by AuSmith » Wed Mar 18, 2009 11:55 pm
2) A cheap proof would be [unparseable or potentially dangerous latex formula]. So, the modulus of a square is the square of the modulus. I say cheap because it takes a lot more work to show that [unparseable or potentially dangerous latex formula] behaves like an exponent. Normally, we define [unparseable or potentially dangerous latex formula], then prove most of the wonderful properties of exponents.
3) I think they have tables for this stuff. You could say [unparseable or potentially dangerous latex formula] and so [unparseable or potentially dangerous latex formula]. At the very least, you're assured [unparseable or potentially dangerous latex formula] has an inverse because it's relatively prime to [unparseable or potentially dangerous latex formula]....... conclude that [unparseable or potentially dangerous latex formula]. Multiply both sides by [unparseable or potentially dangerous latex formula]. Finally, [unparseable or potentially dangerous latex formula].
5) It's just asking you to find the root. [unparseable or potentially dangerous latex formula], so [unparseable or potentially dangerous latex formula].
6) You're compelled to add first. [unparseable or potentially dangerous latex formula]. You then notice that [unparseable or potentially dangerous latex formula] nicely cancels with [unparseable or potentially dangerous latex formula].
[unparseable or potentially dangerous latex formula]