Question

classify root 20 upon root 5 is a rational or irrational number ​

Answer 1

$\huge\underline\mathrm{GIVEN:-}$

• $\sf{\frac{20}{\sqrt{5}}}$

$\huge\underline\mathrm{TO\:PROVE:-}$

Check whether $\sf{\frac{20}{\sqrt{5}}}$ is a rational number or an irrational number

$\huge\underline\mathrm{SOLUTION:-}$

Let us assume, to the contrary, that $\sf{\frac{20}{\sqrt{5}}}$ is rational. That is, we can find co- prime integers p and q (q ≠ 0) such that

∴ ⠀⠀⠀ $\sf{\frac{20}{\sqrt{5}}}$ = $\sf {\frac{p}{q}}$

⚫ Rationalising the denominator:-

$\longmapsto$$\sf{\frac{p}{q} = \large\frac{ 20 \times \sqrt{5}}{\sqrt{5} \times \sqrt {5}}}$

$\longmapsto$$\sf{ \frac{p}{q}=\large\frac{\cancel{20}\sqrt{5}}{\cancel{5}}}$

$\longmapsto$$\sf{4\sqrt{5} = \frac{p}{q}}$

$\longmapsto$$\sf{\sqrt{5} = \frac{p}{4q}}$

➢ Since, p and q are integers so $\sf{\frac{p}{4q}}$ is rational, and so √5 is rational.

✒But this contradicts the fact that √5 is irrational.

So, we conclude that √5 is an irrational.

❛ Hence, $\sf{\frac{20}{\sqrt{5}}}$ is an irrational number ❜

Answer 2

$\large\underline\mathrm{Solution}$

• Let the us to the 20/√5 is rational. that is, we can find co - prime integers p and q (q unequal to 0) such that.

$\implies$ 20/√5 = p/q

Rationalise.

$\implies$ p/q = 20 × √5 / √5 × √5

$\implies$ p/q = 20√5 / 5

$\implies$ p/q = 4√5

$\implies$ √5 = p/4q

Thus, p and q are integers so p/4q is rational, and so √5 is irrational.

So, we conclude that √5 is an irrational.

hence, 20/√5 is an irrational number.