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Old 03-03-2009, 07:02 AM
The Readeemer The Readeemer is offline
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Post Mathematical Induction Question

Hi guys! Anyone know any basic uni maths? This is related to the Principle of Mathematical Induction, and the question I'm having some trouble with is:

If a function f(xy) = f(x) + f(y), where x, y > 0 (zero) then prove by Mathematical Induction that:

f(x1 x2 x3 ... xn) = f(x1) + f(x2) + ... + f(xn) for n>0 (zero), n is an integer.

So far, I've proved that it (obviously) holds for n=1, and assuming it holds for n=k, where k fulfils the conditions for n, but when I assume for k+1, I don't know where to go next for my proof.

Does anyone have any advice on how to prove this? Much appreciated. Thanx =D

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Old 03-03-2009, 10:09 AM
Foo Foo is a male Canada Foo is offline
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Re: Mathematical Induction Question

Just take
>S := x1 x2 x3 ... xk
Note that under your assumption (f holds for k),
>f(S) = f(x1) + f(x2) ... + f(xk)

Consider the series
>f(x1) + f(x2) ... + f(xk) + f(x[k+1])
We see that this is equal to
>f(S) + f(x[k+1])
Which by the definition of f is
>f(S x[k+1])
Substitute in the definition of S to get
>f(x1 x2 x3 ... xk x[k+1])
So
>f(x1 x2 x3 ... xk x[k+1]) =f(x1) + f(x2) ... + f(xk) + f(x[k+1])

Thus, by induction (omitting the proof that f is defined for n=1)
f(x1 x2 x3 ... xn) = f(x1) + f(x2) + ... + f(xn) for n>0 (zero), n is an integer.
Last Edited by Foo; 03-03-2009 at 12:38 PM. Reason: Reply With Quote
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Old 03-05-2009, 03:38 AM
The Readeemer The Readeemer is offline
Goron
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Re: Mathematical Induction Question

Awesome! Thanx for the help and excuse my dumbness! =D

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