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#1 2007-10-30 22:29:11

xXxEmZyxXx
Member
Registered: 2007-10-15
Posts: 12

functions

g: B->A  such that g(f(a))=a   for all aEA
h:B->A such that f(h(b)=b    for all bEB

show that g(b)=h(b)    for all bEB

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#2 2007-10-31 01:43:31

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: functions

Set f(a) = b.  Substitute to get g(b) = g(f(a)), which we know from the definition of g is equal to a.  Thus, g(b) = a.

Do the same for h to get h(f(a)).  We know from the definition of h that f(h(f(a))) = f(a), which tells us that h(f(a)) = a.  Substitute back to get h(b) = a = g(b).


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#3 2007-10-31 01:52:30

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: functions

No more copying our solutions for your homework without understanding what you’re supposed to learn. So tell us honestly first: do you understand the solutions we’ve been giving you? neutral

EDIT: Grr at TheDude for posting before me. Also, I see that TheDude’s proof is almost (but not quite) complete: TheDude is assuming that f is both surjective and injective. This is in fact true but it still has to be stated why.

So, @ xXxEmZyxXx: If you copy TheDude’s solution blindly, you will NOT get full marks for your homework, haha! Now concentrate on understanding rather than copying.

Last edited by JaneFairfax (2007-10-31 02:22:20)

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#4 2007-10-31 02:08:17

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: functions

A much better method is to let c = h(b), where bB. But I will not tell xXxEmZyxXx how to proceed. xXxEmZyxXx must by now have learnt how to do some things by himself/herself. hmm

Well, all right, here’s a hint to help xXxEmZyxXx: f(c) = b. (Why?) ∴ g(b) = g(f(c)) = …?

Last edited by JaneFairfax (2007-10-31 02:20:37)

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#5 2007-10-31 03:44:11

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: functions

Heh, your method is a lot easier.


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