# Homework Help: Looking for Intuitive Reasoning Behind Rational Exponents

1. Oct 10, 2017

### DS2C

Solving these seem fairly simple so far. But I don't know why this works. I asked my instructor and she couldn't give me an intuitive reason as to why.

1. The problem statement, all variables and given/known data
$\sqrt[3]y\cdot\sqrt[5]y^2$

2. Relevant equations
N/A

3. The attempt at a solution

$$\sqrt[3]y\cdot\sqrt[5]y^2$$
$$y^\frac 1 3\cdot y^\frac 2 5$$
$$y^{11/15}$$
SOLUTION = $$\sqrt[15]y^{11}$$

Why does this work?

2. Oct 10, 2017

### PetSounds

You may find this helpful. Scroll to 22-3 if you want information pertinent to your question, but the whole piece is worh reading.

3. Oct 10, 2017

### StoneTemplePython

It kind of depends on what you're looking for by "why" and "intuitive". One simple way to think about exponents that are that are positive integers-- is just the number of times you'd type them in to multiply them on your calculator.

So what would be

$$\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)^{15}= \prod_{k=1}^{15} \big(\sqrt[3]y\cdot\sqrt[5]y^2\big) = \big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)\big(\sqrt[3]y\cdot\sqrt[5]y^2\big)$$

Now use that fact that all these terms are associative and commute -- i.e. get rid of parenthesis and re-arrange terms and simplify this. What do you get?

4. Oct 10, 2017

### DS2C

I can see the equal factor part, but for example, why is the denominator in the exponent the index of the radical and the numerator is the radicands power?
Also, what's that pi symbol looking thing you have there?

5. Oct 10, 2017

### StoneTemplePython

$\prod_{k=1}^{15} \big(\sqrt[3]y\cdot\sqrt[5]y^2\big)$ literally means the multiplication that I show to the right. Are you going to try to simplify the 15 terms and show what you got?

6. Oct 10, 2017

### FactChecker

Given.
Simply changing the notation.

7. Oct 10, 2017

### DS2C

I missed this post, thank you. I'll give that a read!
I'm not sure what you mean. Do you mean to simplify it into $\sqrt[3]y^{15}\cdot\left(\sqrt[5]y^2\right)^{15}$??

8. Oct 10, 2017

### DS2C

One of my confusions lies with the relation between the index of a radical and the power of the radicand to the numerator and denominator of it's exponential form. For example,
$\sqrt[n]a^{m}$ is equal to $a^\frac m n$

9. Oct 10, 2017

### StoneTemplePython

sort of. note that $(a)^{15} = (a^3)^5$ for the same reason that both involve multiplying by a, 15 times.

So you get

$\sqrt[3]y^{15}\cdot\left(\sqrt[5]y^2\right)^{15} = \big(\sqrt[3]y^{3}\big)^5\cdot\left(\sqrt[5]y^2\right)^{15} = \big(\sqrt[3]y^{3}\big)^5\cdot\left(\sqrt[5]y^{30}\right) = \big(\sqrt[3]y^{3}\big)^5\cdot\left(\sqrt[5]y^5 \right)^6$

now look at that last expression. You have a cube root that you are "undoing" cubing process, and you have a fifth root that you are 'undoing' by raising to the 5th power. Hence this gives:

$\big(\sqrt[3]y^{3}\big)^5\cdot\left(\sqrt[5]y^5 \right)^6 = \big(y\big)^5\cdot\left(y \right)^6 = y^5y^6 = y^{11}$

so now we have $y^{11}$ is equal to your original expression raised to the 15th. But your end expression raised to the 15th is
$\big(\sqrt[15]y^{11}\big)^{15} = \big(\sqrt[15]y^{15}\big)^{11} = \big(y\big)^{11} = y^{11}$.

There's some lingering technical nits about whether or not a mapping is invertible that I won't go into, but you should be able to see that both expressions get you to the same place when you raise to the 15th. And the whole idea behind $\sqrt[15]$ is that you had some number you raised to the 15th power, and now you are trying to recover the number that you originally had. Hopefully this is an intuitive way to look at this.

10. Oct 10, 2017

### FactChecker

I don't know if I can help you with intuition on it. I think of them as just different representations of the same operations: $\sqrt[n]a^{m}$ = (am)1/n = am⋅1/n = am/n

11. Oct 11, 2017

### Math_QED

You look after intuition for why we write: $a^{m/n}$?

Then you must first know how this notation is defined and how it relates to roots.

So, let me start with saying that $a^{1/n}$ is defined as the unique number $x$ such that $x^n = a$, i.e. the number $\sqrt[n]{a}$. Then, $(a^{1/n})^m = \sqrt[n]{a}^m = a^{1/n} \dots a^{1/n}$. This is a useful definition as it can be shown it is well defined and it extends the power laws we know from the integers.

12. Oct 12, 2017

### DS2C

Thanks for the responses all. Taking them in as I go. Still working on fully understanding your expalantions. For some reason this concept is coming weirdly to me. Theyre easy to solve, but working on what you're saying to understand the whys.

13. Oct 12, 2017

### epenguin

Think of it this way. If we decide to denote n $x$'s multiplied together (or $x$ multiplie by itself (n-1) times) then we shall have the two laws $x^{a}\times x^{b}=x^{\left( a+b\right) }$ and $\left( x^{a}\right) ^{b}=x^{ab}$.

From this we can see that multiplying certain numbers together, can be done using the operation of addition which is easier. And division into subtraction. (This leads to finding a way to change multiplications and divisions of all numbers into easier additions and subtractions which you may or may not yet have done.)

Now for your question start with $x^{1/n}$ . What does "the n-th root mean? It means "the quantity that multiplied by itself (n-1) times gives you $x$. Now use the above laws to work out what raising $x^{1/n}$ to the n-th power according to the definition gives you.

Last edited: Oct 13, 2017
14. Oct 14, 2017

### scottdave

That Feynman lecture Is cool. Im going to read more.