The Real and Complex Number Systems

An overview of the chapter “The Real and Complex Number Systems” from the famous books of “Principles of Mathematical Analysis” written by Rudin. The authors create a brief introduction of the loopholes of rational numbers, thus introducing real and complex number systems that help fill these holes.

Introduction

P1: Show that $p^2=2$ is not satisfied by any rational $p$
Proof: Suppose $p$ exists, then we can write $p=m/n$ , where m and n are integers that are not both even. If so, $m^2=2n^2$ would also hold, indicating that m is even. If so, m would also be divisible by 4, since m needs to be an integer. If so, n must also be even, to make up for the extra 2. This would mean that both m and n share a common divisor, which is against our initial choice of m and n.

P2: Show that there is no largest rational number that satisfies the condition $p^2<2$ , and similarly no smallest number that satisfies the condition $p^2>2$ .
Proof: Now, suppose $p>0$ , let us define q such that:

$q = p-\frac{p^2-2}{p+2}$

From the above equation, we can also show that

$q^2-2 = \frac{2(p^2-2)}{(p+2)^2}$

If $p^2<2$ , then $q>p$ , and $q^2<2$ . The exact reverse is true when $p^2>2$ .

The purpose of the above two proofs is to show the loopholes in rational numbers. The real number system fills these gaps. This is the principal reason for the fundamental role which it plays in analysis.

Ordered Sets

P3: Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$ , $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$ , then

$\alpha = \sup L$

exists in $S$ , and $\alpha = \inf B$ and exists in $S$ .
Proof: Since $B$ is bounded below, $L$ is not empty. Thus elements in $B$ are always greater than elements of $L$ . Thus, $L$ is bounded above. Suppose the hypothesis holds, and $\alpha = \sup L$ , where $\alpha$ exists in $S$ . If we take a number $\gamma$ less than $\alpha$ , then $\gamma$ will not be a upper bound of $L$ , and hence not belong to $B$ . Thus, it follows that $\alpha$ also serves as the lower bound of $B$ .

The Real Field

P4: If $x \in \mathbb{R}$ , $y \in \mathbb{R}$ , and $x>0$ , then there is a positive integer $n$ such that

$nx>y$

This is called the archimedean property of R.
Proof: Let A be the set of all $nx$ , where $n$ runs through the positive integers. If the above hypothesis were false, $y$ would be an upper bound of A. If so, we could write $\alpha = \sup A$ . If so, we consider another point $\alpha -x$ , where $\alpha -x < \alpha$ since $x>0$ . Then, $\alpha -x < \alpha$ is not an upper bound and could be written as $\alpha -x < mx$ , implying $(m+1)x> \alpha$ . This isn’t possible since $\alpha$ is the upper bound of A.

P5: Between any two real numbers, there is a rational one. If $x \in \mathbb{R}$ , $y \in \mathbb{R}$ , and $y>x$ , then there exists a $p \in Q$ such that $x<p<y$ .
Proof: Since $x<y$ , we have $y-x>0$ . From the previous proof, we can show that $n(y-x)>1$ . From the above equation, we can clearly say that $nx<nx+1<ny$ . Since $nx$ is a real number, there must exist an integer $m$ between $nx$ , and $nx+1$ modifying the above inequality to the following: