Question

$proof \: that \: 3 + 2 \sqrt{5 \: \:} is \: irrationl$
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Answer 1

Given: 3 + 2√5

To prove: 3 + 2√5 is an irrational number.

Proof:

Let us assume that 3 + 2√5 is a rational number.

So, it can be written in the form a/b
3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving 3 + 2√5 = a/b we get,
=>2√5 = a/b – 3
=>2√5 = (a-3b)/b
=>√5 = (a-3b)/2b

This shows (a-3b)/2b is a rational number. But we know that √5 is an irrational number.

So, it contradicts our assumption. Our assumption of 3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Answer 2

let's assume it to be rational

we can write it as $\sf{3+2{\sqrt{5}} = {\dfrac{a}{b}}}$ where a and b are co primes and b ≠ 0 .

Subtracting by 3 , we get

$\sf{2{\sqrt{5}} = {\dfrac{a}{b}} - 3 }$

$\sf{2{\sqrt{5}} = {\dfrac{a-3b}{b}}}$

Dividing by 2 , we get

$\sf {\sqrt{5}} = {\dfrac{a-3b}{2b}}$

Since a and b are integers , $\sf {\dfrac{a-3b}{2b}}$ is a rational number .

We know that $\sf {\sqrt{5}}$ is an irrational number .

So it contradicts our assumption and hence , $\sf {3 + 2{\sqrt{5}}}$ is irrational .