We’ve talked about positive integer powers, negative integer powers. We’ve talked about n-th roots. It’s now time to put all of that together and more and talk about rational powers. Here’s some motivation. How should we define a to the power 1 over n? Well we would like the usual known properties of powers or exponents to hold. So, in particular, we’d like a to the power 1 over n, the whole thing raised to the n, to equal a. Therefore, we are forced really to define a to the power 1 over n as being the n-th root of a, which we do. Thus, the n-th root is actually a fractional power and vice versa.
More generally, for any integers, m and n, where n is in the natural numbers and m can be a negative integer, we define a to the power m over n, which is a typical rational number, to be the n-th root of a to the m. Now we do this for positive numbers, a, only. In fact, if m is negative, we only do it for strictly positive numbers, a, in the base. For a greater than 0, we have defined a to the p for any rational number, p. Some facts that pertain to this definition are now given.
We could prove all of these facts for these general rational numbers, but I immediately give you the same good news I gave you some time ago, these rules visually are exactly the same as they were say for positive integers. That’s good news. We don’t have to memorise a whole set of new rules. Some examples, x to the power of 1/2 is the square root of x. x to the power minus 3/5 turns out to be what I’ve written. Notice that the answer could be rationalised in the denominator if we so insisted. Another example, an exponent of minus 1/2 is undefined. We avoid negative numbers. Why should we avoid these negative bases, you say? Here’s an illustration of what can happen.
Suppose we did define minus 8 to the 1/3 as being minus 2. That’s not totally ridiculous because minus 2 cubed is equal to minus 8. But now, the usual exponential rules that we know so well don’t necessarily work. That is, minus 2 would be the cube root of minus 8, but that would be the same as minus 8 to the power 2 over 6 because, after all, 2 over 6 is the same as 1/3. And now, if we bring in the 2 as we do by the usual exponential rules, we get the cube root, the sixth root rather, of the number 64. Now 64 is 2 to the power 6, and, therefore, we wind up with 2.
And we have proved that minus 2 equals 2. That’s not a good sign. So you see, dangerous to define powers with negative bases, and I suggest that we avoid doing so. What are the graphs of these rational power functions going to look like? Well, by and large, they’re going to have the same general characteristics of some familiar graphs we’ve seen in the past. For example, when q is equal to 1, the function next to the q is an affine function, in fact, going through the origin called the linear function. When q is greater than 1, the graph looks rather like the graph of x squared. It has the increasingness on r plus, for example.
And when q is less than 1, it resembles the graph of the square root of x that we saw earlier. All these functions agree at x equal 1, of course, to give you the value 1. And for negative rational exponents, you get graphs that have this kind of characteristic, there you have decreasing functions of x. And, again, they’re defined only on r plus. It’s useful to know the general nature of these graphs, the monotonicity of the functions, for example, such as the following. Suppose you need to know which of these two numbers is greater than the other. How do you figure that out? Well you could say the following.
The function that to the argument x gives you the power minus 6.72 of x. That’s a strictly decreasing function by the graph that we saw just a moment ago. Why? Because the exponent is a negative number. We also observe that 1/2 is less than pi over 4. How do we prove that? Well we can cross multiply 1 times 4, 2 times pi. That’s equivalent to saying that 4 is less than 2 pi, cross multiplying. Is 4 less than 2 pi? Well that’s the same as saying 2 is less than pi, which is true because we know that pi is 3.14 something, something, and so on.
Now because of the decreasing nature of the negative power function that we have above, we deduce that 1/2 to that negative power is greater than pi over 4 to that same negative power. That’s clear from the graph because it’s a decreasing graph. So we have answered the question regarding the comparison. In fact, if we look at the graph and realise that both 1/2 and pi over 4 are less than 1, we can also add a further conclusion. Both our values here that we’re comparing are greater than 1.