Question

Given that √5 is irrational number. Prove that
6-√5 is irrational number​

Answer 1

$\underline{ \mathfrak{ \: Given:- \: }}$

•√5 is an irrational number.

$\underline{ \mathfrak{ \: To\:Prove:- \: }}$

•6 - √5 is irrational number.

$\underline{ \mathfrak{ \: Proof:- \: }}$

⠀⠀⠀⠀

Let's assume that 6 - √5 is a rational number.

So,It can be written in form of p/q

Where p and q are two co - prime numbers.

⠀⠀⠀⠀

$\leadsto \sf 6 - \sqrt{5} = \dfrac{p}{q}$

$\leadsto \sf \sqrt{5} = 6 - \dfrac{p}{q}$

$\leadsto \sf \sqrt{5} = \dfrac{6q - p}{q}$

Here,

⠀⠀⠀⠀

The LHS i.e. √5 is irrational.

⠀⠀⠀⠀

We observe that LHS is irrational and RHS is rational, which is not possible.

Hence, our assumption that given number is rational is false.

⠀⠀⠀⠀

$\therefore\;{\underline{\sf{ \bf{6 - \sqrt{5}}\;is\; irrational.}}}$

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Answer 2

Given :-

• √5 is an irrational number.

To Prove :-

• 6 - √5 is an irrational number.

Proof :-

Let us assume to the contrary that 6 - √5 is a rational number in the form of p/q where p and q are integers and q ≠ 0.

So

⇒ 6 - √5 = p/q

⇒ 6 - p/q = √5

Here we can see that L.H.S is in the form of p/q which is rational but R.H.S is equal to √5 which is irrational.

So, our contradiction is wrong

So, 6 - √5 is an irrational number.

Hence, proved !