3 ms·
3^2-2^2 =!= 2^2-3^2. (You can exchange a and b in, say a^2+b^2, because 2^2+3^2=3^2+2^2)
by davrosthedalek 2y ago
3^2-2^2 =!= 2^2-3^2.
(You can exchange a and b in, say a^2+b^2, because 2^2+3^2=3^2+2^2)
- wcrossbow 2y agoThis is not what I meant. What is being proved is: a^2-b^2 - (a+b)(a-b) = 0. If you swap a and b you end up with a sign switch on the lhs which is inconsequential.
- davrosthedalek 2y agoThat is not what the proof proves. The proof proves the equivalence how it was originally stated, and assumes for that b<a. Your rewriting is of course true for all a,b and might be used in an algebraic proof. But this transformation is not at all shown in the geometric proof.
- olddustytrail 2y agoDid you think that I meant you can switch them on one side of the equation but not the other? That's not what anyone is saying.
- davrosthedalek 2y agoNo, of course not.
- olddustytrail 2y agoBut that's literally what you just did in your example.
- davrosthedalek 2y agoI did not show the right side at all, so I am not sure how you can make that statement. The point is that a+b is symmetric in a <-> b and a-b is anti-symmetric. Both left and right side are anti-symmetric.