Proof for Rational Exponents

Sep 2015
50
0
Michigan
Hi everyone,

So I've been using rational exponents for a while now, and I began to wonder why do they work in a similar way to integer exponents? Like if you have f(x) = x0.25 if you multiply it my itself 4 times you just get x?

Thanks in advance :)
 
Sep 2012
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434
Washington DC USA
Let's stick to positive numbers to avoid any complexities. And keep it simple to start.

\(\displaystyle \sqrt{9} = 3\), right? Why? Because 3 is the positive number that, when squared, produces 9.

The square root of 16 is 4. Why? Because 4 is the positive number that, when squared, produces 16.

The cube root of 27 is 3. Why? Because 3 is the positive number that, when cubed, produces 27.

The fourth root of 16 is 2. Why? Because 2 is the positive number that, when raised to the fourth power, produces 16.

Now, suppose \(\displaystyle a>0\) and that the fourth root of \(\displaystyle a\) is \(\displaystyle b\)... that is, \(\displaystyle b = \sqrt[4]{a}\), or it could be written as \(\displaystyle b = a^{1/4} = a^{0.25}\).

Then \(\displaystyle b\) is the positive number that, when raised to the fourth power, produces \(\displaystyle a\). So what do you think \(\displaystyle b^4\) is?

So what do you think \(\displaystyle \left(\sqrt[4]{a}\right)^4\) is? What do you think \(\displaystyle \left( a^{1/4} \right)^4\) is? What do you think \(\displaystyle \left( a^{0.25} \right)^4\) is? What do you think \(\displaystyle \left( a^{1/4} \right)\left( a^{1/4} \right)\left( a^{1/4} \right)\left( a^{1/4} \right)\) is?

This behavior is the reason we treat n-th roots as fractional exponents.
 
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Sep 2015
50
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Michigan
Thank you johnsomeone.

What if you have an exponent such as 4/3? How would you rationalize that? How can you prove that first you could take the cubed root and then put the answer to the 4th power, or that you could first take x to the 4th power and then take the cube root of that?
 
Sep 2012
1,061
434
Washington DC USA
It works the same way. \(\displaystyle 8^{4/3} = 16\) because 16 is what you get if you take the number that's the cube root of 8 (i.e. 2), and raise it to the fourth power. Or, equivalently, it's the cube root of 8 to the 4th power (the cube root of \(\displaystyle 64^2 = 4096\)).

Why does it work that way?

Let \(\displaystyle x > 0\) and consider those two ways to compute \(\displaystyle y = x^{4/3}\). Let \(\displaystyle a = x^{1/3}\), so on the one approach, it's \(\displaystyle y = a^4\). But since \(\displaystyle a^3 = x\), get that \(\displaystyle y = a^4 = a^3a = xa\).

Now consider finding \(\displaystyle y = x^{4/3}\) by first raising \(\displaystyle x\) to the 4th power, and then taking the cube root of that. Since \(\displaystyle x^{4} = x^3x\), the cube root of \(\displaystyle x^{4}\) will equal the cube root of \(\displaystyle x^3\) times the cube root of \(\displaystyle x\).

But the cube root of \(\displaystyle x^3\) is just \(\displaystyle x\), and the cube root of \(\displaystyle x\) is \(\displaystyle a\), and so the cube root of \(\displaystyle x^{4} = x^3x\) is \(\displaystyle x\) times \(\displaystyle a\), and so again \(\displaystyle y = xa\).
 
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Deveno

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In general, for integers $a,b > 0$ and $x > 0$:

$x^{\frac{a}{b}} = (x^{\frac{1}{b}})^a = (\sqrt{x})^a$

$= (x^a)^{\frac{1}{b}} = \sqrt{x^a}$

This is because we seek to preserve THIS law of exponents:

$(x^k)^m = x^{km}$

and because $\dfrac{1}{b}\cdot a = a\cdot\dfrac{1}{b} = \dfrac{a}{b}$, because multiplication is COMMUTATIVE.

So it doesn't matter if you take the $b$-th root first, and raise to the $a$-th power, or raise to the $a$-th power first, and then take the $b$-th root.

Things get a little strange when you allow negative exponents, because you have to take reciprocals ($x^{-k} = \dfrac{1}{x^k}$, for $k > 0$).

Things get even stranger if you try to find (rational) exponents of negative real numbers, something best not discussed in polite company.

There's no reason to limit (hah! I kill myself!) exponents to rational values-each exponentiation function $a^x$ can be "extended" to the real numbers by approximating the value of $a^x$ for RATIONAL $x$ NEAR an irrational value.

Thus even things like:

$\sqrt{2}^{\sqrt{2}}$ make sense.
 
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Sep 2015
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Michigan
Thanks to both of you!

Do you know if there is a better way to prove the following : $ √ab = √a √b $ ? When I learned it, I was going by numerical examples, so it was a trial and error proof. Is there anything more rigorous?
 
Feb 2014
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Thanks to both of you!

Do you know if there is a better way to prove the following : $ √ab = √a √b $ ? When I learned it, I was going by numerical examples, so it was a trial and error proof. Is there anything more rigorous?
$\sqrt{x} = y \iff x = y^2.$ That is the definition of the notation.

$\sqrt{a} = c,\ \sqrt{b} = d,\ and\ \sqrt{ab} = e.$

$Assume\ cd \ne e.$

$\therefore (cd)^2 \ne e^2 \implies c^2 * d^2 \ne e^2 \implies a * b \ne ab, which\ is\ false.$

$\therefore cd = e.$

$\therefore \sqrt{a} * \sqrt{b} = \sqrt{ab}.$

Now by weak mathematical induction, you can prove $\sqrt[n]{a} * \sqrt[n]{b} = \sqrt[n]{ab}.$
 
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Deveno

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Thanks to both of you!

Do you know if there is a better way to prove the following : $ √ab = √a √b $ ? When I learned it, I was going by numerical examples, so it was a trial and error proof. Is there anything more rigorous?
You can only prove it for $a,b \in \Bbb R^+_0$ (it turns out it is NOT true for negative numbers, or arbitrary complex numbers, which leads to some humorous "proofs").
 
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Sep 2015
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Michigan
Thanks again!

I guess you can see in the proof given by JeffM that this "trick" only works for positive reals.
 
Feb 2014
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Thanks again!

I guess you can see in the proof given by JeffM that this "trick" only works for positive reals.
Yes. I should have stipulated that the variables a, b, c, d, x, and y are non-negative reals and that n is an integer greater then 1. Sorry about that, and thanks to deveno for making it explicit.