 Irrational Number NCERT Exercise 1.3 to 1.4 Class Ten Mathematics

# Real Number

## Exercise 1.3

Question 1: Prove that sqrt5 is irrational.

Answer: Let us assume the contrary, i.e. sqrt5 is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

sqrt5=a/b

Or, bsqrt5=a

Squaring on both sides, we get;

5b^2=a^2

This means that a2 is divisible by 5 and hence a is also divisible by 5.

This contradicts our earlier assumption that a and b are coprime, because we have found 5 as at least one common factor of a and b.

This also contradicts our earlier assumption that sqrt5 is irrational.

Hence, sqrt5 is irrational.

Question 2: Prove that 3+2sqrt5 is irrational.

Answer: Let us assume to the contrary, i.e. 3+2sqrt5 is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

3+2sqrt5=a/b

Or, 2sqrt5=a/b-3

Since a and b are rational, so a/b-3 is rational and hence, 2sqrt5 is rational.

But this contradicts the fact that 2sqrt5 is irrational.

This happened because of our faulty assumption.

Hence, 3+2sqrt5 is irrational

Question 3: Prove that following are irrationals:

(i) 1/sqrt2

Answer: Let us assume to the contrary, i.e. 1/sqrt2 is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

1/sqrt2=a/b

Or, asqrt2=b

Squaring on both sides, we get;

2a^2=b^2

This means that b2 is divisible by 2 and hence a is also divisible by 2.

This contradicts our earlier assumption that a and b are co-prime, because 2 is at least one common factor of a and b.

This also contradicts our earlier assumption that 1/sqrt2 is rational.

Hence, 1/sqrt2 is irrational proved.

(ii) 7sqrt5

Answer: Let us assume to the contrary, i.e. 7sqrt5 is rational.

There can be two integers a and b (b≠0) and a and b are coprime, so that;

7sqrt5=a/b

Or, a\xx7sqrt5=a

Squaring on both sides, we get;

245b^2=a^2

This means that a2 is divisible by 245; which means that a is also divisible by 245.

This contradicts our earlier assumption that a and b are coprime, because 245 is at least one common factor of a and b.

This happened because of our faulty assumption and hence, 7sqrt5 is irrational proved.

(iii) 6+sqrt2

Answer: Let us assume to the contrary, i.e. 6+sqrt2 is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

6+sqrt2=a/b

Or, sqrt2=a/b-6

Since a and b are rational, so a/b-6 is rational and hence, sqrt2 is rational.

But this contradicts the fact that sqrt2 is irrational.

This happened because of our faulty assumption.

Hence, 6+sqrt2 is irrational proved.

## Exercise 1.4 (NCERT Book)

Question 1: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating decimal expansion:

(i) (13)/(3125), (ii) (17)/(8) (iii) (64)/(455) (iv) (15)/(1600) (v) (29)/(343) (vi) (23)/(2^3\5^3) (vii) (129)/(2^2\5^7\7^5) (viii) (6)/(15) (ix) (35)/(50) (x) (77)/(210)

Answer: (i), (ii), (iv), (vi), (viii) and (ix) are terminating decimal expansion.

(iii), (v), (vii) and (x) are non-terminating repeating decimal expansion.

Question 2: Write down the decimal expansion of those rational numbers in Question 1 which have terminating decimal expansion.

(i) (13)/(3125)=0.00416

(ii) (17)/(8)=2.125

(iv) (15)/(1600)=0.009375

(vi) (23)/(2^3\5^2)=0.115

(viii) (6)/(15)=0.4

(ix) (35)/(50)=0.7

Question 3: The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational , and of the form p/q, what can you say about the prime factors of q?

(i) 43.123456789

Answer: This is a rational number, q has either 2 or 5 or both as prime factor.

(ii) 0.120120012000120000………..

Answer: This is an irrational number.

(iii) 43.123456789

Answer: This is a rational number. In this case, q has 2 or 5 or both as prime factor and has some other factor as well.