Question

prove that √2 is irrational​

Answer 1

Answer:

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Answer 2

Answer:

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Step-by-step explanation:

let √2 be a rational...

then, √2 = p/q (where p and q are two integers, q≠0 and p, q have no common factors.)

squaring both sides :-

(√2)^2 = (p/q)^2

2 = p^2/q^2

p^2 = 2q ......(i)

As 2 divides 2q^2, so 2 divides p^2 but 2 is a prime.

= 2 divides 2q^2, so 2 divides p^2 but 2 is a prime.

= 2 divides p.

Let p=2m, where 'm' is an integer.

Substitute this value of p in (i),

(2m)^2 = 2q^2

4m^2 = 2q^2

q^2 = (2m)^2

As 2 divides (2m)^2, so 2 divides q^2 but 2 is a prime.

2 divides q.

Thus, p and q have a common factor 2. This contradicts that p and q have no common factor.

Hence, √2 is an irrational number.