Rooting is not the inverse of exponentiating!
I read someone spreading this lie again the other day, and it annoyed me. Consider addition whose inverse operation is subtraction. The additive inverse is typically expressed like so: a + (-a) = 0, and we make this shorthand to a - a = 0. All subtraction is, is inversed addition.Consider multiplication, which can be thought of (but isn't necessarily) iterated addition. The multiplicative inverse is expressed like: a * 1/a = 1. In other words, division is multiplication's inverse. (Why is multiplication not necessarily iterated addition? Well, if the numbers are discrete, you're okay. But explain how you can take 2 * pi and add two with itself pi times. 2 * 1.5 is adding two with itself one and a half times (which already sounds strange), in other words take one two, then add it with half of a two. The iterated view isn't necessarily wrong, it's just not helpful after a point.)
Now consider exponentiation, which is often thought of (but isn't necessarily) iterated multiplication. 3^2 is 3 * 3 which is 9, 3^3 is 3 * 3 * 3 which is 27. What's the inverse you ask? Well, for 3^2, take the square root! For 3^3, take the cube root! For x^2 = y, take the square root of (known) x to find (unknown) y! (So long as x isn't negative, 'cause we're afraid of complex numbers.) This is all true so far. For 3^x = y, where x is known, the inverse operation to solve for y is indeed the "xth root", which will always be an actual known number. How about when y is known and x isn't? For 3^x = 27, take the... literal xth root... Wait...
Uh... There's the problem. What does it mean to take the xth root of something when you don't know x? Let's look at what we're actually doing with square rooting and cube rooting: unfortunately it really hurts the view of exponentiation as iterated multiplication.
[math]
3 = \sqrt{9} = \sqrt{3^2} = (3^2)^{\frac{1}{2}} = 3^1
[/math]
It's a known reduction that (a^b)^c = a^(b*c), thus 2 * 1/2 = 1.
What's x^(1/2)? Why, it's the square root of x. x^(1/3) is the cubed root of x. This is why in programming you don't really need a square root function or a cubed root function, just a pow function that can take floats.
Try and phrase it using iterated multiplication. 9^(1/2) means... 9 multiplied with itself one half times? Huh? Half of 9 is 4.5, how can I multiply 9...
You see, the iterated view fails here. (If you can make it work, let me know in the comments.) Also remember that 9^0 = 1.
The inverse of addition is subtraction, and the inverse of multiplication is division, but subtracting is just a form of addition and dividing is just a form of multiplication. Taking a root is just a form of exponentiation (in fractional form), so why shouldn't it be the inverse?
If I give you a problem: 2 * x = 5, you can solve for x. You use the inverse of multiplication and multiply both sides by 1/2 (or divide by 2), and we see x = 5/2. I also give you the problem x^2 = 9, and you what? You take the square root of both sides: you raise each side by a factor of 1/2. (x^2)^(1/2) = x (assume x is known to be positive), and 9^(1/2) = 3. x = 3. (How do you know 3 is the square root of 9? Because you already know 3^2 = 9, thus (3^2)^(1/2) = 3^1, or your calculator told you. How does your calculator tell you? Magic.) You think all's dandy, that you have an inverse for exponentiation. But the inverse operation you used was not on exponentiation itself, it was on squaring. So now I give you the problem I mentioned above: 3^x = 27.
You know the answer is 3, but how to get there? You can try blindly plugging in values, but that's not very helpful. Your calculator doesn't have an "xth root" button on it, and you don't see how (assuming we somehow know x to be non-zero) 3 = 27^(1/x) helps you in finding the value of x. You're stuck. I thought you knew the inverse of exponentiation? No, you don't.
[math]
\log_b(a) = c\ if\ b^c = a
[/math]
The logarithm is the inverse of exponentiation. What the above is read as, is "The log with base b that produces value a is c, if b^c = a."
[math]
\log_3(9) = 2
[/math]
The log with base 3 that produces 9 is 2. In other words, if you have base 3 and raise it to the second power, you get 9. So go back to the problem of 3^x = 27. Take the log (base 3) of both sides. The left side magically becomes x (I'm not going to go into the depths of logarithms), and the right side becomes log base 3 of 27. If you noticed, your calculator has a log button! If you plug it in correctly*, you will get the answer of 3. The log base 3 of 27 is, in actuality, asking what power when applied to 3 will produce 27? We just change the question to: log base 3 of 27? How to answer that question (which is what was asked before, just in a different way) is left as an exercise, but for now use your magic calculator.
Logs are awesome, and are insanely useful. Now can we please stop spreading the myth that rooting is the inverse of exponentiation? That's like saying subtracting three is the inverse of addition.
*To input into your calculator directly, you'll probably have to do log(27)/log(3). This is because the log() function of the calculator is likely in base 10, not in base 3, but log(x)/log(y) where the bases of the log() functions are the same is the same thing as log(x) with base y.
Posted on 2010-01-12 by Jach
Tags: math
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