Question

Prove that under root n is not a rational number, if n is not a perfect square.​

Answer 1

Answer:

Correct Question:

Prove that $\sqrt{n}$is not a rational number, if n is not a perfect square.

Solution:

Let $\sqrt{n}$

be a rational number.

Then, assume $\sqrt{n} = \frac{p}{q}$

[where p, q are coprime and q is not equal to 0].

$\Longrightarrow$ $n = \frac{p {}^{2} }{q {}^{2} }$

By squaring both sides, we get

$\Longrightarrow$ $p {}^{2} = nq {}^{2} \: \: \: \: \: \: \: ......(1)$

Now by using Pythagoras theorem

AC $= \sqrt{(ab) {}^{2} + (b + c) {}^{2} }$

$= \sqrt{(18) {}^{2} + (2.4) {}^{2} } \\ \\ = \sqrt{324 + 576} \\ \\ = \sqrt{9} \\ \\ = 3 \: cm$

Hence, proved.