06-29-2013, 09:32 AM
(06-29-2013, 01:40 AM)Guy1234567890 Wrote: Just as a note, you only have to show that the proposition is true for one case (usually the case where n is the smallest) and then you can immediately show that it is true for the k+1th case.
No, if you you only had to show it was true for one case, then you wouldn't be proving anything. If I had the equation 2+k=4, and showed that it was true when k=2, that doesn't mean it would be true for 3. You'd have to show that the k+1'th term was equivalent to simply substituting k+1 in for k in the original equation. If it works for both k and k+1, since k can be anything, then you've proven that it works for all k's.
(06-28-2013, 10:06 AM)xdot Wrote: The plugin/TeX server seems to ignore multi-line scripts.
Ah, so that's what it was. Fix'd.
Finding products of 2 and/or 3 digit numbers is pretty easy if you just foil them in your head. No memorization required.
