Prove that $5-2\sqrt{3}$ is an irrational number.

Hint: We will first assume that $5-2\sqrt{3}$ is a rational number, then we will equate $5-2\sqrt{3}$ with \[\dfrac{p}{q}\]. In the last step we will do some basic mathematical operations and calculations, if our assumption will be contradicted then $5-2\sqrt{3}$ is an irrational number otherwise it is a rational number.

Complete step-by-step answer:
It is given in the question that we have to prove $5-2\sqrt{3}$ is an irrational number. So, basically a rational number means the number which can be expressed in the form of $\dfrac{p}{q}$ where p and q are integers, $q\ne 0$ and p and q are prime numbers. And the numbers which cannot be expressed in the form of $\dfrac{p}{q}$ are called irrational numbers.
Let us assume that $5-2\sqrt{3}$ is a rational number, it means $5-2\sqrt{3}$ must be equal to $\dfrac{p}{q}$ or $5-2\sqrt{3}=\dfrac{p}{q}$
Transposing 5 from LHS to RHS, we get
$-2\sqrt{3}=\dfrac{p}{q}-5$
Taking LCM, we get,
$-2\sqrt{3}=\dfrac{\left( p-5q \right)}{q}$
Dividing both side with -2, we get
$\sqrt{3}=\dfrac{\left( p-5q \right)}{-2q}$

Since, we know that $\dfrac{p}{q}$ is a rational number, therefore $\sqrt{3}$ must also be a rational number, which is wrong as $\sqrt{3}$ is an irrational number, which totally contradict our assumption.
This is because if $\sqrt{3}$ is an irrational number $5-2\sqrt{3}$ must be an irrational number.
Thus, it is proved that $5-2\sqrt{3}$ is an irrational number.

Note: Student may miss confuse that 5 is a rational number, 2 is a rational number but $\sqrt{3}$ is an irrational number, so, the number of rational number is more than irrational number in the given number $5-2\sqrt{3}$ which is not correct at all. Because of the presence of single $\sqrt{3}$, the complete number $5-2\sqrt{3}$ will become an irrational number.