# For every rational number, there exists sum of two irrational numbers

1. Jan 25, 2010

### ckwn87

1. The problem statement, all variables and given/known data

Prove: For every rational number z, there exists irrational numbers x and y such that x + y = z.

2. Relevant equations

by definition, a rational number can be represented by ratio of two integers, p/q.

3. The attempt at a solution

Is there a way to do this by contraposition?

Would the contraposition be, For all rational numbers x and y, there does not exist an irrational number z such that x + y = z? I can handle from there, but I don't think my contraposition is correct.

2. Jan 25, 2010

### Hurkyl

Staff Emeritus
The question, as stated doesn't have any "if ... then ..." clauses in it. So, you have to rewrite it. You could, for example, sue
For every real z, if (z is rational) then (there exists irrational x and y such that x+y=z​
which you could contrapositive. I'm not sure it helps, though.

Have you tried one the simplest of all techniques -- guess and check?

3. Jan 25, 2010

### ckwn87

Hmm, you're right, I don't think contraposition will help. I'm not sure what you mean by guess and check. Since it says for ALL Z, I'm not sure how I would generalize a sum of two irrational numbers.

4. Jan 25, 2010

### Hurkyl

Staff Emeritus
You can guess for the existentials, though.

(And you can always try specific z's to get an idea before tackling the universal case)

5. Jul 19, 2010

### dimitri151

Prove: For every rational number z, there exists irrational numbers x and y such that x + y = z.
Proof: x=z/2+sqrt(2), y=z/2-sqrt(2), so x+y=z and x, and y are irrational.

6. Jul 20, 2010

### hunt_mat

It looks that simple dimitri.

7. Jul 20, 2010

### dimitri151

It is, isn't it?