1

How can I prove the following:

Prop.: let be $m,n,p \in \Bbb{N}$ and $a \in \Bbb{R}$ then $$a^m\cdot a^n \cdot a^p=a^{m+n+p}$$ ??? I thinked by induction and I must prove:

1) $a^0\cdot a^0 \cdot a^0=a^{0+0+0}$ is true

2) $a^m\cdot a^n \cdot a^p=a^{m+n+p} \Rightarrow a^{m+1}\cdot a^n \cdot a^p=a^{m+1n+p}$ is true

3) $a^m\cdot a^n \cdot a^p=a^{m+n+p} \Rightarrow a^{m}\cdot a^{n+1} \cdot a^p=a^{m+n+1+p}$ is true

4) $a^m\cdot a^n \cdot a^p=a^{m+n+p} \Rightarrow a^{m}\cdot a^{n} \cdot a^{p+1}=a^{m+n+p+1}$ is true

If 1),2),3),4) are true then Prop. is true

Is it correct?

mle
  • 2,287
  • Can you use the property $a^{mb}=a^m\cdot a^b$? Because then you can easily prove that property if you set $b=n+p$. – Hakim Jul 14 '14 at 17:51
  • @Hakin, No I can't... It is a particular exercise! – mle Jul 14 '14 at 17:58
  • What about logarithms? – Hakim Jul 14 '14 at 17:59
  • You might encode the three integers into a single one, and perform induction over the encoded number. The induction step might be tricky, though. – mvw Jul 14 '14 at 18:05
  • @Hakim, I must to use induction... – mle Jul 14 '14 at 18:06
  • 1
    Prove Hakim's property first by induction (note that $m, n\in \mathbb{N}$ which helps). Then your answer follows by using it twice. – Kevin Jul 14 '14 at 18:07
  • @Kevin, Haki,, ok and it is true (thanks for hint :) )... but for curiosity, Can I use the http://math.stackexchange.com/questions/856602/scheme-for-n-dimensional-induction or not? – mle Jul 14 '14 at 18:14
  • 2
    You can, but it'll be really unpleasant. – Danny Smith Jul 14 '14 at 18:26

0 Answers0