contingent of 104 members is to march behind an army band of 96 members in a parade. The two groups are to march in the same number of columns in which they can march?
Answer: Let the number of columns be \( x \) which is the largest number, which should divide both 104 and 96. It means \( x \) should be HCF of 104 and 96.
By using Euclid’s Division Lemma, we have
\( 104 = 96 \times 1 + 8 \)
\( 96 = 8 \times 12 + 0 \)
Thus HCF of 104 and 96 is 8 and 8 columns are required.

Question. If \( d \) is the HCF of 30 and 72, find the value of \( x \) and \( y \) satisfying \( d = 30x + 72y \).
Answer: Using Euclid’s Division Lemma, we have
\( 72 = 30 \times 2 + 12 \) ...(1)
\( 30 = 12 \times 2 + 6 \) ...(2)
\( 12 = 6 \times 2 + 0 \) ...(3)
Thus \( \text{HCF}(30, 72) = 6 \)
Now \( 6 = 30 - 12 \times 2 \)
From (1), \( 6 = 30 - (72 - 30 \times 2) \times 2 \)
\( 6 = 30 - 72 \times 2 + 30 \times 4 \)
\( 6 = 30(1 + 4) - 72 \times 2 \)
\( 6 = 30 \times 5 + 72 \times (-2) \)
\( 6 = 30x + 72y \)
Thus \( x = 5 \) and \( y = -2 \). Here \( x \) and \( y \) are not unique.

Question. If HCF of 657 and 963 is expressible in the form of \( 657x + 963 \times (-15) \), find the value of \( x \).
Answer: Using Euclid’s Division Lemma we have
\( 963 = 657 \times 1 + 306 \)
\( 657 = 306 \times 2 + 45 \)
\( 306 = 45 \times 6 + 36 \)
\( 45 = 36 \times 1 + 9 \)
\( 36 = 9 \times 4 + 0 \)
\( \text{HCF}(657, 963) = 9 \)
Now \( 9 = 657x + 963 \times (-15) \)
or, \( 657x = 9 + 963 \times 15 \)
\( = 9 + 14445 \)
or, \( 657x = 14454 \)
or, \( x = \frac{14454}{657} = 22 \)

Question. Express the HCF/LCM of 48 and 18 as a linear combination.
Answer: Using Euclid’s Division Lemma, we have
\( 48 = 18 \times 2 + 12 \) ...(1)
\( 18 = 12 \times 1 + 6 \) ...(2)
\( 12 = 6 \times 2 + 0 \)
Thus \( \text{HCF}(18, 48) = 6 \)
Now \( 6 = 18 - 12 \times 1 \)
From (1) \( 6 = 18 - (48 - 18 \times 2) \)
\( 6 = 18 - 48 \times 1 + 18 \times 2 \)
\( 6 = 18 \times (2 + 1) - 48 \times 1 \)
\( 6 = 18 \times 3 - 48 \times 1 \)
\( 6 = 18 \times 3 + 48 \times (-1) \)
Thus \( 6 = 18x + 48y \), where \( x = 3, y = -1 \).
Here \( x \) and \( y \) are not unique.
\( 6 = 18 \times 3 + 48 \times (-1) \)
\( = 18 \times 3 + 48 \times (-1) + 18 \times 48 - 18 \times 48 \)
\( = 18(3 + 48) + 48(-1 - 18) \)
\( = 18 \times 51 + 48 \times (-19) \)
\( 6 = 18x + 48y \), where \( x = 51, y = -19 \