Talk:Indeterminate (variable)

Unclear definition
The definition "a quantity that is not known, and cannot be solved for" is not clear.
 * Chaitin's constant is not known, and cannot be solved for. Does that mean it is an indeterminate?
 * The text implies that an indeterminate is not a variable. This does not fit with the common use of the term "variable" as I know it.
 * In the text "the polynomial $$X-1$$", X is not known and cannot be solved for. In the text "the polynomial equation $$X-1 = 0$$", X is not known but can be solved for. Does that mean X is an indeterminate in the polynomial by itself but not in the polynomial used in the polynomial equation?  --Lambiam 17:39, 24 July 2008 (UTC)


 * I agree with the criticism; this definition should be rewritten. In fact, I would say an indeterminate is a symbol that is not used to designate any (other) known or unknown quantity: it is neither an unknown (initially unspecified value, but which with some good luck can be solved for), nor a parameter (value supposed to be given and fixed in the problem at hand, but not explicitly known), nor a formal variable (as the bound variable in some syntactic construction like a summation, or the variable used for the argument in a function definition; in both cases such a variable stands for many different values at once). Note that parameters nor formal variables are known, and cannot be solved for either, showing again the defect of the current definition.
 * What distinguishes indeterminates is that by standing for nothing else, they become bona fide values in their own right. So for instance by constructing the polynomial ring Z[X], the symbol X acquires the status of indeterminate (but many other formal constructions can introduce indeterminates as well). This is a rather fundamental conceptual step that ought to be described properly in this article. But to answer your last question: indeed, if a polynomial is used in a polynomial equation, its indeterminates lose their status, and become unknowns; regarding $$X-1 = 0$$ as an equation of formal polynomials it just states an unconditional falsehood (both members are distinct polynomials, period). Marc van Leeuwen (talk) 10:51, 2 August 2008 (UTC)


 * I find this entry confusing. In case 2, doesn't "2 + 3x = a + bx" actually mean "'2 + 3x' = ˹Φ + ψx˺" where the funny brackets are quasi-quotes and Φ and ψ are meta-variables?Adam.a.a.golding (talk) 01:57, 21 December 2009 (UTC)

Getting back to why (-1)-(-1)^2=(-1)-1=-2, isn't a concern; we need to remember that for our function, the modulo 2 property is a requirement of the Domain. The numerical output structure is the Range (which can be defined as something else). Our target output of 0 needs to make sense for what we define the Range to be. THIS is the reason why doing modulo 2 on such output is not correct. I believe what the author of the equation example tried to do was to say "hey.. there's some function that generates an intended result value with a source of unknown x (the structure of x is defined by some Domain. We are talking about particular Numbers here, not a fancy matrix) which generates more than 1 solution, but this unknown x is not an indeterminate X; because an indeterminate (with a structure defined by the Domain) requires that the function will ALWAYS generate the intended result value regardless of what value of X is." 60.240.106.190 (talk) 00:52, 11 February 2015 (UTC)
 * The final example given in the section "Polynomials" (regarding 0-0^2=0, 1-1^2=0, with modulo 2) seems to lack an additional note for it in the main page which might cause some misunderstanding? We can find that another modulo 2 value of '-1' gives the form (-1)-(-1)^2=(-1)-1=-2, which results in the answer of -2 which in modulo 2 also equals 0 (yes, there is a small mistake in doing this which i'll get to soon, but the statement is not incorrect). There is a bit of trickiness in seeing that binding a range limit (ie. Must be an integer, and must have value equivalence after applying modulo 2) to some unknown x; is trying to give an example too complicated which a student might get confused (because it's perfectly fine for the intermediate X to be limited as well -- say 'must only be an integer').

why is x the unknown
Perhaps it is worth to mention, why we use in general x as unknown. http://www.ted.com/talks/terry_moore_why_is_x_the_unknown.html --77.186.34.177 (talk) 21:12, 2 July 2012 (UTC)

definition is unclear
The following defines an indeterminate in terms of other undefined concepts: symbol, variable, placeholder, and in terms of what it is not. Can we do any better?

In mathematics, and particularly in formal algebra, an indeterminate is a symbol that is treated as a variable, but does not stand for anything else but itself and is used as a placeholder in objects such as polynomials and formal power series. In particular, it does not designate a constant or a parameter of the problem, it is not an unknown that could be solved for, and it is not a variable designating a function argument or being summed or integrated over; it is not any type of bound variable.--345Kai (talk) 22:07, 3 September 2018 (UTC)