



Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
The concept of rational exponents, extending the definition of exponents beyond natural numbers to the set of rational numbers. The author explains how negative exponents and the multiplication of exponents work, and introduces the concept of square roots and their unique properties. The document also discusses the process of calculating square roots and testing for rationality.
Typology: Study Guides, Projects, Research
1 / 7
This page cannot be seen from the preview
Don't miss anything!




1 Exponent Review 1 1.1 xn^ · xm^................................. 2 1.2 x
n xm^...................................^2 1.3 (xn)m^................................. 3
2 Further Extensions: Rational Exponents 3
3 Exploring Square Roots 4 3.1 Counting Square Roots....................... 4 3.2 Calculating Square Roots...................... 5 3.3 Testing Rationality of
x...................... 6
4 Conclusions 7
Let us begin by reviewing exponents and what they mean to us so far. We have defined exponents through repeated multiplications, as follows:
xn^ = x︸ · x · x ·︷︷... · x · x︸ n times
With this definition, we have a few standard properties of exponents we can utilize.
We can see that
xn^ · xm^ = x︸ · x · x ·︷︷... · x · x︸ n times
· x︸ · x · x ·︷︷... · x · x︸ m times = x︸ · x · x ·︷︷... · x · x︸ n+m times xn^ · xm^ = xn+m
Thus, exponents are additive when we multiply two terms with the same base.
n xm
This is similar to the previous example, using division instead of multiplication:
xn xm^
n times ︷ ︸︸ ︷ x · x · x ·... · x · x x︸ · x · x ·︷︷... · x · x︸ m times xn xm^
= xn−m
since we can cancel m instances of x in the numerator.
This leads to two questions. What if n = m? In this case, we get per- fect cancelations from both the numerator and the denominator, resulting in 1 for an answer. Using this formula, we have x^0 = 1 because of these perfect cancelations, almost regardless of the value of x.^1
Our second question: what if m > n? Then xn−m^ results in a negative exponent. This is fine, provided we understand how to interpret those negative exponents. For example:
x^2 x^3
x · x x · x · x
x · x x^ ·^ x^ ·^ x^
x
= x−^1
Thus, a negative exponent for x means that factor needs to appear in a denominator. (If the factor of x with a negative exponent is already in the
(^1) I say “almost” because x 6 = 0 as it appears in the denominator. 0 0 is undefined.
which makes perfect sense: rational exponents allow us to reverse, or undo a previous exponent. The concept was developed far earlier than this notation, which is why we also see notation like
x = x
(^12) , or more generally, n
x = x
(^1) n . Note the small n in the n
“radical” notation: it is assumed to be 2 when omitted. When n = 2, it is called a “square root,” when n = 3 it is a “cube root” and when it is higher we simply call it the nth root (e.g. fourth root, fifth root, etc.)
This begs a further question: what happens if we take the square root of something and arrive at an answer which doesn’t have an integral exponent? For example, 2 (^12) ?
3 Exploring Square Roots
Let us assume that r^2 = a, indicating that r =
a. We need to determine if this is a unique value or not. We will show that 0 has a unique square root and that all other numbers have two square roots.
We know that, of x · y = 0, then either x = 0 or y = 0. If x =
0, then x^2 = x · x = 0, and thus x = 0. Therefore, 0 is its own square root.^2
Let us assume that a > 0. Let us also assume x^2 = a and y^2 = a. We now prove that there are only two possibilities for x and y: either x = y, or x = −y. We prove this by contradiction: assume x > y. By the axiom of inequality, if x > y, then a = x^2 > xy > y^2 = a, which is a contradiction if x and y are both positive: a would need to be strictly greater, and not equal to itself. If x > 0 and y < 0, then the part that fails is the xy > y^2 step: the left hand side is negative, while the right hand side is positive. To repeat the logic allowing for negatives, we must instead demand that |x| ≥ |y|, and our transformed statement becomes a = x^2 ≥ |xy| ≥ y^2 = a. This statement may be valid if and only if we drop the inequalities and x^2 = |xy| = |x| |y| = y^2. Demanding either x^2 = |x| |y| or |x| |y| = y^2 means |x| = |y|, so that either x = y or x = −y. This is, of course, based on the assumption that x > y; we can repeat the statements for y > x and arrive at the same conclusion. (^2) If x = x (^2) , then x is said to be idempotent. We will deal with idempotents in more detail later. For now, suffice it to say that 0 and 1 are the only idempotents in an algebraic field.
We can now calculate the positive square root of a positive number, provided we are allowed one assumption: infinite processes are valid. We have already seen that they have questionable validity under the rational number system, but, as the next few lessons will demonstrate, an algebra that allows the existence of infinite processes allows for significantly many more tools in our mathematical toolkit.
Let us again begin by searching for the number r > 0 which satisfies r^2 = a when a > 0. Let us begin with the first approximation x 1 ≈ r, and develop a means to refine this approximation. If x 1 < r, then (x 1 )^2 < r^2 = a by the axiom of inequality. Thus, we also have x 1 · r < r^2 = a. If we can find a y 1 such that x 1 · y 1 = a, then we will know that x 1 < r < y 1. Similarly, if x 1 > r, then we would need to find a y 1 such that x 1 · y 1 = a and y 1 < r < x 1. We can find the corresponding y 1 the same way in both cases: if
x 1 · y 1 = a
then y 1 = a x 1
for all x 1 > 0. If our first approximation is exactly correct, then x 1 = y 1 = r. If not, then either x 1 < r < y 1 or y 1 < r < x 1. Thus, a better approximation of the square root would be a number between both x 1 and y 1. If x 1 > y 1 , then we can use the axiom of inequality to show that
x 1 =
(x 1 + x 1 ) >
(x 1 + y 1 ) >
(y 1 + y 1 ) = y 1
In the case where x 1 < y 1 , we have instead
x 1 =
(x 1 + x 1 ) <
(x 1 + y 1 ) <
(y 1 + y 1 ) = y 1
In either case, 1 2
(x 1 + y 1 ) =
x 1 + a x 1
is between x 1 and y 1 , just as r is, and so we define our next approximation for r as
x 2 =
x 1 +
a x 1
Generally speaking, we can define
xn+1 =
xn +
a xn
to our contradiction: if q = 2n and p = 2k, then pq could have been written as k n , and thus was not originally in lowest terms.^ We can continue this process infinitely many times should we so choose, but as pq needs to be finite to be a
valid representation of
2, we are forced to conclude that there is no such valid representation of
2, and
2 is not a rational number.
4 Conclusions
If we continue to restrict ourselves to the rational numbers, we will be unable to define fractional exponents for every possible base, although we can do so for certain specific cases. In future lessons, we will find other situations in which we need to move beyond the rational numbers to add new tools to our mathematical toolkit. In lesson 32, we will arrive at the extended system, known as the real number system, and we will get there by adding a single axiom to our set of