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: 
Consider $r$ rational and assume

$$\sqrt 2 + \sqrt 3 + \sqrt 5=r $$

then

$$\iff \sqrt 2 + \sqrt 3)=r - \sqrt 5$$

$$5+ 2\sqrt 6=r^2 + 5-2r \sqrt 5$$

$$2\sqrt 6 + 2r \sqrt 5=r^2$$

$$24+20r^2+8r \sqrt{30}=r^4$$

$$8r \sqrt{30}=r^4-24-20r^2+$$

$$\sqrt{30}=\frac{r^3}8-\frac 3 r-\frac52 r$$

then it suffices to show that $\sqrt {30}$ is not rational (which is true since $30$ is not a perfect square).
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.