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.
Hello, Colette!
Is there a typo in the problem? . . . The statement is wrong.
Suppose that and are defined by
and for each natural number
Show that
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!
Ahh there's a -2 that cropped up this time.
Let
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.
So let be the codomain of g so that .
So in other words, the function g spits out something in but f can only handle things from . So doesn't make sense because .
(Note still can make sense, but it's certainly not possible that )
Hello, Colette!
Ah, much better . . .
Suppose that and are defined by
and for all natural numbers .
Show that
Obviously, these two expression are equal for only specific values of .
As it turns out, there are no real values of for which the two composites are equal.