Question

prove that 6-4 root 2 is an irrational number​

Answer 1

Ｔｏ ｐｒｏｖｅ

6-4√ 2 ᴀs ɪʀʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ

Let's assume that 6-4√2 is a rational number

◉ Rational number

♦The numbers which can be shown in p/q . where q≠0

♦ Which are co primes.

So, if we assumed that the given number is rational so it will fulfill the above property.

Then we can show it in p/q form.

So,

6-4√2 =p/q.

( where q≠0, and 6, 4,p,q are integers and p and q are co primes)

-4√2 =( p/q)-6

4√2. = 6-(p/q)

$\sqrt{2} = \frac{p - 6q}{4q}$

because p and q are integers so p-6q/4q will be rational so √2 is also a rational number

But it contradict the fact that√2 is irrational number.

So our hypothesis is wrong

Sᴏ, ᴡᴇ ᴄᴀɴ ᴄᴏɴᴄʟᴜᴅᴇ ᴛʜᴀᴛ 6-4√2 ɪs ᴀɴ ɪʀʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ.

Hence proved