What is the meaning of $x$ raised to any non-positive value? We know that $x^{-a} = \dfrac{1}{x^a}$ and $x^0 = 1$, but where does that come from? What is the proof? Why is this true? What about $x$ raised to a fraction, like say $\frac{1}{3}$? How do you multiply $x$ $\frac{1}{3}$ times?
Asked
Active
Viewed 148 times
2
-
It is its inverse: $$0\neq x\in\Bbb R\implies x^{-1}:=\frac1x$$ – DonAntonio Aug 07 '16 at 14:24
-
I have edited the question. Everybody knows that. – jimpix Aug 07 '16 at 14:24
-
1For beginners: there is no proof. It is a definition. For more advanced students: by means of other things, like limits...but relying, again, on certain definitions. – DonAntonio Aug 07 '16 at 14:26
-
Not a beginner here. – jimpix Aug 07 '16 at 14:27
-
Then remember limits, logarithmic functions and etc. – DonAntonio Aug 07 '16 at 14:27
-
1To go from say $x^3$ to $x^2$, that is, to lose $1$ in the exponent, we have to divide by $x$. Therefore to go from $x^0$ to $x^{-1}$ we have to divide $x^0$ by $x$ which gives $1/x$. – Zubzub Aug 07 '16 at 14:28
-
How does (-1)x=-x make sense? – Jacob Wakem Aug 07 '16 at 14:44
-
A related question. – Lucian Aug 07 '16 at 22:15
-
Because i told you toooooo. – Jacob Wakem Oct 27 '16 at 04:29
2 Answers
9
From that definition, it doesn't make sense. It's what we mathematicians call a generalisation. The original definition of powers only allow natural numbers to be exponents.
However, one can ask "Is it possible to come up with a definition of $x^{-a}$ that plays well along with the old definition?" By which we mean that $x^{b+c}=x^bx^c$ and $(x^b)^c=x^{bc}$ still holds. And it turns out that there is exactly one definition that works.
In the same way you get $x^{1/n}=\sqrt[n]x$. Not because the old definition says so, but because it's the only thing it can be if we want it to work.
Arthur
- 199,419
8
We know that $x^{a+b}=x^ax^b$ since an $(a+b)$ amount of $x$'s is the same as an $a$ amount of $x$'s times a $b$ amount of $x$'s.
From this, we have
$$x^0=x^{a-a}=x^ax^{-a}$$
Thus, we have
$$x^{-a}=\frac{x^0}{x^a}$$
and once you show $x^0=1$, your done.
As for fractions, show that
$$x^{\frac ab}=y$$
For some value of $y$ which we are trying to solve for.
We see this is the same as
$$x^{\frac ab}=\underbrace{x\times x\times x\times\dots x}_{a/b?}=y$$
However, we see that by multiplying both sides on itself $b$ times, we have
$$\underbrace{x\times x\times x\times\dots x}_{a/b}=y\implies\underbrace{x\times x\times x\times\dots x}_{a}=y^b$$
and so,
$$\underbrace{x\times x\times x\times\dots x}_{a/b}=\sqrt[b]{\underbrace{x\times x\times x\times\dots x}_{a}}$$
$$x^{\frac ab}=\sqrt[b]{x^a}$$
Simply Beautiful Art
- 74,685