SOLUTION: Prove that there exist some n
such that
(x^n)^2=x^n for all n>= 1.
r is idemotent ring where x^2= x
I have more this one is just to check who is qualified to help me
Algebra ->
Sequences-and-series
-> SOLUTION: Prove that there exist some n
such that
(x^n)^2=x^n for all n>= 1.
r is idemotent ring where x^2= x
I have more this one is just to check who is qualified to help me
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Question 872663: Prove that there exist some n
such that
(x^n)^2=x^n for all n>= 1.
r is idemotent ring where x^2= x
I have more this one is just to check who is qualified to help me Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! r is idempotent ring where x^2 = x
Prove there exists some n such that
(x^n)^2 = x^n for all n>= 1
now for n = 1, we have
x^2 = x which is true for x a member of r
for n+1 we need to show
(x^n+1)^2 = x^n+1
(x^n+1)^2 = (x^n * x)(x^n * x)
(x^n+1)^2 = (x^n)^2 * x^2
(x^n+1)^2 = x^n * x
(x^n+1)^2 = x^n+1
