Proving a Bunch of Statements Been doing practices problems so...
Consider the following statements: 
(a) Define what it means for a real number to be rational and for a real number to be irrational. 
Answer: I assume that this just want the definition of rational and irrational numbers so I won't go into detail what I put. 
(b) Prove that the sum and that the product of two rational numbers is rational. 
Answer: Let $\alpha,\beta\in\mathbb{Q}$ Let $\alpha=\frac{A}{B}$ and $\beta=\frac{C}{D}$ with $A,B,C,D\in\mathbb{Z}$  
Then $\alpha+\beta=\frac{A}{B}+\frac{C}{D}=\frac{AD+BC}{BD}\in{Q}$ 
Also, $\alpha\beta=(\frac{A}{B})(\frac{C}{D})=\frac{AC}{BD}\in\mathbb{Q}$
(c) Prove or disprove. If $x$ and $y$ are positive irrational numbers, then $xy$ is also an irrational number. 
Let $x=\sqrt{m}$ and $y=\sqrt{n}$ with $m,n\in\mathbb{N}$ 
If $m=n$, then $\sqrt{m}\sqrt{n}=\sqrt{mn}=\sqrt{m^2}=\sqrt{n^2}=m=n$ 
This result is not irrational. 
However, if $m\neq{n}$, then $\sqrt{mn}$ would be considered irrational. 
Therefore, this statement is true conditionally.
(d) Prove or disprove that, if $x+y$ is irrational then $x$ and $y$ are irrational. 
Let $x=\sqrt{m}$ and $y=\sqrt{n}$ with $m,n\in\mathbb{N}$.
Assume $m=n$. Then $\sqrt{m}+\sqrt{n}=\sqrt{m}+\sqrt{m}=\sqrt{n}+\sqrt{n}=2\sqrt{m}=2\sqrt{n}$ 
and we know a rational $(2)$ times an irrational $(\sqrt{m}$ or $\sqrt{n})$ is irrational. Therefore, this proves this statement.
(e) Prove or disprove that, if $xy$ is irrational and $y$ is irrational then $x$ is rational.  
I have NO idea for this one. Help on this one would be awesome.
As for the others, are they okay? Or can I improve? Or do they need fixing? Thank you.
 A: Your proof of b) is correct.
To c), "is true conditionally" doesn't make any sense, since the problem doesn't have any conditions. The simple answer is "The statement is wrong", since $\sqrt{2}\cdot\sqrt{2} = 2$ is a counterexample.
Your proof of d) is horribly wrong and broken. In fact, it is so horribly wrong that I sorta doubt you did a) correctly. A correct proof would be a counterexample (for example $\sqrt{2}+1$ is irrational even though not both summands are irrational), since the statement is wrong.
e) Again, a simple counterexample suffices. Example: $\sqrt{2}\cdot\sqrt{3} = \sqrt{6}$ is an irrational product of two irrational numbers.
A: For (c), it is misleading to say

this statement is true conditionally.

You give a counterexample to the statement, so the statement is false as stated.
Your solution for (d) does not work. Not every irrational number can be written as $\sqrt{n}$ for $n\in\mathbb{N}$; you must prove (or disprove) that the statement holds for any choice of irrational $x,y$. In this case, the statement is false, as rational $+$ irrational $=$ irrational.
For (e), it is enough to prove or disprove that the product of irrational numbers may be irrational. You can find an example with radicals in this case.
