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Algebra of functions

  1. Sep 25, 2007 #1
    is the set of all functions a ... (blank), i don't know what the correct classification is, but it should contain the property of the multiplicative inverse. that's basically what the question is. does the set of all (really all) functions contain these two properties:

    [tex] f=f[/tex]
    and
    [tex] f \frac{1}{f} =1[/tex]

    for example for polynomial functions this is true

    [tex]f(x)=x[/tex]
    [tex]f(x)\left(\frac{1}{f(x)}\right)=\frac{x}{x}=1[/tex]

    at least so i've been taught since the beginning of time.

    but the question is does this hold true for ALL functions. basically what kind of set is the set of all functions.
     
  2. jcsd
  3. Sep 25, 2007 #2

    Hurkyl

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    (1) What kind of functions? I bet you have a specific domain and range in mind...
    (2) What is the definition of the multiplicative inverse of a function?
    (3) Think about your example some more...
     
  4. Sep 25, 2007 #3

    Hurkyl

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    Incidentally, in the world of algebra, sets are characterized by the fact that they have no operations defined on them. A set only contains elements, it doesn't have an additional structure like a multiplication operation.

    So what you meant to say is that you were asking about the algebra whose underlying set is the set of functions and has a multiplication operator. (By the way, how is that operator defined?)
     
  5. Sep 25, 2007 #4
    1. no no specific domain in mind, this is a question about all domains, all functions, everything.
    2. the definition would seem to be as i've described it to be in my example. if it is not please correct me.

    i wasn't sure if it was a question pertaining to the algebra over the set all functions or the field over the set of all functions or w/e.

    again if i go with what has been taught to me then the multiplication operator is defined as simply the multiplication of the outputs of each function.

    you have to understand this is my first steps into the world of real math. so functions as mappings, fields, algebras, operators etc is all very new to me.
     
  6. Sep 25, 2007 #5

    matt grime

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    field


    Are you sure you mean functions? From where to where?

    No, you haven't. 1/x is not a polynomial.
     
  7. Sep 25, 2007 #6
    thx2u
    quite true, it is a rational function and now i am completely confused about how a polynomial function multiplied by a rational function is 1,the identity.
     
  8. Sep 25, 2007 #7

    matt grime

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    What? Polynomials are rational functions too. Rational functio fields - the basis of algbraic geometry.
     
  9. Sep 25, 2007 #8
    i didn't realize this.

    you should have just told me that polynomials are a subset of the rationals and that the identity i put up is actually for the field of rationals.
     
  10. Sep 25, 2007 #9

    matt grime

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    Many apologies. I will work on my mind reading skills. Is there anything else you're ignorant of that I should magically guess I need to tell you?
     
  11. Sep 25, 2007 #10
    considering i didn't realize that a field contains the property of multiplicative inverse and the last statement in post #4, which you might notice is immediately before your first post in this thread, it should be quite obvious that i'm ignorant of many things.

    if you don't have the patience to explain things in terms i can understand at least refrain from confusing me.
     
  12. Sep 25, 2007 #11

    matt grime

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    I did not confuse you. You confused yourself by not understanding the definitions. Don't blame me if you haven't learnt the definitions. If you state something to be a rational function I presume that means you understand what they are (you introduced the phrase). If you didn't understand from the definition of rational function (the ratio of two polys) that this clearly means the polys are a subset - polynomial divided by 1 - then that is not my fault, again. I take it you understand that integers are rational numbers. This is no different.

    (If you'd used 'could have told me' rather than 'should have told me' when telling me off, I'd've not gotten anywhere near as annoyed).
     
    Last edited: Sep 25, 2007
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