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.

\(\displaystyle \sqrt{9} = 3\), 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

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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?

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 \(\displaystyle y = xa\).

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

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

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.

$\sqrt{x} = y \iff x = y^2.$ That is the definition of the notation.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?

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

$Assume\ cd \ne e.$

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

$\therefore cd = e.$

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

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

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").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?

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.Thanks again!

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