and so on, forever
The value keeps getting smaller and smaller, becoming half its former size every time you decrease the integer power by 1, but always remaining positive. This sequence of values is said to converge. In this case it “approaches 0.”
Reciprocal-of-Integer Powers
Now that we’ve seen what happens when a number is raised to an integer power, let’s find out what goes on when a number is raised to a power that is the reciprocal of an integer.
Integer roots are reciprocal-of-integer powers
Suppose we take some number or quantity a, and raise it to the power 1/p where p is a positive integer. We write this as
a1/p
We can surround the exponent with parentheses for clarity. If we do that to the above expres-sion, we get
a(1/p)
In this case, the parentheses are not technically necessary because the whole ratio is written as a superscript anyway.
When we take a reciprocal-of-integer power of a quantity, the result is often called a root.
If you have a number and raise it to the power 1/p, it is the same thing as taking the pth root of that number. If p is a positive integer, then the pth root of a quantity is something we must multiply by itself p times in order to get that quantity.
The square root
If the general formulas above confuse you, it can help if we look at an example. We know that 52= 5 × 5= 25
The second power is often called the square, so we can say, “5 squared equals 25.” By defini-tion then
251/2= 5
We would say, “The square root of 25 is equal to 5.” In general, for any two numbers a and b, and for any positive integer p, we can say this:
If ap=b, then b1/p=a
The reason the 2nd power is called the square and the 1/2 power is called the square root can be explained in terms of the dimensions and area of a perfect geometric square. For any geomet-ric square, the interior area is equal to the 2nd power of the length of any one of the edges, as shown in Fig. 8-1. That’s why the 2nd power is called the square. Looking at it the other way, the length of any one of the edges is equal to the 1/2 power of the interior area. That’s why the 1/2 power is called the square root. In the figure, the radical notation for square root is shown, in addition to the 1/2 power notation. The radical consists of a surd symbol (√) with a line extending over the top of the quantity of which the square root is taken.
The cube root
Now let’s see what happens when p= 3, so 1/p= 1/3. We can easily figure out what happens when we raise a number, say 4, to the 3rd power:
43= 4 × 4 × 4= 64
Interior area = A
Length of edge = s
Length of edge = s
A=s2 s=A1/2 and
= A
Figure 8-1 The area of a geometric square is equal to the 2nd power, or square, of the length of any edge. Therefore, the length of any edge is equal to the 1/2 power, or square root, of the area.
Reciprocal-of-Integer Powers 113
114 Powers and Roots
The third power is often called the cube. We can say, “4 cubed equals 64.” Now if we go with the reciprocal power and work backwards, we get
641/3= 4 This can be read as, “The cube root of 64 equals 4.”
The 3rd power is called the cube and the 1/3 power is called the cube root because of the relationship between the edges and the interior volume of a geometric cube. For any perfect cube, the volume is equal to the 3rd power of the length of any edge (Fig. 8-2). Going the other way, the length of any edge is equal to the 1/3 power of the volume. The figure also shows the radical notation for the cube root. The fact that the radical refers to the cube root, rather than the square root, is indicated by the small numeral 3 in the upper-left part of the radical symbol.
Higher roots
Whenp is a positive integer equal to 4 or more, people write or talk about the numerical pow-ers and roots directly. That’s because geometric hypercubes having 4 dimensions or more are not commonly named. A 4-dimensional hypercube is technically called a tesseract, but you should expect incredulous stares from your listeners if you say “2 tesseracted is 16” or “The tesseract root of 81 is 3.”
Here are some examples of higher powers and roots. You can check the larger ones on your calculator if you like. 3rd power, or cube, of the length of any edge.
Therefore, the length of any edge is equal to the 1/3 power, or cube root, of the volume.
56= 15,625 so 15,6251/6= 5
−37= −2,187 so (−2,187)1/7= −3 (−5)9= −1,953,125, so (−1,953,125)1/9= −5 64= 1,296 so 1,2961/4= 6
(−6)4= 1,296 so 1,2961/4= −6 ... What?
The radical notation can be used for any integer root. For the 1/n power, a small numeral n is placed in the upper left part of the radical symbol. If you use this notation, you must be sure that the radical symbol extends completely over the quantity of which you want to take the root. If you use the fractional notation, parentheses, brackets, and braces should be used to define the quantity of which you want to take the root.
Are you confused?
Now you will ask, “Can 6 and −6 both be valid 4th roots of 1,296?” The answer is “Yes.” Both 6 and −6 will work here:
6 × 6 × 6 × 6= 1,296 and
(−6) × (−6) × (−6) × (−6)= 1,296
If you multiply any negative number by itself an even number of times, you’ll get a positive number.
Therefore, if you have some number a and its additive inverse −a, and then you raise both of those num-bers to an even positive integer power p, you will get
(−a)p=ap
every time! If we call (−a)p or ap by some other name such as b, then the pth root of b is ambiguous. That would mean, for example,
161/4= 2 and −2
811/4= 3 and −3
15,6251/6= 5 and −5
It could even mean something as simple, and yet as troubling, as 11/2= 1 and −1
Mathematicians get around this problem by saying that whenever “two numbers at once” are the result of a reciprocal power, the positive value is the correct one, unless otherwise specified. That means
161/4= 2
811/4= 3
Reciprocal-of-Integer Powers 115
116 Powers and Roots
15,6251/6= 5
11/2= 1
You can indicate that you want to use the negative value by placing minus signs like this:
−(161/4)= −2
−(811/4)= −3
−(15,6251/6)= −5
−(11/2)= −1
Sometimes you will actually want to let either the positive or the negative value be used. In cases of that sort, you should throw a plus-or-minus sign (±) into the mix, like this:
±(161/4)= ±2
±(811/4)= ±3
±(15,6251/6)= ±5
±(11/2)= ±1
Here’s another possible confusion-maker. Always pay special attention to where the parentheses are placed if you see a negative number raised to a power. Also, be careful if there are no parentheses at all. If there’s any doubt, it’s best to place extra parentheses in an expression so everyone knows exactly what it means.
For example,
(−2)3= (−2) × (−2) × (−2)= −8 and
−23= −(23)
= −(2 × 2 × 2)
= −8
In contrast to this,
(−2)4= (−2) × (−2) × (−2) × (−2)= 16 but
−24= −(24)
= −(2 × 2 × 2 × 2)
= −16
Negative reciprocal powers
We still have not explored what happens when we raise a number to a negative reciprocal-of-integer power. You can probably figure out the meaning of an expression such as 125−1/3, or
125 to the −1/3 power. You take the 1/3 power of 125, which is 5, and then take the reciprocal of that, which is 1/5. Mathematically, it goes like this:
125−1/3= 125−(1/3)
= 1/(1251/3)
= 1/5
Even roots of negative numbers
What happens when you take an even root of a negative number? The simplest example of this sort of problem is the square root of −1, but there are plenty of others. What can you multiply by itself to get −1? Nothing that we’ve defined yet! What is the 1/4 power of 16? Again, noth-ing we know of so far.
Mathematicians have defined quantities like this. We will explore them in Chap. 21.
They’re called imaginary numbers. They have some fascinating properties. Unlike division by 0 or the 0th root of 0, even roots of negative numbers can be “tamed.” They are commonly used in science and engineering.
Here’s a challenge!
State the rule for negative reciprocal powers in general terms, where a is the base (the number to be raised to the power) and p is a positive integer.
Solution
The power to which we want to raise the base is −1/p, where p is some positive integer. (We know that
−1/p will be negative, because a negative divided by a positive always gives us a negative.) If we use the method from the above example where we evaluated 125−1/3, then we have
a−1/p=a− (1/p)= 1/(a1/p)