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# Prove that the following are irrationals : (i) $\frac{1}{\sqrt{2}}$ (ii) $7\sqrt{5}$ (iii) $6+\sqrt{2&space;}$

CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/NCERT Solutions/Exercise 1.3

(i) Assume that is Rational. Hence, is rational as a and b are integers. Hence is rational which contradicts the fact that is irrational. Therefore, our assumption was incorrect and is irrational. (ii) Assume that 7 is rational. where a and b are some integers. therefore, a and 7b are ratonal... read more

(i) Assume that $\frac{1}{\sqrt{2}}$ is Rational.

Hence, $\frac{1}{\sqrt{2}}&space;=\frac{a}{b},&space;\sqrt{2}=\frac{b}{a}$

$\frac{b}{a}$  is rational as a and b are integers.

Hence $\sqrt{2}$ is rational which contradicts the fact that $\sqrt{2}$ is irrational.

Therefore, our assumption was incorrect and $\frac{1}{\sqrt{2}}$ is irrational.

(ii) Assume that 7$\sqrt{5}$ is rational.

$7\sqrt{5}&space;=&space;\frac{a}{b}$ where a and b are some integers.

therefore, $\sqrt{5}=&space;\frac{a}{7b}$

a and 7b are ratonal and so is $\sqrt{5}$

But this contradicts the fact that $\sqrt{5}$ is irrational.

Hence our assumption was incorrect and $7\sqrt{5}$ is rational.

(iii)  Assume $6+\sqrt{2}$ as rational

$6+\sqrt{2}=&space;\frac{a}{b}&space;,&space;\sqrt{2}=&space;\frac{a}{b}&space;-&space;6$

Since a and b are integers, $\frac{a}{b}-6$ is also rational and $\sqrt{2}$ should be rational as well.

This contradicts the fact that $\sqrt{2}$ is irrational and hence our assumption was incorrect.

$6+\sqrt{2}$ is irrational.

Whenever you find some prime numbers in root you can conclude that they are irrational.

1) 1⁄√2 Rationalise this number 1 *√2⁄√2 *√2 = √2/2 √2 is irrational number and 2 is rational number division of rational number by irrational number is irrational . Hence 1/√2 is an irrational number. 2)7√5 7 is rational number √5... read more

1) 1⁄√2

Rationalise this number 1 *√2⁄√2 *√2

= √2/2

√2 is irrational number and 2 is rational number division of rational number by irrational number is irrational . Hence  1/√2 is an irrational number.

2)7√5

7 is rational number √5 is an irrational number product of rarional and irrational number is irrational number.

3)6+√5

6 is a rational number √5 is irrational number sum of rational and irrational number is an irrational number.

Tutor

(i)Let is a rational number=. Squaring both sides ½=, So b^2=2a^2.So b^2 is divisible by 2 and b is divisible by 2. Let b=2c where c is an integer. Then b^2=4c^2, Hence 2a^2=4c^2. and a^2=2c^2.so a^2 is divisible by 2 and a is divisible by 2.This shows a and b have a common factor 2. But for a... read more

(i)Let is a rational number=. Squaring both sides ½=, So b^2=2a^2.So b^2 is divisible by 2 and b is divisible by 2. Let b=2c where c is an integer. Then b^2=4c^2, Hence 2a^2=4c^2. and a^2=2c^2.so a^2 is divisible by 2 and a is divisible by 2.This shows a and b have a common factor 2. But for a rational number a/b there cannot be a common factor other than 1.So this contradicts 1/√2 is rational. So it is an irrational number.

(ii) Let 7√5 is a rational number=a/b.

then√5=a/7b is a rational number. But we know√5 is irrational. So 7√5 is irrational.

(iii)Let 6+√2 is a rational number=a/b

then√2=(a/b)-6=(a-6b)/6 is a rational number.

So 6+√2 is irrational.

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