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.

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- August 8th 2006, 10:40 AMkodirlDiscrete 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.

- August 8th 2006, 11:15 AMThePerfectHackerQuote:

Originally Posted by**kodirl**

Then,

Also,

Thus,

Thus, saying that they are no equal is wrong. - August 9th 2006, 03:37 AMkodirlThanks
Is that the actual answer and how it should be wrtten?

- August 9th 2006, 04:53 AMOReillyQuote:

Originally Posted by**kodirl**

- August 11th 2006, 02:21 AMColetteHey 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???? ;)

- August 11th 2006, 05:26 AMSoroban
Hello, Colette!

Is there a typo in the problem? . . . The statement is*wrong.*

Quote:

Suppose that and are defined by

and for each natural number

Show that

We have: . . . . These are equal!

- August 11th 2006, 06:11 AMColetteoops 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

- August 11th 2006, 07:02 AMSoltras
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 ) - August 11th 2006, 08:19 AMSoroban
Hello, Colette!

Ah, much better . . .

Quote:

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.