Prove that √2 is an irrational number.

2
by Manishtiwari

2014-09-02T19:43:08+05:30

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Let √2 be a rational number p/q, where p and q are integers. p and q are  relatively prime to each other, coprime. no common factors among them.

√2 = p / q
p² = 2 q²

p * p = 2 * q * q
p and q are co-prime. no common factors. Then 2 on RHS must be a factor of p.
then p = 2 k

2k * 2k = 2 * q * q
2 k * k = q * q

2 on LHS must be a factor of q on RHS.  So q = 2 * m.
But we have started √2 = p/q with p and q such that they are co-prime. They have no common factors. But if √2 is rational, they have a common factor 2.
So the assumption is wrong.
√2 cannot be a rational number.

2014-09-02T19:57:10+05:30
Suppose √2 is a rational number
:.
2/1 =a/b
a=
2b
squaring on both sides
a
²= 2b²
a²/2=b²
2 divides a²⇒2 divides a
:. as 2 divides 'a' if we multiply 2 by a number we will get a
let that number be c
:. a=2c
(2c)²=2b²          (a²=2b²)
4c²= 2b²
b²=4c²/2
b²=2c²
c²=b²/2
⇒2 divides b²⇒2 divides b
this is a contradiction that they are co-prime
hence our assumption is wrong
:. √2 is irrational