For any positive integer `n`, √(`n`) is either irrational or integral. The proof of this is fairly simple, but it's a good example of an elementary proof by contradiction.

Proof: Assume √(`n`) = `a`/`b`, where `a` and `b` are relatively prime and `b` ≠ 1. (In other words, assume √(`n`) is a nonintegral rational number.)
From here, square both sides to achieve `n` = `a`^{2}/`b`^{2}. Then multiply both sides by `b`^{2} to get `a`^{2} = `nb`^{2}.
From this we can conclude that, because `a`^{2} is an integer multiple of `b`^{2}, `b`^{2} | `a`^{2} (`b`^{2} divides `a`^{2}).

Now we must get a little sidetracked and go on an exploration of elementary number theory. It so happens to be true that, for any integers `m` and `n`, `m`^{2} | `n`^{2} implies that `m` | `n`. The proof of this is rather more complex than can be gotten into now -- perhaps that's a node for another day. Another fact that we'll soon need to use is that, if `m` | `n` and `m` and `n` are relatively prime, `m` equals one.

Anyway, back to our proof. Since we have concluded that `b`^{2} | `a`^{2}, we also know that `b` | `a`. Since we also know that `a` and `b` are relatively prime, we can conlude that `b` = 1. However, above we assumed that `b` ≠ 1 in order to make `a`/`b` nonintegral; thus, we have reached a contradiction and have proven that √(`n`) cannot possibly equal any rational nonintegral number. Since all real numbers are irrational, rational, or integral, this leaves the irrationals and the integers. (Yes, the integers are technically a subset of the rationals, so saying that a number is a "rational or an integer" is like saying a shape is a "rectangle or a square"; but for the purposes of this proof we had to differentiate.)

Thus, the square root of any positive integer is either an integer or an irrational number. **QED.**