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: 
If $\sqrt{2}+\sqrt{3}$ is rational then so too is $\sqrt{2}-\sqrt{3}$ because $(\sqrt{2}+\sqrt{3})\cdot (\sqrt{2}-\sqrt{3}) = 2 - 3 = -1$

But adding the two terms, $(\sqrt{2}+\sqrt{3})+ (\sqrt{2}-\sqrt{3}) = 2\sqrt{2}$ which is irrational. Two rational numbers cannot sum to an irrational number so we have a contradiction. Therefore $\sqrt{2}+\sqrt{3}$ is irrational. We can say $\sqrt{2}+\sqrt{3}$ = I and come to the same result/conclusion for I$ + \sqrt{5}$. 

In this case we reach the assumption that I$^2-5$ is rational. 

But I$^2-5= (\sqrt{2}+\sqrt{3})^2-5 = 5+2\sqrt{6}-5 = 2\sqrt{6}$ which is irrational and another contradiction. Hence $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational.
No, this solution is incorrect I'm afraid. You need $i\_2\sqrt{2}+i\_2\sqrt{3}+i\_2\sqrt{5}=i\_1$, that's true, but that doesn't mean that $i\_2\sqrt{2}$ needs to be rational. Also, if it *did* need to be rational, that doesn't mean that $i\_2$ must be $\sqrt{2}$. In can also be $3\sqrt{2}/5$, for example.