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Proof of function

  1. Jul 11, 2009 #1

    Mentallic

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    I'm interested in getting a better answer to showing/proving a graph is a function than the usual "horizontal line test" response. Is the use of logic the best tool for proving a function? Because that line test method is basically that.
    Also, translating the function by illustrating it on the x-y plane isn't always so simple without a calculator, so other methods may be necessary.
     
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  3. Jul 11, 2009 #2

    Office_Shredder

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    Depends on what you want to do. Are you trying to show a graph is representative of a function (in which case you don't need to worry about trying to draw it) or are you trying to show that an equation of the form f(x,y) = 0 i.e. an equation with x's and y's can be written so that y is a function of x?
     
  4. Jul 11, 2009 #3

    Mentallic

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    I'm trying to show the second that you mentioned :smile:

     
  5. Jul 11, 2009 #4

    Office_Shredder

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    What you really need to show is:

    Given a value of x, there is only one possible value of y(x). Example for each:

    x2 + y2 = 1

    If x = [itex] \frac{1}{\sqrt{2}} [/itex] then y can be plus or minus [itex] \frac{1}{\sqrt{2}}[/itex] so this does not give a function (and drawing the circle we see this is obvious). More generally, if y2 = 1-x2 and w2 = 1-x2 for the same value of x, then y2 = w2 so y= +/- w and in general y and w do not have to be equal

    If on the other hand we have 2y-x-1 = 0 Then if 2y-x-1 = 0 and 2w-x-1 = 0 for a set value of x, then 2y=x+1=2w and we can see y=w is necessary. Hence for each value of x there is only a single value of y that satisfies this, and y can be defined as a function of x.

    There's also another possible concern about the domain. y is a function of x over the real numbers by definition also means for each x, there exists a possible value of y. If your possible function was [itex]y = \sqrt{1-x^2}[/itex] (taking only positive values of square root) unlike in the circle case this satisfies the condition for each x there is at most one value of y. But by definition, this is not a function over the real numbers because outside of the interval [-1,1] y is not defined. Instead it's only a function over that interval. Generally this isn't the condition you're worried about though (since what the domain of the function is probably isn't the question at hand)
     
  6. Jul 11, 2009 #5

    Mentallic

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    So it is necessary to make the y variable of an equation f(x,y) the subject so as to know if it is the form [tex]y^2=y(x)[/tex] or maybe other operative functions that result in more than one value such as [tex]sin(y)=y(x)[/tex] or others that exist (which I wouldn't know about).
    Is it possible to show the equation is a function or not without re-arranging the equation? Possibly because re-arrangement would be too difficult or impossible with elementary functions.
     
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