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Let [itex]f: \mathbb{R} \rightarrow \mathbb{R}[/itex] satisfy [itex]f(x+y) = f(x)f(y) \ \forall x,y \in \mathbb{R}[/itex].

Let a = f(1) > 0. Show that [itex]f(r)=a^r \ \forall r \in \mathbb{Q}[/itex].

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This question actually has multiple parts to it. I have already proved the following:

f(0)=1

f(-x) = 1/f(x) for all x in R

f(x)>0 for all x in R.

[itex]f(n)=a^n \ \forall n \in \mathbb{N}[/itex]

[itex]f(n)=a^z \ \forall z \in \mathbb{Z}[/itex]

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Let r=p/q p,q in Z, q not 0.

To solve the problem I have been trying to show that [itex]f(1/q) = a^{1/q}[/itex] which I believe would lead to the solution, since we have the multiplicative property, and if not at least get a better idea of how to get the whole thing. However, this does not seem to be going anywhere, it feels like I need something else. I just had an idea of developing some type of composition property, and maybe that will get it. If you think this might work feel free to ignore this post, if not, any ideas are welcome. Not sure if that is good, basically I am thinking of trying to get some kind of identity involving f(xy) and then maybe something might happen? I suppose [itex]f(xy) = (f(x))^y[/itex].

Thanks!

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# Function with multiplicative property

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