1. ## 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.

2. 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.

3. ## Thanks

Is that the actual answer and how it should be wrtten?

4. Originally Posted by kodirl
Is that the actual answer and how it should be wrtten?
It should be writen as PerfectHacker wrote it!

5. ## 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????

6. 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: . $\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}" 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!

7. ## 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

8. 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 $f \circ g = g \circ f$)

9. 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
$

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.