Question

√2-8 is an irrational number​

Answer 1

Let us assume that root2 - 8 is a rational number

root2 - 8 = a/b ( a,b are co-primes)

root2 = a/b + 8

root2 = a+8b/b

We know that an irrational is never equals to a rational . This leads to contradiction . Our assumption must be wrong .

Therefore , root2 - 8 is an irrational number.

Answer 2

$let \: us \: assume \: \sqrt{2} - 8 \: is \: a \: rational \: number \\ so \: \sqrt{2} - 8 = \frac{p}{q} \\ \sqrt{2 } q - 8q = p \\ \sqrt{2} q = p + 8q \\ \sqrt{2} = \frac{p +8q}{q}$

so LHS =RHS

so √2 is irrational number as we know so√2-8 is also an irrational number.

hope it helps....

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