I know this is Nth duplicate of this question, as I've seen it before, and I've been reading some of the answers as well. However, being really a total beginning with proofs, I'd appreciate if someone could comment on the following.
The task is to prove that there are distinct integers $m$ and $n$ such that $$\frac{1}{m} + \frac{1}{n}$$ is an integer.
I tried to approach this from different angles. First I came up with a proof by contradiction (I think this is what it is) and then by a direct example. I'm not sure if either is actually correct and if one should be preferred over the other (I think proof by a direct example is preferable when an existential quantifier is involved).
So my question is are either of the following attempts correct? And in case both are, which one is preferable?
Claim: T there are two distinct integers $m$,$n$ such that $1/m+1/n$ is an integer.
Proof: By contradiction.
- Write out the original claim: $$(\exists{m,n} \in \mathbb{Z})(m \ne n \wedge \frac{1}{m} + \frac{1}{n} \in \mathbb{Z})$$
- Assume the original claim is false, so its negation $$(\forall{m,n} \in \mathbb{Z})(m = n \vee \frac{1}{m} + \frac{1}{n} \notin \mathbb{Z})$$ must be true.
- Let m = 1 and n = -1, then $$\frac{1}{m} + \frac{1}{n} = \frac{1}{1} + \left(-\frac{1}{1}\right) = 1 - 1 = 0$$
- But now we shown that (2) which we assumed to be true is false.
Conclusion: Assuming the negation of the original claim to be true we have derived a contradiction, so the original claim must be true.
Proof: By an example.
- Let m = 2 and n = -2;
- Then $$\frac{1}{m} + \frac{1}{m} = \frac{1}{2} + \left(-\frac{1}{2}\right) = 0$$
Conclusion: There are two distinct integers m and n, namely m=1 and n=-1, such that 1/m+1/n is an integer.