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 = a2/b2. Then multiply both sides by b2 to get a2 = nb2.
From this we can conclude that, because a2 is an integer multiple of b2, b2 | a2 (b2 divides a2).
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, m2 | n2 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 b2 | a2, 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.