X^(n-1) = 0, what does x equal ?

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In the reals ab=0 if and only if one of a or b is zero.
 
but its a^b = 0
soz, n is a natural number
 
matt wouldn't have said that unless it applied to your problem.

is there any way of writing your problem in the form a.b=0?
 
not that i know of? pls explian
 
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What does x^m mean? (For m a natural number)
 
x*x m times

x^2 = xx
x^3 = xxx
x^m = x...?
 
So you do know how to make a power of x into a product of two (or more) numbers, all of which are x. Now, apply the first thing I posted.
 
x^(n-1) = 0

x*x*x*x = 0?
 
If you multiply any collection of (real) numbers together and get zero one of them must be zero, that is an underlying and very important fact about them, it is how you factor polynomials, remember?
 
There are actually two answers to your question. For some values of n,
x= ___ . For other values of n, there is no such x.
 
matt grime said:
In the reals ab=0 if and only if one of a or b is zero.
Somewhat unrelated, but that's because the Reals are a principle ideal domain isn't it? I'm vaguely trying to remember my 'Groups, Rings and Modules' course from 2 years ago.
 
No. A principal ideal is one in which any ideal is generated by a single element. The reals, being a field, only have trivial ideals anyway (ie 0 and R). This has nothing to do with zero divisors. You're thinking of an integral domain. There are non principle ideal domains that are integral (eg Z[x,y]: the ideal (x,y) is not principle), and there are principal ideal domains that are not integral like Z/4Z.

As far as I'm concerned, the fact that there are no zero divisors in R comes first, therefore it implies they are an integral domain, rather than they are an integral domain therefore there are no zero divisors. Small point, and wholly semantic.
 
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Now I'm having second thoughts. A PID has no zero divisors because it is required to be an integral domain as well. Domain being, apparently, synonymous with integral domain, though that isn't necessarily universal: are domains presumed commutative?
 
In my algebra class we assumed domains to be commutative. But we didn't assume rings to have unity. Strange, in my opinion.
 

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