Prove that for every rational number z and every irrational number x, there exists a unique irrational number such that x+y=z This is a homework assignment, please tell me if my proof is correct!
Prove that for every rational number z and every irrational number x, there exists a unique irrational number such that x + y = z.
Assume a and b are integers with GCD (a,b) =1, and that c is an irrational number.
There exists a number d = (a/b) + c.
Assume that d is rational. Then d - (a/b) would be rational, what is a contradiction because c is irrational. Therefore, d is irrational.
To prove uniqueness we can use the fact that the addition of any two real numbers has only one result, then d is unique. Q.E.D.
 A: You don't need to restrict the integers forming the ratio to having a greater common denominator of 1.   Just assert that they exist.

Any $z$ that is a rational number can be expressed as the ratio of two integers.  (For strictness we require the denominator integer to be non-zero.)
Any $x$ is an irrational number cannot be so expressed as the ratio of two integers.
For any real numbers, $x$ and $z$, there is only a unique number $y$ where $x+y=z$.
If this $y$ were rational it could be expressed as the ratio of some two integers.  For any sum $x+y$ which can also be expressed as the ratio of two integers, it would then follow that $x$ could be expressed as the sum of two integers.  (By reason the product of any two integers is an integer.)
By contraposition: for any irrational $x$, and any rational $z$, the number $y$ where $x+y=z$, must be both unique and irrational.
A: You have proven that the sum of a rational number and an irrational number is irrational, but you have not proven the main result.
If $x$ is irrational and $z$ is rational, what (possibly) irrational number $y$ would you wager is such that $x+y=z$? Why might this be irrational? (Hint: Use what you just proved.)
Why must it be unique?
