How do I understand fractional exponents like 1/2, 3/2, 1/5, etc?
Like $y$ raised to a power 2 is $y.y$ and when the power is increased or decreased by one, it means we multiply the number by another $y$ or divide it by $y$ respectively.
But how do I think about fractional exponential powers by this same logic?
Thank you.
On
The exponents follow the rule
$$(a^b)^c=a^{bc}$$
meaning that the power of a power is the power to the product of the exponents.
Then assume a rational power $p/q$. We write
$$a^{p/q}=b$$ and raise both members to the $q^{th}$:
$$a^{(p/q)q}=a^p=b^q.$$
So $b$ is the number such that when raised to the $q^{th}$ power yields $a^p$. In other words, it is the $q^{th}$ root of $a^p$.
If you didn't learn the roots yet, understand that this is the "inverse" operation of exponentiation. For instance,
$$3^5=243\iff243^{1/5}=\sqrt[5]{243}=3.$$
Very often, roots are irrational numbers,
$$1.24573093\cdots^5=3\iff 3^{1/5}=\sqrt[5]3=1.24573093\cdots$$
On
"how do I think about fractional exponential powers by this same logic?"
Carefully.
You have to expand the definition.
We notice that $b^n*b^m = \underbrace{b*...*b}_{n}\underbrace{b*...*b}_{m}=\underbrace{b*...*b}= b^{n+m}$ for positive integers $n$ and $m$ and we take $b^{n+m} = b^nb^m$ as a rule.
Then we think. Well what that means $b^0*b^m = b^{0+m} = b^m$. That would mean $b^0 =1$, doesn't it? Well, in one sense it does not make sense that we can't mulitple $b$ times itself zero times. That's meaningless.
But that is no longer what the definition of $b^n$ means. It means that for $n\in \mathbb Z$ and $n \ge 2$ but... for $n\in \mathbb Z$ and $n \le 1$ we are going to have it mean.... whatever it has to be in order for the rule $b^{n+m} = b^n*b^m$ to work.
That means: $b^1 = b$ and $b^0 = 0$ and for a negative integer $-n < 0$ then $b^{-n} = \frac 1{b^n}$ (becuase we must have $b^{-n}*b^n = b^{-n+n} = b^0 = 1$.
Okay, but we also have the rule $(b^n)^m = \underbrace{b^n*...*b^n}_{m} = b^{\underbrace{n+....+n}_m} = b^{nm}$.
So what does that make $b^{\frac 1m}$ have to be?
Well, $(b^{\frac 1m})^m = b^{\frac 1m*m} = b^1 = b$. So that means $b^{\frac 1m}$ has to be $\sqrt[m]{b}$, right?
And that's that. $b^{\frac mn} = (\sqrt[n]{b})^m$.
That's it.
....
Except we have to be careful. $x^2 = 4$ has two possible answer $-2,2$ so which will it be. ANd $x^2 = -4$ has none.
So we add a caveat rule: we only feel safe talking about $b^{\frac mn}$ if $b \ge 0$ if $n$ is even and in that case the value is positive. We can talk about $(-4)^{\frac mn}$ if $n$ is odd (in which case the value is negative) but for most of the time, we should avoid it. Many mathematicians will simply say $b^q;q\in \mathbb Q$ is only defined if $b\ge 0$.
And for $x \in \mathbb R$, we say $b^x$ is only defined fro $b\ge 0$. Unless we are doing complex analysis where we accept the square roots of negative numbers. In that case, things get very different. But it's still a matter of defining things by what the must be in order for the rules to work.
On
Let us start with $1/2$
If I want to find $16^{1/2}$, I expect to find a number whose square is $16.$
That is if $$x= 16^{1/2}$$ we like to have $x^2=16$
Well we have two such real numbers, namely $4$ and $-4$
We denote them by $ \sqrt {16} =4$ and $- \sqrt {16} =-4$
Sometimes you do not get a real solution at all, for example $(-16)^{1/2}$ is not a real number, because when you square a real number the answer is always non-negative.
Sometimes you get only one solution for example $27^{1/3}=3$ because 3^3=27 and no other real number satisfies $x^3=27.$
Once you find $a^{1/n}$ then $a^{m/n}$ is found by $(a^{1/n})^m$
For example $$27^{2/3} = (27^{1/3})^2 = 3^2 =9 $$
You may practice with some fractional powers and check your answers with a calculator to build confidence.
Notice that, for example, if $x^{\frac{1}{2}}=a$, then by squaring both sides we get $x=(x^{\frac{1}{2}})^2=a^2$.
In other words $\sqrt{x}=a$.
Following the same logic you can get that for every $n \in \mathbb{N}$, $x^{\frac{1}{n}} = \sqrt[n]{x}$
After you accepted that, by exponentiation rules, you know that for every $p,q \in \mathbb{N}$: $$ x^{\frac{p}{q}} = x^{p \cdot \frac{1}{q}} = (x^p)^{\frac{1}{q}} = \sqrt[q]{x^p} $$
So in general, raising to a rational number $\frac{p}{q}$ means to take the $p$-th power and the $q$-th root.
And, of course, it also generalize to negative rational numbers.