Question

Given that √5 is irrational prove that 3+2√5 is irrational​

Answer 1

AnswEr :

☯ Let's consider that 3+2√5 is an rational number.

So, it can be written in the form of p/q. Where q is not equal to 0.

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Therefore,

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$\implies\sf 3 + 2\sqrt{5} = \dfrac{p}{q} \\\\\\:\implies\sf 2 \sqrt{5} = \dfrac{p}{q} - 3 \\\\\\:\implies\sf \sqrt{5} \dfrac{1}{2} \bigg(\dfrac{ p}{q} -3 \bigg)$

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Here, we can see that $\bf\dfrac{1}{2} \bigg(\dfrac{p}{q} - 3 \bigg)$ is rational. Therefore, √5 should rational. But it arises contradiction because √5 is an irrational number.

Therefore, our assumption is wrong.

$\therefore$ 3+2√5 is an Irrational number.

⠀⠀⠀⠀⠀⠀ ⠀⠀⠀Hence Proved!

Answer 2

Answer:

Prove that 3+2√5 is irrational

Given:3 + 2√5

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

Proof:

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

Soit can be written in the form a/b

3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving3 + 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 But √5 is an irrational number.

so it contradictsour assumption.

Our assumption of3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved