Question

Prove that following are irrationals:
(i) 7root5
(ii)6+root2
(iii)1/root2

Answer 1

Step-by-step explanation:

1.) 7√5

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Let 7√5 be a rational number.

So,

7√5 = p/q.

(p and q are co-prime number and q ≠ 0)

=> √5 = p/7q

As, we can see that p/7q ia rational so √5 should also be rational. But this contradict the fact that √5 is irrational.

So, by this we can say that 7√5 is irrational number.

2.) 6√2.

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Let us assume that 6+√2 is rational.

That is , we can find coprimes a and b (b≠0) such that

=> 6+√2 = a/b

=>√2= a/b -6

=> √2 = a-6b/b

Since , a and b are integers , \frac{a-6b}{b} is rational ,and so √2 is rational.

But this contradicts the fact that √2 is irrational.

So, we conclude that 6+√2 is irrational

3.) 1/√2

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To prove 1/√2 is irrational

Let us assume that √2 is irrational 

1/√2 = p/q (where p and q are co prime)

q/p = √2

q     = √2p

squaring both sides

q²   = 2p²                                                  .....................(1)

By theorem 

q is divisible by 2

∴ q = 2c ( where c is an integer)

 putting the value of q in equitation 1

2p² = q² = 2c² =4c²

p² =4c² /2 = 2c²

p²/2 = c² 

by theorem p is also divisible by 2

But p and q are coprime

This is a contradiction which has arisen due to our wrong assumption

∴1/√2 is irrational