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Math Help - Discrete maths

  1. #1
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    Discrete maths

    suppose that f:N->N and g:N->N are defined by f(n)=nto the power of 3 and g(n)=n to the power of 2 for each natural number n.Show that gof not equal to fog.
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  2. #2
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    Quote Originally Posted by kodirl
    suppose that f:N->N and g:N->N are defined by f(n)=nto the power of 3 and g(n)=n to the power of 2 for each natural number n.Show that gof not equal to fog.
    f=\{ (n,n^3)|n\in \mathbb{N} \}
    g=\{ (n,n^2)|n \in \mathbb{N}\}
    Then,
    fg=f(n^2), n\in \mathbb{N}=n^6, n\in \mathbb{N}
    Also,
    gf=g(n^3), n\in \mathbb{N}=n^6, n\in \mathbb{N}
    Thus,
    gf=fg
    Thus, saying that they are no equal is wrong.
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  3. #3
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    Thanks

    Is that the actual answer and how it should be wrtten?
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  4. #4
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    Quote Originally Posted by kodirl
    Is that the actual answer and how it should be wrtten?
    It should be writen as PerfectHacker wrote it!
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  5. #5
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    Wink Hey you guys!!!!!

    Suppose that f:N->N and g:N-> are defined by f(n)=n^3 and g(n)=n^2 for each natural number n.Show that gof is not equal fog????
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  6. #6
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    Hello, Colette!

    Is there a typo in the problem? . . . The statement is wrong.


    Suppose that f\!:\!N \to N and g\!:\!N \to N are defined by
    f(n)=n^3 and g(n)=n^2 for each natural number n.

    Show that g\circ f \neq f\circ g

    We have: . n^3)^2\:=\:n^6 \\ \\ f\circ g\:=\:f(g(n)) \:=\:f(n^2)\:=\n^2)^3\:=\:n^6\end{array}" alt="\begin{array}{ccc}g\circ f\:=\:g(f(n)) \:=\:g(n^3)\:=\n^3)^2\:=\:n^6 \\ \\ f\circ g\:=\:f(g(n)) \:=\:f(n^2)\:=\n^2)^3\:=\:n^6\end{array}" /> . . . These are equal!

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  7. #7
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    oops sorry i left out part of the q

    Suppose that f:N->N and g:N-> are defined by f(n)=n^3 and g(n)=n^2-2 for each natural number n.Show that gof is not equal fog???? thats the correct version
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  8. #8
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    Ahh there's a -2 that cropped up this time.

    Let f(n) = n^3,\,g(n)=n^2-2

    Well the domain for both is the natural numbers, but I noticed that the you left off the codomain for the function g. I don't know why, but I would venture to guess because its codomain has to include -1, which isn't in the natural numbers.

    g(1) = 1^1-2 = -1

    So let D be the codomain of g so that g:N \to D.

    So in other words, the function g spits out something in D but f can only handle things from N. So f \circ g doesn't make sense because D \not\subseteq N.

    (Note g \circ f still can make sense, but it's certainly not possible that f \circ g = g \circ f)
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  9. #9
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    Hello, Colette!

    Ah, much better . . .


    Suppose that f:N\to N and g:N \to N are defined by
    f(n)=n^3 and g(n)=n^2-2 for all natural numbers n.

    Show that g\circ f \neq f\circ g

    g\circ f\;=\;g(f(n))\;=\;g(n^3)\;=\;(n^3)^2 - 2\;=\;n^6 - 2

    f\circ g\;=\;f(g(n))\;=\;f(n^2-3) \;=\;(n^2-2)^3  \;=\;n^6 - 6x^4 + 12n^2 - 8<br />

    Obviously, these two expression are equal for only specific values of n.
    As it turns out, there are no real values of n for which the two composites are equal.

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