Question

show that 2-root3 is an irrational number​

Answer 1

Answer:

As we know that ✓3 is an irrational number and 2 is a rational number.

And subtraction of an irrational number from a rational number gives us an irrational number implies that 2-✓3 is an irrational number.

Answer 2

$\star \: {\underbrace{\rm{To \: Prove}}} \: \star$

• $\rm{2 - \sqrt{3} }$ is an irrational number.

$\star \: {\underbrace{\rm{Proof}}} \: \star$

Let's assume that $\rm{2 - \sqrt{3} }$ is an rational number.

So,

$\implies\sf{ \frac{a}{b} = 2 - \sqrt{3} \: ... \: (where \;a \; and \; b \; are \: co \: prime \; numbers)}$

On further simplifying,

$\implies\sf{ \frac{a}{b} - \frac{2}{1} = - \sqrt{3} }$

$\implies\sf{ \frac{a - 2b}{b} = - \sqrt{3}}$

$\implies\sf{ \sqrt{3} = - \frac{a - 2b}{b}}$

$\sf{ \sqrt{3} = - \frac{a - 2b}{b}}$ is in form $\sf{ \frac{p}{q} ,}$ hence it is a rational number.

But as we know that,

$\implies\sf{ \sqrt{3} \: is \: rational \: number}$

We know that,

$\implies$ rational ≠ irrational

Thus, if means that $\implies\sf{\sqrt{3}}$ is irrational and that is a contradiction because we have proven that it is irrational.

Henceforth;

$\implies\sf{2 - \sqrt{3} \: is \: irriational \: number.}$

$\therefore$ Hence proved!