Question

prove that 3+2 root 5 is irrational where root 5 is irrational​

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

Hence proved

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Answer 2

Answer:

Let us assume on the contrary that

$3 + 2 \sqrt{5}$is rational. Then their exit co-prime positive integers$a$and $b$such that,

$3 + 2 \sqrt{5} = \frac{a}{b}\\ \\ 2 \sqrt{5} = \frac{a}{b} - 3 \\ \\ \sqrt{5} = \frac{a - 3b}{2b} \\ \\ \sqrt{5} \: is \: rational \\ \\ (∵ a,b \: are \: integers \:∴ \frac{a - 3b}{2b} \: is \: rational)$

This contradicts the fact that $\sqrt{5}$is irrational. So, our supposition is incorrect. Hence,$3 + 2 \sqrt{5}$is an irrational number.