Question: The question is prove $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is an irrational number.

I started by assuming the opposite that $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is a rational number. I stated that a rational number is a number made by dividing two integers. So I set $\sqrt{2} + \sqrt{3} + \sqrt{5} = i\_1/i\_2$, where $i\_1$ and $i\_2$ are two integers. I multiplied $i\_2$ onto both sides and got $i\_2\sqrt{2} + i\_2\sqrt{3} + i\_2\sqrt{5} = i\_1$. I then said that in order to turn an irrational number such as $\sqrt{2}$ into a rational number you can multiply, $\sqrt{n}\sqrt{n}=n$. Meaning $i\_2$ would have to hold the value of $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ which is impossible. So it is an irrational.

I think I made a mistake somewhere but I am not sure.

Answer: 
Let $x=\sqrt 2+\sqrt 3+\sqrt 5$ and $y=x-\sqrt 5=\sqrt 2+\sqrt 3.$. We have $$0=(y-\sqrt 2-\sqrt 3)(y-\sqrt 2+\sqrt 3)=(y-\sqrt 2)^2-(\sqrt 3)^2=y^2-2y\sqrt 2-1.$$
So $y^2-1=2y\sqrt 2.$ Squaring this and re-grouping, we have $y^4-10y^2+1=0.$ Substituting $x-\sqrt 5$ for $y$ in this, expanding and re-grouping, we obtain $$A=B\sqrt 5 \quad \text {where}\quad A=x^4+4x^2-24\quad \text {and}\quad B=20x^3.$$

Now if $x$ is rational then $A$ and $B$ are rational with $B=20x^3\ne 0, $ implying $\sqrt 5=A/B\in \Bbb Q, $ which is false. So $x$ cannot be rational.
The sum of two irrational numbers can be rational, so you cannot go from your hypothesis that $i\_2\sqrt2 + i\_2\sqrt 3+i\_2\sqrt 5$ is rational to the claim that each of them is separately rational. An example would be $3-\sqrt 2$ and $\sqrt 2$ are both irrational but their sum is $3$. 

For the sum of two square roots, like $\sqrt 2 + \sqrt 3$ you can use the fact that the square of a rational is rational. You can assume it is rational and square it to get $5+2\sqrt 6$. As you are down to one square root you can follow the usual proof that $\sqrt 2$ is irrational.