Can sumone pls show me how to do this question.
Q.Suppose that f:N->N and g:N->N are defined by f(n)=n^3 and g(n)=n^2 for each natural number n.Show that g o f is not equal to f o g.

2. Originally Posted by dcapdogg
Can sumone pls show me how to do this question.
Q.Suppose that f:N->N and g:N->N are defined by f(n)=n^3 and g(n)=n^2 for each natural number n.Show that g o f is not equal to f o g.
It can't be done, because they are equal:

$
(f\circ g)(x)=f(g(x))=f(x^2)=(x^2)^3=x^6
$

$
(g\circ f)(x)=g(f(x))=g(x^3)=(x^3)^2=x^6
$
.

RonL

3. ## oops i left a bit out

g(n)=n^2-2

4. Originally Posted by dcapdogg
g(n)=n^2-2
In that case:

$
(f\circ g)(x)=f(g(x))=f(x^2-1)=(x^2-2)^3$
$=x^6 - 6x^4 + 12x^2 - 8
$

$
(g\circ f)(x)=g(f(x))=g(x^3)=(x^3)^2-2=x^6-2
$

and so they are not the same function.

RonL

5. Originally Posted by CaptainBlack
In that case:

$
(f\circ g)(x)=f(g(x))=f(x^2-1)=(x^2-2)^3$
$=x^6 - 6x^4 + 12x^2 - 8
$

$
(g\circ f)(x)=g(f(x))=g(x^3)=(x^3)^2-2=x^6-2
$

and so they are not the same function.

RonL
I find it fascinating that the last person that posted the same question left out the same detail.

-Dan

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