Can someone check my proof?

[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!

[mathman]

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

[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.

[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).

[Tobias Funke]

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

[mathman]

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

Your version is several lines shorter than the original.