Base case: n = 1. Trivial because it says that d = 0.
Suppose d^n implies that d = 0. Then assume
d^{n + 1} = d^n * d = 0.
Because D is an integral domain, either d^n = 0 or d = 0. In the first case, we can apply the induction hypothesis and have d = 0. In the second case, d = 0 and we are done. So
you ought to to apply induction for a perfect evidence, yet i think of the assumption is this: d^n = d(dddd.... n-a million circumstances) which suggests that d = 0 or dddd... (n-a million circumstances) = 0 on the grounds that D is an necessary area (so it has no divisors of 0). In the two case, we are able to discover that d = 0, on the grounds that we are able to be conscious this above approach to the case dddd.... (n-a million circumstances).
Comments
Proof by induction on n.
Base case: n = 1. Trivial because it says that d = 0.
Suppose d^n implies that d = 0. Then assume
d^{n + 1} = d^n * d = 0.
Because D is an integral domain, either d^n = 0 or d = 0. In the first case, we can apply the induction hypothesis and have d = 0. In the second case, d = 0 and we are done. So
d^{n + 1} = 0 implies d = 0.
you ought to to apply induction for a perfect evidence, yet i think of the assumption is this: d^n = d(dddd.... n-a million circumstances) which suggests that d = 0 or dddd... (n-a million circumstances) = 0 on the grounds that D is an necessary area (so it has no divisors of 0). In the two case, we are able to discover that d = 0, on the grounds that we are able to be conscious this above approach to the case dddd.... (n-a million circumstances).