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.
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