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wikipedia says for positive integer exponents its defined by the initial condition

$b^1=b$ and the recurrence relation $b^{n+1} = b^n*b$ and for negative integers its defined as $a^{-n}=\frac{1}{a^n}$.

so is this definition of $a^x$ for all integers, as you are clearly describing the nature of this notation for all integer $x$, but $0$?

also is $a^{-n}=\frac{1}{a^n}$ like an axiom as its just taken to be true or can you prove it. Thanks (hopefully this is clearer than my earlier question)

Rhys Hughes
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James Doherty
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1 Answers1

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Simply $$b^0=1$$ $$b^{n+1}=b(b^n)$$ $$b^{-n}=\frac{1}{b^n}$$ In terms of proving your last statement, consider that $\frac{b^n}{b^n}=1=b^0$ and $b^xb^y=b^{x+y}$. Then $x+y=0 \iff y=-x$ and you can establish that $b^{-n}=\frac{1}{b^n}$

Rhys Hughes
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  • So ... have we defined $0^{-1}$ here? – GEdgar May 19 '18 at 18:48
  • I am simply working off the constraints the OP presented, which are $a \in \Bbb R$, $a\ne 0$ and $x \in \Bbb Z$. – Rhys Hughes May 19 '18 at 18:51
  • so you proved the bottom identity using the first two, but where did they come from? – James Doherty May 19 '18 at 18:54
  • Thanks but i don't want to understand how to prove anything i would like to know in a general sense: a^x has no meaning until you define it so i believe you done that with your first two statements for 0,1,2.... But my question is where did this definition come from? – James Doherty May 19 '18 at 19:20
  • It's not accurate to say that "$a^x$ has no meaning until you define it". In fact, it has exactly the intuitive meaning that it had when you first encountered it in school: "$a^x$ means $a$ multiplied by itself $x$ times". But the mathematician's job is to express this not in an English sentence but instead as a rigorous mathematical definition. – Lee Mosher May 19 '18 at 22:01
  • @JamesDoherty $b^n=\underbrace{b\cdot b\cdot b\cdot ... \cdot b}_{n\text{times}}$. It then follows that $b^n=\frac 1b(b^{n+1})$, and taking $b=0$, we get that $b^0=\frac{b}{b}=1$. – Rhys Hughes May 19 '18 at 22:04
  • Thanks but does that definition you have given for b^n only hold for positive integers? have they not defined b^0 as 1 so that the properties fit? – James Doherty May 19 '18 at 23:36
  • Based off my product, it's not immediately clear that $b^0=1$, so I used the product to show that it is. – Rhys Hughes May 19 '18 at 23:47
  • is the correct way to think about this function not in parts. b^0=1 is just a definition and this is used in the recurrence relation a^(x+1)=a^x*a^1 to define the function for all integers non negative. Then you have to define what the function means for negative indices as well using the third equation? – James Doherty May 19 '18 at 23:53