Question

5. Prove that the following numbers are irrational.
(1) 3 - √5
(ii) √2 + √5​

Answer 1

Answer:

they can be written as p/q form

Answer 2

Answer:

Hence $3-\sqrt{5}$ and $\sqrt{2}+\sqrt{5}$ are irrational numbers.

Step-by-step explanation:

Given numbers are $3-\sqrt{5}$ and $\sqrt{2}+\sqrt{5}$.

Let us assume $3-\sqrt{5}$ is a rational number.

So it can be written in the form $\frac{a}{b}$.

$3-\sqrt{5}=\frac{a}{b}$

Here a and b are coprime numbers and $b\neq 0$.

Solving

$3-\sqrt{5}=\frac{a}{b}$

we get,

$\sqrt{5}=\frac{a}{b}+3$

$\sqrt{5}=\frac{a+3b}{b}$

This shows that $\sqrt{5}$ is a rational number.

But we know that $\sqrt{5}$ is an irrational number, which contradicts our assumption.

Our assumption $3-\sqrt{5}$ is a rational number is incorrect.

$3-\sqrt{5}$ is an irrational number.

Let us assume that $\sqrt{2}+\sqrt{5}$ is a rational number.

A rational number can be written in the form $\frac{p}{q}$ where p,q are integers and $q\neq 0$.

$\sqrt{2}+\sqrt{5}=\frac{p}{q}$

Squaring on both sides, we get

$(\sqrt{2}+\sqrt{5})^2=(\frac{p}{q})^2$

$2+5+2\sqrt{10}=\frac{p^2}{q^2}$

$7+2\sqrt{10}=\frac{p^2}{q^2}$

$2\sqrt{10}=\frac{p^2}{q^2}-7$

$\sqrt{10}=\frac{p^2-7q^2}{2q^2}$

p,q are integers then $\frac{p^2-7q^2}{2q^2}$ is a rational number.

Then $\sqrt{10}$ is also a rational number.

But this contradicts the fact $\sqrt{10}$ is an irrational number.

Our assumption is incorrect.

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