Prove a statement (discrete math)?

Question

"There are distinct integers m and n such that 1/m + 1/n is an integer."

The only thing I can think of is the values of m and n are +1 and/or -1 for the sum of these fractions to give an integer. Other than that I am lost formulating the proof.

Thanks much for your help!

Answer 1

You might also take 1/2 + 1/2 = 1. Or 1/a + 1/(-a) = 0 for any nonzero a. But notice for this kind of an existence theorem, you only need one example. Therefore, your example is a perfectly legitimate proof.

Answer 2

1/m + 1/n = (m+n)/(mn)
If m+n = 0, then you have a solution.
Therefore, you have infinite number of solutions, since as long as m = -n = any integer other than zero, you have one solution.

Answer 3

p implies q AND q implies r If q is actual then r additionally must be actual. If q is fake, we don't what r is. If p is actual then q additionally must be actual. If p is fake, we don't care what q is. If p is actual then q is actual. by using fact q is actual if p is actual, then r is likewise actual if p is actual. subsequently, p implies r. actual for all p, q, r.