Exponentiation

Exponentiation is repeated multiplication, just as multiplication is repeated addition. In this course, we only deal with integer exponents. Fractional exponents are possible, but they are trickier. In Intermediate Algebra you'll learn to use any rational exponents, but people usually don't talk about irrational exponents without using calculus (although you don't really need calculus even for them).

Multiplication

Multiplication by a natural number is repeated addition, as you surely already know. I want to review first how that works, so that it will be easier to understand exponentiation. For example, 3 · (2/7) means (2/7) + (2/7) + (2/7), which works out to 6/7. In other words, 3 · (2/7) means 2/7 added to itself 3 times. (This is where the phrasing ‘3 times’ for ‘3 ·’ comes from!) There's really nothing special about 2/7 here, so the general rule for multiplication by 3 is that 3 · a means a + a + a. And this works for natural numbers besides 3, of course, as in the following table:
This expression …means …
4 · a, a + a + a + a;
3 · a, a + a + a;
2 · a,a + a;
1 · a,a.
Of course, this table goes as far as you like upwards, so it describes multiplication by any natural number.

Can it go any further downwards? That is, can we extend it to describe multiplication by any whole number (including zero) or even any integer (including the negative integers)? The pattern that I've written so far really doesn't continue; at each stage, we remove one more addition of a, one more ‘+ a’ disappears, but the last step, 1 · a, doesn't really have any addition, and there is no more plus sign to remove. After all, a added to itself one time isn't really a added to anything at all!

However, I can tweak this table slightly to allow the pattern to continue one more step. We know that 0 · a is supposed to be 0, and you might think that this is what you get when you remove everything from the last line of the table. This really only works, however, if you think that a calculation begins with 0. In the case of repeated addition, that really is true, so let's make it explicit: to multiply a by (say) 3, start with 0, then add a to that 3 times: 0 + a + a + a instead of simply a + a + a. Then the table looks like this:

This expression …means …
4 · a, 0 + a + a + a + a;
3 · a, 0 + a + a + a;
2 · a, 0 + a + a;
1 · a,0 + a;
0 · a,0.
Now the pattern continues perfectly down to 0 · a, removing one more addition of a at each step.

Can we continue this to multiplication by negative integers? Not exactly, but we can start a similar, reverse pattern by changing addition to the inverse operation of subtraction. The final table looks like this:

This expression … means …or … for short
4 · a, 0 + a + a + a + a, a + a + a + a;
3 · a, 0 + a + a + a, a + a + a;
2 · a, 0 + a + a, a + a;
1 · a, 0 + a,a;
0 · a,0,0;
(−1) · a, 0 − a,a;
(−2) · a, 0 − a − a, a − a;
(−3) · a, 0 − a − a − a, a − a − a;
(−4) · a, 0 − a − a − a − a, a − a − a − a;
This table goes on forever in both directions, showing how to multiply by any integer.

I've put a new column to the right, giving the simplified expression that you would usually use. (Why write in all those zeroes if you don't need them?) But the middle column is the one that really shows the pattern: start with 0, then add or subtract as necessary.

Definition

OK, if exponentiation is repeated multiplication, then we should just be able to almost copy everything above, or in fact only the last table. But some of the changes we have to make to copy correctly can be tricky, so I'm going to go through things a little slowly.

As the first example, consider (2/7)3, which means (2/7) · (2/7) · (2/7), which works out to 8/343. Again, there's really nothing special about 2/7; in general, the rule is that a3 means a · a · a. Here, a is called the base and 3 is the exponent; the result a3 is the power, or in full ‘the 3rd power of a’. You can also read this as ‘a raised to the 3rd power’, or simply ‘a to the 3rd’, or even ‘a cubed’ (a nickname that comes from geometric applications).

Here is the table (analogous to the first table above) for raising numbers to a natural exponent:

This expression …means …
a4, a · a · a · a;
a3, a · a · a;
a2, a · a;
a1,a.

What is a0 then? In other words, how do we extend this table one more line to make the analogue of the second table above? You might think that a0 is 0, just like 0 · a is 0, on the grounds that you start calculating at 0. But in fact that does not work for repeated multiplication; when you multiply, you really start calculating at 1!

To convince yourself of this, just imagine that you replace 1 with 0 in the table below. Then all of the results would simply be 0, which is not correct; only if you use 1 instead can you get the same answers as before.

This expression …means …
a4, 1 · a · a · a · a;
a3, 1 · a · a · a;
a2, 1 · a · a;
a1,1 · a;
a0,1.
In other words, 1 is the analogue for multiplication of 0 for addition. What is the analogue for multiplication of subtraction? It is division, although there is an extra complication: you cannot divide by 0. Therefore, you cannot raise 0 to a negative exponent.

Here is the final table, analogous to the final table in the previous section:

This expression … means …or … for short;
a4, 1 · a · a · a · a, a · a · a · a;
a3, 1 · a · a · a, a · a · a;
a2, 1 · a · a, a · a;
a1, 1 · a,a;
a0,1,1;
a−1, 1/a,1/a;
a−2,1/a/a, 1/(a · a);
a−3,1/a/a/a, 1/(a · a · a);
a−4, 1/a/a/a/a, 1/(a · a · a · a);
Again, the middle column shows the general pattern: start with 1, then multiply or divide as necessary.

To form the usual (short) expressions in the right column, I can use rules for dividing fractions. For example, 1/a/a, or 1/a divided by a, means 1/a multiplied by the reciprocal of a, or (1/a) · (1/a), which works out to (1 · 1)/(a · a), or 1/(a · a).

Tricky bits

Notice that 0−1, 0−2, and so on are all undefined. This is because their definitions involve division by 0, which is undefined; and this is because the reciprocal of 0 is undefined; and this is ultimately because you can never get 1 by multipying any real number by 0.

Sometimes people say that 00 is also undefined, but this isn't really correct. The definition works perfectly well to produce 00 = 1. Of course, you can always make an exception in a definition if you want something to be undefined, and in fact mathematicians argued about this for almost 200 years. But in the last 50 years or so, people have begun to actually use 00, and they find that 00 has to be defined as 1 for a bunch of other things (mostly in the mathematical field of combinatorics) to work.

Incidentally, this is how such arguments usually get settled: peeople go back and forth for a while, getting nowhere as long as it's purely theoretical; but once somebody finds a way to use the ideas, then it's obvious what the right answer has to be. Sometimes it takes the textbooks a while to catch up, however; the textbook for this course still says that 00 is undefined, but in doing so it is behind the times!

Notice that the sign (positive or negative) of the exponenet has nothing to do with the sign of the power. For example, 22 = 2 · 2 = 4, while 2−2 = 1/(2 · 2) = 1/4, and these are both positive. This is because a negative exponent simply tells you to divide instead of multiplying, which has nothing to do with whether your result is negative.

On the contrary, if the base is negative, then the sign of the power depends on the parity (even or odd) of the exponent. For example, (−2)2 = (−2) · (−2) = 4, while (−2)3 = (−2) · (−2) · (−2) = (−2) · 4 = −8. This is because multiplying (or dividing) by a negative number always switches the sign, so as you raise a negative number to a power, you keep going back and forth between negative and positive until the sign of the final answer depends on whether the number of times you flip is even or odd.

When you raise a fraction to a power, you can do the numerator and denominator separately. For example, (2/3)2 = (2/3) · (2/3) = (2 · 2)/(3 · 3) = 4/9, and you get the same result if you do (22)/(32). Similarly, my original example (2/7)3 can be worked out simply as (23)/(73) = (2 · 2 · 2)/(7 · 7 · 7) = 8/343. This is because multiplying fractions amounts to simply multiplying the numerators and denominators separately. Furthermore, when repeatedly multiplying the same numbers (as we do in exponentiation), you'll never have to reduce the fraction as long as it was reduced originally.

When raising a fraction to a negative exponent, the easy thing to do is to take the reciprocal first. For example, (2/3)−2 = (3/2)2, since 3/2 is the reciprocal of 2/3, so (2/3)−2 is simply 9/4. This is because raising to a negative exponent means division, which is simply multiplication by the reciprocal. As another example, (1/2)−3 = 23, because 2 is the reciprocal of 1/2, so (1/2)−3 works out to 8. Conversely, 2−3 is 1/8.

Also, you should be careful with the order of operations; unless grouping symbols (like parenthese) get in the way, exponentiaton always comes first. For example, −42 means −(4 · 4), which is −16, while (−4)2 means (−4) · (−4), which is 16. Also, 2/32 means 2/(3 · 3), which is 2/9, while (2/3)2 means (2/3) · (2/3), which is 4/9.


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