Question

Prove that 2√3 / 5 is an irrational

Answer 1

Answer:

Step-by-step explanation:

Please see the attached file,

Answer 2

Prove that $\sf{\dfrac{2 \sqrt{3}}{5}}$ is irrational.

___________________

Proof:

Assume that $\sf{\dfrac{2 \sqrt{3}}{5}}$ is rational.

$\sf{\dfrac{2 \sqrt{3}}{5} = \dfrac{a}{b} \: - - (a \: and \: b \: are \: integers)}$

$\longrightarrow \: \sf{\dfrac{2 \sqrt{3}}{5} = \dfrac{a}{b}}$

$\longrightarrow \: \sf{2 \sqrt{3} = \dfrac{a}{b} \times 5}$

$\longrightarrow \: \sf{2 \sqrt{3} = \dfrac{5a}{b}}$

$\longrightarrow \: \sf{\sqrt{3} = \dfrac{5a}{2b}}$

$\sf{\dfrac{5a}{2b}}$ is a rational number as a and b are integers.

So, the above mathematical statement says $\sf{\sqrt{3} = \dfrac{5a}{2b}}$. Which says $\sf{\sqrt{3}}$ is rational. This is a contradiction to the fact that $\sf{\sqrt{3}}$ is irrational.

This contradiction was arisen due to wrong assumption.

Hence proved, $\sf{\dfrac{2 \sqrt{3}}{5}}$ is irrational.