Question

Prove that the given number is irrational
$\sf3 + 2\sqrt{5}$

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

Answer 2

First things first, we know that $\sqrt{5}$ is irrational. We need proof.

Proof

It is a real number such that $2<\sqrt{5}<3$ and $(\sqrt{5} )^2=5$, so first, let's assume it is a rational number.

$\dfrac{p}{q}=\sqrt{5}\implies \dfrac{p^2}{q^2} =5\Longleftrightarrow p^2=5q^2$ ...[I]

So, $p$ is a multiple of 5.

∵[I] $p=5k\implies p^2=25k^2$ ...[II]

∵[II] $25k^2=5q^2\Longleftrightarrow 5k^2=q^2$, so $q$ is a multiple of 5.

So, $p$ and $q$ are both divisible by 5, even if we choose coprime numbers. This is a contradiction.

Such rational numbers do not exist, so the assumption was wrong.

So it is an irrational number.

Now we can prove $3+2\sqrt{5}$ is irrational.

Solution

$3+2\sqrt{5}$ is a real number.

Let $\dfrac{m}{n} =3+2\sqrt{5}$.

Then eventually we get $\sqrt{5} =\dfrac{m-6n}{2n}$.

Is $\dfrac{m-6n}{2n}$ a rational number? Yes. Is $\sqrt{5}$ a rational number? No.

Hence a contradiction and our assumption was wrong, so $3+2\sqrt{5}$ is not a rational number.

So $3+2\sqrt{5}$ is an irrational number.

Additional Information

We are solving based on this fact:

'If a real number is not a rational number, it is an irrational number.'

Let's see what we can prove if we ignore 'real number'.

Assume $\dfrac{p}{q} =\sqrt{2} i$. Then $\dfrac{p^2}{q^2} =-2$ so $p^2=-2q^2$. ...[I]

We see $p$ is divisible by $-2$.

∵[I]  $p=-2k \implies p^2=4k^2$...[II]

∵[II] $4k^2=-2q^2\Longleftrightarrow -2k^2=q^2$, so $q$ is divisible by $-2$.

Hence our assumption that $\sqrt{2} i$ is a rational number fails. Then, $\sqrt{2} i$ is an irrational number, or a real number.