Prove that \[3+2\sqrt{5}\] is an irrational number.

Hint: Use the method of contradiction to prove that \[3+2\sqrt{5}\] is an irrational number. Start assuming that \[3+2\sqrt{5}\] is a rational number of the form \[\dfrac{a}{b}\], here a and b are integers and \[b\ne 0\]. Then we will show that this assumption is not correct, by using the fact that \[\sqrt{5}\] is an irrational number and hence we just have to show that \[3+2\sqrt{5}\] is not in the form of \[\dfrac{a}{b}\] a rational number.

Complete step-by-step answer:
In the question, we have to prove that \[3+2\sqrt{5}\] is an irrational number. So, applying the method of contradiction, we will first assume that \[3+2\sqrt{5}\]is a rational number. Now, it is known that a rational number is of the form \[\dfrac{a}{b}\] where \[b\ne 0\], also a and b are integers. So, here we will assume that \[3+2\sqrt{5}\]is of the form \[\dfrac{a}{b}\]. So, we have:
\[\begin{align}
  & \Rightarrow \dfrac{a}{b}=3+2\sqrt{5} \\
 & \Rightarrow \dfrac{a}{b}-3=2\sqrt{5} \\
 & \Rightarrow \dfrac{a-3b}{b}=2\sqrt{5} \\
\end{align}\]
So here we can see that the left hand side is a rational number but the right hand side is not a rational number.
Because \[\sqrt{5}\] is an irrational number. So, this contradicts our assumption that \[3+2\sqrt{5}\] is a rational number. So, \[3+2\sqrt{5}\] is an irrational number. Hence proved.

Note: Another method to prove that \[3+2\sqrt{5}\] is an irrational number is as follows.
Since, we know that the addition of an irrational number with the rational number will also be an irrational number. So, here we \[\sqrt{5}\]as an irrational number. So, adding this irrational number to 3, will again give an irrational number. Hence, proven that \[3+2\sqrt{5}\] is an irrational number.