Prove that difference and quotient of $\left(3+2\sqrt{3}\right)\mathrm{and}\left(3-2\sqrt{3}\right)$ are irrational.

Irrational Number: Any number which cannot be expressed in the form of $\frac{p}{q}$, where p and q are integers, $q\ne 0$.

We have to prove that difference of $\left(3+2\sqrt{3}\right)\mathrm{and}\left(3-2\sqrt{3}\right)$ are irrational.

Step 1: Find the difference of $\left(3+2\sqrt{3}\right)\mathrm{and}\left(3-2\sqrt{3}\right)$.

$\left(3+2\sqrt{3}\right)-\left(3-2\sqrt{3}\right)=3+2\sqrt{3}-3+2\sqrt{3}=4\sqrt{3}$

which is an irrational number.

Step 2: Find the quotient of $\left(3+2\sqrt{3}\right)\mathrm{and}\left(3-2\sqrt{3}\right)$.

$\begin{array}{rcl}\frac{3+2\sqrt{3}}{3-2\sqrt{3}}& =& \frac{3+2\sqrt{3}}{3-2\sqrt{3}}\times \frac{3+2\sqrt{3}}{3+2\sqrt{3}}\\ & =& \frac{{\left(3+2\sqrt{3}\right)}^2}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\\ & =& \frac{9+12+12\sqrt{3}}{9-12}\\ & =& \frac{21+12\sqrt{3}}{-3}\\ & =& -\left(7+4\sqrt{3}\right)\end{array}$

which is an irrational number.

Therefore, the difference and quotient of $\left(3+2\sqrt{3}\right)\mathrm{and}\left(3-2\sqrt{3}\right)$ are irrational.