Why does $a^0 = 1$ and $a^{-p} = {\frac{1}{a^p}}$ if $(a \not = 0)$ and $p \not= 0$?
How can we prove these formulas?
Asked
Active
Viewed 59 times
-1
-
Are you looking for formal proofs or intuitions about why these statements are true? – scoopfaze Jul 02 '19 at 02:39
-
Based on your previous question, is $p$ supposed to be in $\Bbb{R} \setminus { 0}$? Also, $a^{-p} = \frac{1}{a^p}$, not $a^{\frac{1}{p}}$. – Theo Bendit Jul 02 '19 at 02:49
-
I think of $a^0$ like an empty product, which is $1$ – J. W. Tanner Jul 02 '19 at 03:03
2 Answers
2
This is just an intuition test of why the properties should be defined like this:
The fundamental property of the exponential functions is the following: $b^x\cdot b^y=b^{x+y}$.
The other properties are derived in such a way that the mentioned property is satisfied. For example: since we want the equation to be fulfilled $$b^0\cdot b^x=b^{0+x}=b^x$$ we must define $b^0:=1$. Since we want the following equation to be fulfilled $$b^{-x}\cdot b^x=b^{-x+x}=b^0=1$$ we must define $b^{-x}:=1/b^x$. Since we want the following equation to be fulfilled $$\underbrace{b^{1/n}\cdot b^{1/n}\cdots b^{1/n}}_{n\, \textrm{factors}}=b^1=b$$ we must define $b^{1/n}:=\sqrt[n]{b}$. Similarly, we must to define $b^{m/n}:=(\sqrt[n]{b})^m.$
azif00
- 20,792
1
$$a^{0} = a^{n-n} = \frac{a^{n}}{a^{n}} = 1$$
Also,
Do you mean
$$ a^{-n} = \frac{1}{a^{n}}? $$
DDS
- 3,199
- 1
- 8
- 31