(A twist in a classical question) Sum and product of two irrational numbers is rational? So I know that it is possible for the sums and products of irrational numbers to be rational. But, the only instances I know of that happening is when a certain combination of additive or multiplicative inverses of the irrational numbers in question are used.
My question is, given an irrational number $p$, can you multiply or add an irrational number $q$ to it so that their sum/product is a rational number given that $q$ cannot be written as any combination involving either $-p$ or $p^{-1}$?
 A: If $p+q = n/m$ then $q = n/m - p$ which is "a combination" involving $-p$. Same is true for the product.
A: Let $a$ be irrational. If $b$ is some irrational number such that $a+b$ is rational then you may write $a+b=p/q$, where $p,q$ are integers. But then $b=p/q-a$. 
Similarly, if $b$ is some irrational number such that $ab$ is rational then you may write $ab=p/q$, and so $b=p/q*a^-1$.
So, depending on precisely what you mean by "combination of ..." the above observations will probably give you an answer. 
A: The answer is no. This is just a basic property of adding numbers. Let's say you have a situation where two irrationals, $P$ and $Q$, sum to a rational, R. Then $P + Q = R$. No matter what the situation is, $Q = R - P$ will be true. Each irrational can be written as the resulting rational, $R$ minus the other irrational being added. 
Like I said, this is just a basic property of adding numbers, but the extra complexity of rationals and irrationals probably makes this harder to see. You can talk about this same thing with any numbers. Given the number $2$, can you find a number to add to it to make $5$? Well, $2 + (5 - 2) = 5$. The second number can always be written as a combination of the first number and the result. Nothing special is going on with the irrationals. What is happening is that in the simple situation where we're just dealing with integers, $5 - 2$ is not left as a combination of the result and the first number, it is just simplified to $3$. 
Now consider the situation $\sqrt{2} + (1 -\sqrt{2} ) = 1$. What is $1 - \sqrt{2} $? It simplifies to some irrational number, but we don't have a name, or a particular symbol for this irrational number (like the name/symbol for $5 - 2$ is $3$). So the only way this irrational number is identified out of the space of all other irrational numbers is by leaving it as $1 - \sqrt{2} $.
