4.5 Real Numbers
One of the properties of the integers that does not extend to the rational numbers is the least-upper-bound property (3.4). In order to prove this, we first need to prove a series of small lemmas (a smaller version of a theorem that is used to prove a larger theorem).
Lemma 4.7 If \(n\) is an integer, then \(n\) can be written as \(2k\) for some integer \(k\) or \(2j+1\) for some integer \(j\), but cannot be written both ways. (In other words, an integer is even or odd, but not both.)
Proof. We will use the technique of a proof by contradiction and assume that there is an integer \(n\) such that \(n=2k\) and \(n=2j+1\) for some integers \(j\) and \(k\). Then \(2j=2k+1\) and so the equivalent equation \(2(j-k)=1\) is true. However, since \(2\) does not have a multiplicative inverse in the integers, \(j-k\) cannot be an integer. This means that our assumption is false and our lemma is proven.
Lemma 4.8 If \(n\) is an integer such that \(n^2\) is even, then \(n\) is even.
Proof. When trying to prove that something does not satisfy a certain property, then it is best to use a proof by contradiction or to prove the contrapositive.
We will prove the contrapositive (\(\sim q \rightarrow \sim p\)) of the statement (\(p \rightarrow q\)), which is logically equivalent to the original statement. Let \(n\) be an odd integer, i.e. \(n=2j+1\) for some integer \(j\). Then \(n^2=(2j+1)^2=4j^2+4j+1=2(2j^2+2j)+1\) and so \(n^2\) is odd and the original statement is proven.
Lemma 4.9 There does not exist a rational number whose square is \(2\).
Proof. We will again use the technique of a proof by contradiction. Assume that there is a rational number whose square is \(2\). We will denote the rational number by \(\frac{a}{b}\) where \(a\) and \(b\) are not both even. (This can be accomplished by finding an equivalent representation without 2 as a common factor.) Since its square is \(2\), we have that \(\left(\frac{a}{b}\right)^2 =2\) and equivalently that \(a^2 = 2 b^2\). Since \(a^2\) is even, the previous lemma tells us that \(a\) is even. So there exists an integer \(k\) such that \(a=2k\) and so \(a^2=4k^2\). We then have that \(4k^2=2b^2\) and so \(b^2=2k^2\). This means that \(b^2\) is even and consequently that \(b\) is even. This contradicts our original assumption and so no such rational number exits.
We will now look at the set \[S = \left\{ \frac{a}{b}\in \mathbb{Q} \middle \vert \frac{a}{b} >0 \: \mbox{ and } \: \left(\frac{a}{b}\right)^2 \geq 2\right\}\] which is bounded below in its definition and does not have a least element. Therefore, the rational numbers do not satisfy the least-upper-bound property.
Since the rational numbers do not satisfy the least-upper-bound property, the rational numbers have “holes” that need to be filled, and so are not complete. Building upon the work of Bertrand (1840) and in a parallel vein to Heine (1872), Dedekind (1872) developed a construction of the real numbers from the rational numbers using what are now referred to as Dedekind cuts using equivalence classes of such cuts. Another method of construction of the real numbers from the rational numbers follows the path of Cauchy (1821) and defines the real numbers as equivalence classes of Cauchy sequences of rational numbers. As each of these methods requires the development of a great deal of mathematical machinery, we will refer to other textbooks in analysis for the details. With Rudin (1976) providing an excellent treatment of Dedekind cuts in the Appendix of Chapter 1. Chapter I of Thurston (1956) provides an detailed treatment of the construction of the Cauchy numbers to generate the real numbers through equivalence classes.
Because the real numbers form a complete number system in that it satisfies the least-upper-bound property, for any bounded set of real numbers, \(S\), we can define \(\sup(S)\) to the be the least element of the set \(\{ x \in \mathbb{R} \vert x\geq a, \forall a\in S\}\) and \(\inf(S)\) is the greatest element of the set \(\{x\in \mathbb{R} \vert x\leq a, \forall a\in S\}\). These two attributes of a set play an important role in the development of calculus.
Since \(\sqrt{2}\) is not rational, but we know it is a real number because of the completeness of the reals, and so there are real numbers that are not rational. We define these numbers to be irrational.
Definition 4.2 A real number, \(a\), is called rational if there exist integers \(p\) and \(q\) such that \(a=\frac{p}{q}\). A real number that is not rational is called irrational.
Note: sometimes irrational numbers are defined as real numbers that cannot be written as a fraction of integers.
Related Content Standards
- (HSN.RN.3) Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a non-zero rational number and an irrational number is irrational.
We have already shown that the sum and product of two rational numbers are rational, but we would like to explore how rational and irrational numbers interact with each other.
Theorem 4.8 Let \(u\in \mathbb{Q}\) and \(v\in (\mathbb{R}-\mathbb{Q})\). Then \(u+v\notin \mathbb{Q}\) and if \(u\neq 0\), \(uv\notin \mathbb{Q}\).
Proof. Let \(u\in \mathbb{Q}\) and so there exist integers \(p\) and \(q\) such that \(u=\frac{p}{q}\). Let \(v\in (\mathbb{R}-\mathbb{Q})\). We will now prove that \(u+v\) is not rational using a proof by contradiction. If \(u+v\) is rational, then there exist integers \(m\) and \(n\) such that \(u+v=\frac{m}{n}\). This means that \[v= \frac{m}{n}-u = \frac{m}{n}-\frac{p}{q} = \frac{mq-np}{nq}\] making it a rational number, contradicting the assumption that it is irrational.
The proof that \(uv\) is irrational is very similar.
One of the important properties to notice about the rational and irrational numbers is that they are dense in the real numbers in that between any two real numbers one can find both a rational and an irrational number.
Theorem 4.9 (Density of the Rational Numbers) Let \(x,y\in \mathbb{R}\) be any two real numbers where \(x<y\). Then there exists a rational number \(q\in \mathbb{Q}\) such that \(x<q<y\).
Proof. Since \(x<y\), we know that \(y-x>0\). So there exists a positive natural number \(n\in \mathbb{N}\) such that \(0<\frac{1}{n}<y-x\). Equivalently, \(0<1<ny-nx\) and \(nx+1<ny\). From properties of the integers we know that there exists an integer \(N\in \mathbb{Z}\) such that \(N-1\leq nx<N\), or equivalently \(N\leq nx+1 < N+1\). Therefore, \[nx<N\leq nx+1 < ny\] and so \(x<\frac{N}{n} <y\). So \(q=\frac{N}{n}\) is a rational number such that \(x<q<y\).
Theorem 4.10 (Density of the Irrational Numbers) Let \(x,y\in \mathbb{R}\) be any two real numbers where \(x<y\). Then there exists an irrational number \(v\in \mathbb{R}\setminus\mathbb{Q}\) such that \(x<v<y\).
Proof. Since \(x\) and \(y\) are real numbers, \(\frac{x}{\sqrt{2}}\) and \(\frac{y}{\sqrt{2}}\) are also real numbers such that \(\frac{x}{\sqrt{2}}<\frac{y}{\sqrt{2}}\). By the density of the rational number, there exists a \(q\in \mathbb{Q}\) such that \[\frac{x}{\sqrt{2}} < q < \frac{y}{\sqrt{2}},\] or equivalently \(x<q\sqrt{2}<y\). Since \(q\in \mathbb{Q}\) and \(\sqrt{2}\) is irrational, \(q\sqrt{2}\) is irrational and so if \(v=q\sqrt{2}\), \(v\in \mathbb{R}\setminus\mathbb{Q}\) such that \(x<v<y\).
Since the operations of addition and multiplication on the real numbers are associative and commutative, for each natural number \(n\) we can define \(a^n\) recursively and we can prove the following properties using the same techniques as in Section 4.1, and Section 4.3 using limit ideas from the definition of the real numbers.
Theorem 4.11 Let \(a,b \in \mathbb{R}\) and let \(m,n\in \mathbb{Z}\).
\(a^0=1\) and \(a^1=a\)
If \(a>1\) and \(n>0\), then \(a^n >1\). If \(a>1\) and \(n<0\), then \(0<a^n<1\).
\(a^{-n} = \frac{1}{a^n}\)
\(a^m\cdot a^n = a^{m+n}\), \((ab)^n=a^n\cdot b^n\), and $ (am)n = a^{mn}$
If \(0<a<b\) and \(m\in \mathbb{N}\), then \(0<a^m<b^m\).
If \(a>1\) and \(m<n\), then \(a^m<a^n\).
With these properties, we are able to define define algebraic numbers, which is a number system between the rational numbers and the real numbers.
Definition 4.3 A real number, \(a\), is called algebraic if \[n_0 + n_1 a + n_2 a^2 + \cdots + n_m a^m =0\] for some integers \(n_0, n_1, n_2, \ldots, n_m\).
For all rational numbers in the form \(\frac{p}{q}\), we know that \[q \cdot \left(\frac{p}{q}\right)^1 + (-p) = 0\] and so all rational numbers are algebraic numbers. We can also see that \(\sqrt{2}\) is an algebraic number since \((\sqrt{2})^2-2=0\). We say that the algebraic numbers are a number system because they include the additive identity, \(0\); the multiplicative identity, \(1\); and are closed under addition and multiplication.
We will prove in Section 5.4 that there are an infinite number of real numbers that are not algebraic. In fact, most real numbers are not algebraic. We will call any real number that is not algebraic a transcendental number. Two such examples of transcendental numbers are \(e\) and \(\pi\).
In the early 1930’s, Gel’fond (1934) and Schneider (1935) independently solved Hilbert’s seventh problem from his 1900 address to the International Congress of Mathematics showing that if \(a\) and \(b\) are algebraic numbers with \(a\neq 0,1\) and \(b\) irrational that \(a^b\) is a transcendental number. Since we have such a limited knowledge of these transcendental numbers, it is evident that the continuum of the real numbers is a very deep and complicated concept that we are only beginning to understand.
We include a graphical representation of how these different sets of numbers fit together. All of these sets of numbers are subsets of the complex numbers, and a set above another set represents an inclusion relationship.
4.5.1 Properties of Exponents
Let \(a>0\) be a real number. For each integer \(n>0\), we define \[S_n= \left\{ x\in \mathbb{Q}\middle \vert x>0 \mbox{ and } x^n>a\right\}\] and so we can define \[\sqrt[n]{a}=a^{\frac{1}{n}} := \inf(S_n).\]
Related Content Standards
- (8.EE.2) Use square root and cube root symbols to represent solutions to equations of the form \(x^2=p\) and \(x^3=p\), where \(p\) is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that \(\sqrt{2}\) is irrational.
If \(a>0\) is a real number, \(m,n,p,q\) are integers, \(n>0\), and \(q>0\) such that \(\frac{m}{n}=\frac{p}{q}=r\), we have that \(mq=pn\). So \[\left(\left(a^{\frac{1}{n}}\right)^m\right)^{nq} = a^{mq} = a^{pn} = \left( \left( a^{\frac{1}{q}} \right)^p \right)^{nq}.\] This means that we can define \(a^r:=\left(a^m\right)^{\frac{1}{n}}\).
Related Content Standards
- (HSN.RN.1) Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents.
- (HSN.RN.2) Rewrite expressions involving radicals and rational exponents using the properties of exponents.
We can now prove various exponential properties of real numbers with rational exponents. In order to prove a generalization of Theorem 4.11 to rational exponents we will only need to prove the corresponding properties for rational numbers of the form \(\frac{1}{n}\) and it directly follows from \((a^n=b^n) \Leftrightarrow a=b\) for real numbers \(a\) and \(b\) and positive integers \(n\). As such, we will omit the majority of these proofs from this text. However, will include one lemma and proof in order to understand the flavor of the proofs.
Lemma 4.10 Let \(a\in \mathbb{R}^+\). For all \(r,s\in \mathbb{Q}\), \(a^{r+s}=a^r\cdot a^s\).
Proof. Let \(a\in \mathbb{R}^+\) and let \(r,s\in \mathbb{Q}\) such that \(r=\frac{m}{n}\) and \(s=\frac{p}{q}\) for some integers \(m\), \(n\), \(p\), and \(q\).
Since \(q\neq 0\) and \(n\neq 0\), \[\left(a^{r+s}\right)^{qn} = \left( a^{\frac{mq+pn}{qn}}\right)^{qn} = a^{(mq+pn)}\] and \[\left(a^r\cdot a^s\right)^{qn} = \left(a^{\frac{mqn}{n}}\right)\cdot \left(a^{\frac{pqn}{q}}\right) = a^{(mq+pn)}.\] Therefore, \(a^{r+s}=a^r\cdot a^s\).
For a real number \(y\), define \[S_y= \left\{ a^r \middle \vert r \leq y\right\}.\] Furthermore, define \[a^y = \sup(S_y).\] Using arguments from analysis and the properties of limits, one can extend the previous properties of exponents to real valued exponents as given in Theorem 4.12. Due to the mathematical machinery necessary for these proofs, we will refer the reader to an analysis textbook.
Theorem 4.12 Let \(a\) and \(b\) be positive real numbers and let \(x,y\in \mathbb{R}\).
\(a^0=1\) and \(a^1=a\)
\(a^x >1\) for all \(x>0\), and \(0<a^x<1\) for all \(x<0\)
\(a^{-x} = \frac{1}{a^x}\)
\(a^x\cdot a^y = a^{x+y}\), \((ab)^x=a^x\cdot b^x\), and \((a^x)^y = a^{xy}\)
If \(0<a<b\) and \(x>0\), then \(a^x<b^x\).
If \(a>1\) and \(x<y\), then \(a^x<a^y\).
4.5.2 Exercises
If \(n=2\), \(a^\frac{1}{2}\) is often written as \(\sqrt{a}\), but for \(n=\{3,4,5, \ldots\}\), \(a^{\frac{1}{n}}\) is written as \(\sqrt[n]{a}\). This difference in notation can cause many problems for students. What are some other notation differences in the secondary curriculum that have similar situations?
What can be said about the sum and product of two irrational numbers?
Suppose that \(a\in \mathbb{R}\) with \(a>0\). If \(n\) is an even positive integer, show that there are exactly two real numbers that are solutions to the equation \(x^n=a\).
Show that \(\sqrt{2}+\sqrt{3}\) is an algebraic number by finding a polynomial for which it is a zero.
Write out a detailed proof that \(\sqrt{3}\) is irrational using the techniques of the proof that \(\sqrt{2}\) is irrational. For which numbers can this process be generalized?