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Im trying to prove that for every $x\in\mathbb{Q}\ \mbox{the following is true:}\ \ \ \ $ $g\left(s+t\right)=g\left(s\right)\cdot g\left(t\right)\Rightarrow$ $g\left(x\right)=g\left(1\right)^{x}$

I've managed to prove that it's true for every $x\in\mathbb{N}$ really easy by induction:

$g\left(x+1\right)=g\left(x\right)\cdot g\left(1\right)\overset{\text{i.h}}{=}g\left(1\right)^{x}\cdot g\left(1\right)=g\left(1\right)^{x+1}$

but im having trobule to extend it to $\mathbb{Q}$

any advices ?

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    Now, try extend it to $\mathbb{Z}$. – 2017-01-19
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    After that, proof that holds for $1/q,\ q\in\mathbb{Z},q\neq0$. – 2017-01-19
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    Finally, you can proof for all $\mathbb{Q}$. – 2017-01-19
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    Curiosity: If we have the function continuos, then we can proof the property for $\mathbb{R}$! – 2017-01-19

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Hint: $$g(1)=g(3/3)=g(1/3+1/3+1/3)=g(1/3)^3$$

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    In this case, he has to proof first that the property holds for all $1/q$, with $q\in\mathbb{Z},\ q\neq0$. – 2017-01-19