# Can someone check my proof?

1. Feb 2, 2010

### iamsmooth

$$\forall q \in \textbf{Q}, \exists r \in\textbf{Q}$$ so that $$q + r\in \textbf{Z}$$ (Q is set of all rational numbers, and Z is set of all Integers)

Proof:

let q be an arbitrary rational number
thus, $q=\frac{a}{b}$ for some integers a and b, and b is not 0
let $r = \frac{b-a}{b}$ where $b-a,b\in\textbf{Z}$, b is still not 0
$$q + r = \frac{a}{b} + \frac{b-a}{b}$$

$$= \frac{a+b-a}{b}$$

$$= \frac{b}{b}$$

$$=1$$ and 1 is an integer

End of proof

I'm not sure if I was redundant with anything, or if I forgot to say anything. I think I only need to find one example since the second quantifier says there exists, which I think means I only need to show one algebraically for an arbitrary rational number. Also, I can take advantage of the fact that an integer is an integer, so I don't have to define it I guess...

I'm in a first year discreet mathematics course. If there's anything wrong with my proof, please let me know.

Thanks, appreciate it!

Last edited: Feb 2, 2010
2. Feb 2, 2010

### mathman

Are Q and Z restricted to > 0? If not r = -q will always work.

3. Feb 2, 2010

### iamsmooth

Yeah, -r would work, but I was wondering if my proof works as well. I realized that -r would work afterwards, but I already wrote down my version which I think works out algebraically, but yeah I wanted to confirm before I submit this.

4. Feb 3, 2010

### mathman

Your proof is correct, but as a mathematician I can tell you it is awful. Simple proofs are always preferred over complicated ones.

A more interesting case is restricting Q and Z to be positive. Then something like your proof might be needed.
(Hint: replace b-a by nb-a, where n is sufficiently large).

5. Feb 3, 2010

### Tobias Funke

Your proof is fine. I agree with mathman in spirit, but can you honestly say q+(1-q)=1 is "complicated"?

6. Feb 4, 2010

### mathman

Your version is several lines shorter than the original.