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:

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

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

RonL

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:

$\displaystyle (f\circ g)(x)=f(g(x))=f(x^2-1)=(x^2-2)^3$$\displaystyle =x^6 - 6x^4 + 12x^2 - 8 \displaystyle (g\circ f)(x)=g(f(x))=g(x^3)=(x^3)^2-2=x^6-2 and so they are not the same function. RonL 5. Originally Posted by CaptainBlack In that case: \displaystyle (f\circ g)(x)=f(g(x))=f(x^2-1)=(x^2-2)^3$$\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

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