Question

prove thate √2 is irrational number​

Answer 1

Look after the answer in image

Answer 2

Answer:

Let us assume that  √2 is rational

√2=a/b where a and b are co prime

√2b=a

squaring both sides

2b^2=a^2 -----------------(1)

2 divides a^2, therefore 2 divides a (theorem 1.3, Let p be a prime number. If p divides a^2 then p divides a, where a is a positive integer)

So we can write a=2c, for some integer c

Substituting value of a in (1)

2b^2 = (2c)^2

2b^2=4c^2

b^2=2c^2

2 divides b^2, therefore 2 divides b (theorem 1.3, Let p be a prime number. If p divides a^2 then p divides a, where a is a positive integer)

Therefore a and b have at least 2 as common factor

This contradicts the fact that a and b are co prime

This contradiction has arisen because of our incorrect assumption

Thus √2 is irrational

Hope this helps :)