Question

Prove that √2 is an irrational number by contradiction method .

Answer 1

let √2 be a rational number then
√2 = p/q
squaring both the sides we get
2=p^2/q^2
(2p)^2=q^2 {equation 1}
this implies that q^2 is divisible by 2 and then can also be said that q is divisible by 2
hence can be written as
q=2k where k is an integer
squaring both sides
q^2 = (2k)^2
from equation 1
(2k)^2=(2p)^2
and
p^2 = 2k^2
hence we can say 2 is the common factor in p and q and this is a contradiction to the fact that p and q are co prime numbers
hence √2 cannot be expressed as p/q
hence √2 is an irrational number.

Answer 2

HI !

To prove :-

√2 is irrational

Proof :-
Let us assume that √2 is rational 

Let ,
√2 = p/q , where p and q are integers and q≠ 0 , p and q are co prime numbers.

Squaring both sides ,

2 = p²/q²

2q² = p²

2 divides p²

2 divides p             ----> [1]

p = 2m

2q² = (2m)²

2q² = 4m²

q² = 2m²

2 divides q²

2 divides q            ----> [2]

From 1 and 2 , 2 divides p and q .
2 is a common factor of p and q .
This is a contradiction. p and q are not co prime
Hence ,√2 is irrational.
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