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I was wondering if someone can tell me what the logic behind converting fractional exponents to radicals is? For example, the exponent 1/2 is a square root, 1/3 is a cube root, and 2/3 is the cube root of x squared and so on.

Can someone explain this in a very easy way for me? Thanks :P

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  • $\begingroup$ $\left(x^{\frac{a}{b}}\right)^b=x^a$ so $x^{\frac{a}{b}} = \sqrt[b]{x^a}$ if $x$ real and positive $\endgroup$ – Henry Dec 1 '14 at 7:07
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Intuitive explanation

Let's start with a simpler example: numbers.

All the positive integers can be written as long sums of "1's": $2 = 1+1$, $3 = 1+1+1$, $10 = 1+1+1+1+1+1+1+1+1+1$. Each of these numbers counts a certain number of 1's, or units.

Now for fractions. When we first start learning about fractions, we learn to think about them using objects from everyday life. Here's I'll use pizzas and slices. Let's stick to "slices" of size $1/n$. If you have $1/2$ of a pizza, you could also say that if you had two slices of size $1/2$, then you would be able to form the whole pizza: $1/2 + 1/2 = 1$. The same works for $1/5$ of a pizza: you need 5 of those slices in order to form a whole pizza, and we see this since $1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 1$. So if you have slices of size $1/n$, then you need $n$ slices in order to make the whole pizza.

What does this have to do with exponentiation? As we saw above, we can say that $x^2 = x \cdot x$, $x^3 = x \cdot x \cdot x$, et cetera. So $x^n$ counts how many $x$'s need to be multiplied in order to get $x^n$--$n$. In a sense, it's a lot like the repeated addition of 1's to form larger numbers, except you're multiplying $x$'s instead of adding 1's.

Similarly, let's say we want to divide $x$ up with respect to multiplication. In other words, we want to find "slices" of $x$ that can be multiplied by themselves a certain number of times that then return $x$. This is where fractional exponents come in. One half of an $x$ can be denoted by $x^{1/2}$, one third $x^{1/3}$, one sixth $x^{1/6}$. Then, just like with normal fractions, we can write: $$x^{1/2} \cdot x^{1/2} = x^{1/2 + 1/2} = x^1 = x$$ This is true by the laws of exponents. We can do the same thing for $x^{1/3}$: multiplying it by itself three times, we get $$x^{1/3} \cdot x^{1/3}\cdot x^{1/3} = x^{1/3 + 1/3 + 1/3}= x^1 = x$$ So in general, $x^{1/n}$ is the number that you have to multiply $n$ times by itself in order to get back $x$.

Remembering $\sqrt[n]{x}$, this sounds awfully familiar. If we take the square root of $x$, we know that $\sqrt{x} \cdot \sqrt{x} = x$. With the cube root, we have to multiply it by itself three times in order to get $x$: $\sqrt[3]{x} \cdot \sqrt[3]{x} \cdot \sqrt[3]{x} = x$. In general, we have to multiply $\sqrt[n]{x}$ by itself $n$ times in order to get $x$ back.

But wait a second...this is exactly the same principle we derived for $x^{1/n}$. So (intuitively at least), we can say that $x^{1/n} = \sqrt[n]{x}$.

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$$\sqrt{x^2}=x$$ $$(x^{2})^\frac{1}{2}=x^{(2)(\frac{1}{2})}=x$$ So you can see how raising something to the $\frac{1}{n}$ power is the same thing as that root/radical.

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$$x^{a/b} = x^{1/b \,\cdot\, a} = ({x^{1/b})}^a = \left(\sqrt[b]{x}\right)^a$$

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  • $\begingroup$ As a piece of diet advice, the bth root is actually edible. Most mathematicians don't know this. Thought you might want to know. $\endgroup$ – Mats Granvik Feb 25 '17 at 12:12

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