Question

how will we prove that
$\sqrt{6}$
is irrational ?

Answer 1

Here is your answer ⤵⤵⤵⤵

Let √6 be a rational number.

√6 = a/b ( where a and b are Co prime integers and b is not equal to 0)

Squaring both sides

6 = a²/b²

=> 6b² = a²

=> a² is divisible by 6

=> a is divisible by 6.

a = 6c

squaring both sides

a² = (6c)²

=> a² = 36c²

also, a² = 6b²

=> 6b² = 36c²

=> b² = 6c²

=> b² is divisible by 6

=> b is divisible by 6

Which is a contradiction as a & b are Co prime integers.

=> Our assumption is wrong.

=> √6 is an irrational number.

HOPE IT HELPS YOU ☺☺ !!!

Answer 2

Answer:

Step-by-step explanation: