Question

prove 9 - root6 is irrational(proving irrational numbers)

proper working

Answer 1

Step-by-step explanation:

Let us assume that 9-root6 is a rational number (we are having an idea that it is an irrational but we are assuming like that so if our results which are true get wrong our this assumption is wrong but if our true results come as correct our assumption is also true)*only for understanding

Let's Begin

let us assume

$9 - \sqrt{6}$

is a rational so it would be of the form p/q where p and q both are integers (because of the definition of a rational number)

$9 - \sqrt{6} = \frac{p}{q}$

$\sqrt{6} = \frac{9q - p}{q}$

it is implying that root 6 is a rational because all the numbers in numerator and denominator are integers and by the definition of rational numbers root 6 is a rational number

but root 6 is an irrational as

$\sqrt{6} = \sqrt{3} \times \sqrt{2}$

and both root 3 and root2 are irrationals (this is a proven result)

so we can see that our previously proven results which are coming out as false implying that our assumption has a mistake so aur assumption that 9-root6 is a rational is wrong and we arrive at the result that 9-root6 is an irrational

HOPE THIS HELPS AND IS EASY TO UNDERSTAND IF IT IS THEN PLEASE MARK ME THE BRAINLIEST⭐⭐⭐⭐⭐

Answer 2

Answer:

Assume 9 - √6 as a rational number

∴ 9 - √6 = a / b      { ∵ rational no. can be represented in the form of p/q }

- √6 = a/b - 9

-2√3 = a/b - 9

-√3 = ( a/b - 9 ) / 2

√3 = ( -a / b + 9 ) / 2

( -a / b + 9 ) / 2  i.e RHS is rational but √3 i.e LHS is irrational but an irrational no. can't be equal to a rational no.

Since this case is not possible ∴ our assumption is wrong.