question_answer

Prove that \[3+2\sqrt{3}\] is an irrational number.

Answer:

Let us assume to the contrary, that \[3+2\sqrt{3}\] is rational.

So that we can find co-prime positive integers a and \[b(b\ne 0),\] such that \[3+2\sqrt{3}=\frac{a}{b}\]

Rearranging the equation, we get

\[2\sqrt{3}=\frac{a}{b}-3=\frac{a-3b}{b}\]

\[\sqrt{3}=\frac{a-3b}{2b}=\frac{a}{2b}-\frac{3b}{2b}\]

\[\sqrt{3}=\frac{a}{2b}-\frac{3}{2}\]

Since a and b are integer, we get \[\frac{a}{2b}-\frac{3}{2}\] is rational and so \[\sqrt{3}\] is rational.

But this contradicts the fact that \[\sqrt{3}\] is irrational.

So we conclude that \[3+2\sqrt{3}\] is irrational.                            Hence Proved.