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

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Discussion Overview

The discussion revolves around the equation x^(n-1) = 0, specifically exploring the implications of this equation in the context of real numbers and the properties of exponents. Participants examine the conditions under which x can equal zero and the mathematical principles that apply to this scenario.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning
  • Historical

Main Points Raised

  • Some participants assert that in the reals, the product ab = 0 if and only if one of a or b is zero.
  • There is a clarification that x^m, where m is a natural number, represents x multiplied by itself m times.
  • One participant suggests that the equation x^(n-1) = 0 can be expressed as a product of x's, prompting further exploration of the implications.
  • Another participant notes that there are two potential answers to the original question, depending on the value of n, indicating that for some values of n, x can equal zero, while for others, it cannot.
  • There is a discussion about the properties of the reals as a principal ideal domain and the relationship to zero divisors, with some participants debating the definitions and implications of these mathematical concepts.
  • Concerns are raised about the assumptions made in algebra classes regarding commutativity and the presence of unity in rings.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the equation and the properties of the reals, particularly regarding the definitions of ideal domains and the conditions under which x can equal zero. The discussion remains unresolved with multiple competing perspectives presented.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the definitions of mathematical terms, the scope of the equation, and the conditions under which the statements hold true. Some mathematical steps and implications remain unresolved.

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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
 
Last edited:
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?
 
  • #10
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?
 
  • #11
oh yeah... thanks :)
 
  • #12
There are actually two answers to your question. For some values of n,
x= ___ . For other values of n, there is no such x.
 
  • #13
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.
 
  • #14
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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  • #15
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?
 
  • #16
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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