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Ch17, 8.4: Prove "in any integral domain, only 1 and -1 are their own multiplicative inverses"
I don't quite get the solution; seems like the proof assumes "only 1 and -1 are their own multiplicative inverses". Please correct me if I am wrong! But I think I do have another proof:
Instead of solving a*a=1, we could let a=1+x, then solve (1+x)(1+x)=1, which expands to 1+x+x+x*x = 1, so x(1+1+x)=0. Integral domains do not have divisors of zero (theorem 2 on page 174), so at least one of x and 1+1+x must be zero (page 173), so either x=0 or x=-1-1; a=1 or a=-1.
The text was updated successfully, but these errors were encountered:
you're right! He didnt use the properity of integral domains.
We can also proof by contradiction.assume $t \ne -1$ and $t \ne 1$,use $t*t = 1$ to construct $(t-1)(t+1) = 0$
Same idea!
Ch17, 8.4: Prove "in any integral domain, only 1 and -1 are their own multiplicative inverses"
I don't quite get the solution; seems like the proof assumes "only 1 and -1 are their own multiplicative inverses". Please correct me if I am wrong! But I think I do have another proof:
Instead of solving
a*a=1
, we could leta=1+x
, then solve(1+x)(1+x)=1
, which expands to1+x+x+x*x = 1
, sox(1+1+x)=0
. Integral domains do not have divisors of zero (theorem 2 on page 174), so at least one ofx
and1+1+x
must be zero (page 173), so eitherx=0
orx=-1-1
;a=1
ora=-1
.The text was updated successfully, but these errors were encountered: