question_answer

Show that \[3\sqrt{7}\] is an irrational number.

Answer:

Let us assume, to the contrary, that \[3\sqrt{7}\] is rational.

That is, we can find co-prime a and \[b(b\ne 0)\] such that \[3\sqrt{7}=\frac{a}{b}\]

Rearranging, we get \[\sqrt{7}=\frac{a}{3b}\]

Since 3, a and b are integers, \[\frac{a}{3b}\] can be written in the form of \[\frac{p}{q}\], so \[\frac{a}{3b}\] is rational, and so \[\sqrt{7}\] is rational.

But this contradicts the fact that \[\sqrt{7}\] is irrational.

So, we conclude that \[3\sqrt{7}\] is irrational.                    Hence Proved.