For the case a^2 = a (a belongs to R), I already showed that a+a = 0 for all a in R
What can you say a ring in which for all a in R, a^3 = a and more generally, for all a in R, a^n = a where n >= 4?
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For the case a^2 = a (a belongs to R), I already showed that a+a = 0 for all a in R
What can you say a ring in which for all a in R, a^3 = a and more generally, for all a in R, a^n = a where n >= 4?
well, ifQuote:
Originally Posted by knguyen2005
for all
then
because for any
that's jsut trivial! a non-trivial (even hard probably) question is this:
supposeis a ring with identity element and
is the smallest integer for which
what is the smallest integer
for which
?
for example ifwould it be possible to have
or
?
Thanks for a quick reply
If we use the formula http://www.mathhelpforum.com/math-he...e9129494-1.gif
Then for the case n = 2, we have : 2a = 0 implies that a + a = 0. So the formula works
when n = 3, 6a = 0 hence 3a+3a = 0
But when I do the calculation:
Let a in R, a^3 = a ad so (-a)^3 = (-a) by hypothesis
But (-a)^3 = (-a)(-a)(-a) = (a^2)(-a)
Thus, (-a) = (a^2)(-a) hence a^2 = 1.
What am I doing wrong here?
Why did I not get the result 6a = 0 when n = 3?
Thanks again
how did you get that? also we can't cancelQuote:
Originally Posted by knguyen2005
out because
I knew my calculation was wrong, I just want to do another way to get to the answer.
If I dont want to use the formula http://www.mathhelpforum.com/math-he...e9129494-1.gif when n = 3. How am I supposed to get to the answer 6a = 0 ? I mean can we do it another way instead of the formula u gave?
Thank you so much for your valuable time
Quote:
Originally Posted by knguyen2005
so
and thus