4 ms·
Doesn't this allow one to prove x=y for any x, y? x/0 = x(1/0) = x*0 = 0, so x/0 = 0 for all x. So x/0 = y/0. Multiply both sides by 0: x = y.
by istjohn 9mo ago
Doesn't this allow one to prove x=y for any x, y?
x/0 = x(1/0) = x*0 = 0, so x/0 = 0 for all x.
So x/0 = y/0.
Multiply both sides by 0: x = y.
- deleted 9mo ago[deleted]
- deleted 9mo ago[deleted]
- rnhmjoj 9mo agoNo, because x/y is just an arbitrary operation between x and y. Here you're assuming that 1/x is the inverse of x under *, but it's not.
- orbifold 9mo agoI mean in a normal math curriculum you would define only the multiplicative inverse and then there is a separate way to define fraction, if you start out with certain rings. It is kind of surprising to me that they did a lazy definition of division.
- Smaug123 9mo agoWhat theorem did you use that allowed you to multiply both sides by $0$? (That theorem had conditions on it which you didn't satisfy.)