# Please help....

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• Aug 17th 2006, 02:33 AM
dcapdogg
Please help....
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. :confused:
• Aug 17th 2006, 02:40 AM
CaptainBlack
Quote:

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. :confused:

It can't be done, because they are equal:

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

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

RonL
• Aug 17th 2006, 02:44 AM
dcapdogg
oops i left a bit out
g(n)=n^2-2
• Aug 17th 2006, 03:23 AM
CaptainBlack
Quote:

Originally Posted by dcapdogg
g(n)=n^2-2

In that case:

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

\$\displaystyle
(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
• Aug 17th 2006, 04:04 AM
topsquark
Quote:

Originally Posted by CaptainBlack
In that case:

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

\$\displaystyle
(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
• Aug 17th 2006, 04:38 AM
ThePerfectHacker
User banned (7 days).
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