# Proof involving quantifiers and rationality/irrationality

1. Jan 22, 2009

### Oxygenate

I have no idea how to begin this proof.

1. The problem statement, all variables and given/known data
Prove that for every rational number z, there exist irrational numbers x and y such that x + y = z.

3. The attempt at a solution
I can't think of even a way to start this proof...it's just quite obvious that the sum of two irrational numbers will equal a rational number, somehow.... Please give me some pointers to start this proof with.

2. Jan 22, 2009

### epenguin

"it's just quite obvious that the sum of two irrational numbers will equal a rational number" that, which is the statement converse of what you are asked, is not obvious and not true in general.

For the problem, you do know of a rational number A and an irrational x one that is less than it. Can you prove that (A - x) is also irrational?

Then you can do it for any pair of rationals.

3. Jan 22, 2009

### Oxygenate

Okay, so suppose A is irrational number. Then let x = 1/2z – a. And let y = 1/2z + a. Then x + y = z, and x and y are irrational.

4. Jan 22, 2009

### epenguin

Yes, understand you mean x = z/2 – a. And let y = z/2 + a

But you have to prove explicitly that if z is irrational, so is z/2 + a etc.

5. Jan 22, 2009

### Oxygenate

Okay, so:

Let y = z - x. Then suppose y is rational, then m/n - x = a/b, in which case x = m/n - a/b = (mb - na) / (nb), which is a rational number. But this is a contradiction to the fact that x is an irrational number. Thus the statement (for every rational number z, there exist irrational numbers x and y such that x + y = z) is true.

I essentially just took your idea and developed it. Thanks for your help. :)