Prove that the following $\sqrt{3}+\sqrt{5}$ is irrational.

Hint: In this question, we will prove that $\sqrt{3}$ and $\sqrt{5}$ are irrational separately by contradiction method. After that we will apply the property that the sum of two irrational numbers is always an irrational number.

Complete step-by-step solution -
In this question, we have been asked to prove that $\sqrt{3}+\sqrt{5}$ is irrational. For that, we will first prove that $\sqrt{3}$ is irrational.
Let $\sqrt{3}$ be a rational number, and we know that a rational number can be expressed in the form of $\dfrac{p}{q}$, where $q\ne 0$. So, we get our number as, $\sqrt{3}=\dfrac{p}{q}$, where $p$ and $q$ are co-prime to each other, that is $p$ and $q$ have no common factors and $q\ne 0$. Now we will square both sides of the equation,
$\begin{align}
  & {{\left( \sqrt{3} \right)}^{2}}={{\left( \dfrac{p}{q} \right)}^{2}} \\
 & \Rightarrow 3=\dfrac{{{p}^{2}}}{{{q}^{2}}} \\
\end{align}$
By cross multiplying the above equation, we get,
$3{{q}^{2}}={{p}^{2}}\ldots \ldots \ldots \left( i \right)$
From the above equation, we can see that 3 is a factor of ${{p}^{2}}$. If 3 is a factor of ${{p}^{2}}$, then 3 must be a factor of $p$. So, we can write, $p=3m$. So, we get, ${{p}^{2}}=9{{m}^{2}}$. So, by substituting it in equation (i), we get,
$\begin{align}
  & 3{{q}^{2}}=9{{m}^{2}} \\
 & \Rightarrow {{q}^{2}}=3{{m}^{2}} \\
\end{align}$
We can see here that 3 is a factor of ${{q}^{2}}$, which means that 3 is a factor of $q$. But we assumed that $p$ and $q$ have no common factors, so it is a contradiction and hence our assumption is wrong. Therefore, we can say that $\sqrt{3}$ is irrational.
Now, we will prove that $\sqrt{5}$ is irrational. For that let us consider that $\sqrt{5}$ is a rational number, and we know that a rational number can be expressed in the form of $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime. So, we get our number as, $\sqrt{5}=\dfrac{a}{b}$, where, $b\ne 0$ and $a$ and $b$ are co-prime, which means that $a$ and $b$ have no common factors. Now we will square both sides of the equation,
$\begin{align}
  & {{\left( \sqrt{5} \right)}^{2}}={{\left( \dfrac{a}{b} \right)}^{2}} \\
 & \Rightarrow 5=\dfrac{{{a}^{2}}}{{{b}^{2}}} \\
\end{align}$
By cross multiplying the above equation, we get,
$5{{b}^{2}}={{a}^{2}}\ldots \ldots \ldots \left( ii \right)$
This means that 5 is a factor of ${{a}^{2}}$, which means that 5 is a factor of $a$, so we can represent $a$ as, $a=5n$. So, by substituting it in equation (ii), we get,
$\begin{align}
  & 5{{b}^{2}}={{\left( 5n \right)}^{2}} \\
 & \Rightarrow 5{{b}^{2}}=25{{n}^{2}} \\
 & \Rightarrow {{b}^{2}}=5{{n}^{2}} \\
\end{align}$
We can see from here that 5 is a factor of ${{b}^{2}}$, which means that 5 is a factor of $b$ also. But we assumed that $a$ and $b$ have no common factors. So, our result contradicts our assumption.
Therefore, we can say that $\sqrt{5}$ is irrational.
We know that the sum of two irrational numbers is always an irrational number, which means that $\sqrt{3}+\sqrt{5}$ is also irrational.
Hence, we have proved the expression given in the question.

Note: In this question, the most important point that we have to remember is that we have to prove $\sqrt{3}$ and $\sqrt{5}$ are irrational separately. Also we need to remember that for proving the same, we will use the contradiction method by assuming them as rational numbers and then after expressing them in the $\dfrac{p}{q}$ form, we have to prove that $p$ and $q$ are not coprime.