# Show that the fol

(i) Let assume that is rational

Therefore it can be expressed in the form of , where p and q are integers and q≠0

Therefore we can write = √2 =  is a rational number as p and q are integers. Therefore √2 is rational which contradicts the fact that √2 is irrational.

Hence, our assumption is false and we can say that is irrational.

(ii) Let assume that 7√5 is rational therefore it can be written in the form of where p and q are integers and q≠0. Moreover, let p and q have no common divisor > 1.

7√5 = for some integers p and q

Therefore √5 =  is rational as p and q are integers, therefore √5 should be rational.

This contradicts the fact that √5 is irrational.

Therefore our assumption that 7√5 is rational is false. Hence 7√5 is irrational.

(iii) Let assume that 6 + √2 is rational therefore it can be written in the form of where p and q are integers and q≠0. Moreover, let p and q have no common divisor > 1.

6 + √2 = for some integers p and q

Therefore √2 = - 6

Since p and q are integers therefore - 6 is rational, hence √2 should be rational. This contradicts the fact that √2 is irrational. Therefore our assumption is false. Hence, 6 + √2 is irrational.

(iv) Let us assume that 3 - is rational

Therefore 3 - can be written in the form of where p and q are integers and q≠0

3 - =  -3 =  = Since p, q and 3 are integers therefore is rational number

But we know is irrational number, Therefore it is a contradiction.

Hence 3 - is irrational

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<span lang="EN-USRS Aggarwal - Mathematics