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.
You need to login to perform this action.
You will be redirected in
3 sec