Talk:Indeterminate (variable)

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Unclear definition[edit]

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 is not known and cannot be solved for. In the text "the polynomial equation ", 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 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)

  • 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').

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." (talk) 00:52, 11 February 2015 (UTC)

why is x the unknown[edit]

Perhaps it is worth to mention, why we use in general x as unknown. -- (talk) 21:12, 2 July 2012 (UTC)