01 Irrational Numbers
Section 1.1 gives a proof of the Fundamental Theorem of Arithmetic and uses it to show that various real numbers are irrational. The proof of this theorem is not important for this course, and the theorem itself is only used to prove the existence of irrational numbers. The exercises in Section 1.1 explore some other facts about irrational numbers. (Note that the historical background information in this section, and elsewhere in the textbook, is not required for this course. But it can be interesting to read it anyway.)
Mathematicians work with various "number systems." The set of natural numbers, $\N,$ is the most basic. This set contains the positive integers: $\N=\{1,2,3,\dots\}.$ Adding the negative integers and zero, we get the set of integers: $\Z=\{\dots,-2,-1,0,1,2,\dots\}.$ Two natural numbers can be added but not always subtracted. In some sense, $\Z$ is invented to make subtraction possible; when you subtract a natural number from a smaller natural number, the result is not a natural number, but it is a negative integer. We say that the natural numbers are closed under addition but not closed under subtraction. The set of integers is closed under both addition and subtraction.
Simlarly, the set of integers is not closed under division. The rational numbers are invented to make division possible (except of course for division by zero). The set $\Q$ of rational numbers contains all quotients $\frac{n}{m},$ where $m$ and $n$ are integers and where the denominator, $m,$ is not zero. Of course, the same rational number can be represented by many different fractions; for example $\frac{15}{3}=\frac{5}{1}=\frac{-30}{-6}=\frac{105}{21}=\cdots.$ If you want to be fancy, a rational number can be defined as an equivalence class of fractions where $\frac{a}{b}$ is equivalent to $\frac{c}{d}$ if and only if $ad=bc.$ Every non-zero rational number can be represented uniquely as a fraction $\frac{p}{q}$ where $q$ is a positive integer and $p$ and $q$ have no common factors.
The set of real numbers, $\R,$ is different. If the motivation for inventing $\Z$ and $\Q$ is arithmetic, the motivation for inventing $\R$ is more geometric. The set of real numbers is identified with a geometric object, a "line." There is a one-to-one correspondence between real numbers and points on a line. To get a correspondence, you have to decide which two points correspond to the numbers 0 and 1, but once you do that, the correspondence is determined. This gives the familiar picture of the real number line:
It is not immediately obvious that you need anything more than the rational numbers to account for all of the points on the line. However, the geometry of the line leads us to numbers that are not rational. That is, if we define the real numbers to be the points on a line, then there are real numbers that are not rational. Real numbers that are not rational are called irrational.
The original geometric proof of this fact used a square whose sides have length 1. According to the Pythagorean theorem, the diagonal of that square has length $\sqrt{1^2+1^2},$ or $\sqrt 2.$ But $\sqrt 2$ cannot be a rational number. However, it is certainly a real number. You can locate that number on the real line geometrically simply by laying the diagonal on the line; if one endpoint of the diagonal is at zero, then the other endpoint is at $\sqrt 2.$ The well-known proof that $\sqrt 2$ is irrational is given in the textbook. In fact, we can use the Fundamental Theorem of Arithmetic to get a large supply of irrational numbers:
Theorem: Let $n$ be a positive integer greater than 1, and suppose that $n$ has prime factorization $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ (where $\seq[p]k$ are distinct primes, and $\seq[e]k$ are integers greater than zero). Then $\sqrt n$ is rational if and only if every $e_i$ is an even number.
Proof: Suppose that every $e_i$ is an even number. Then $n=p_1^{2i_1}p_2^{2i_2}\cdots p_k^{2i_k}$ where $\seq[i]k$ are integers. So $\sqrt n = p_1^{i_1}p_2^{i_2}\cdots p_k^{i_k},$ which is a rational number. (Note that this square root is in fact an integer, not just a rational number. The square root of an integer can only be an integer or an irrational number.)
Conversely, suppose that one of the exponents, say $e_i$, is odd. Suppose, for the sake of contradiction, that $\sqrt n$ is rational. Write $\sqrt n=\frac{a}{b}.$ where $a$ and $b$ are integers. Then $n=\frac{a^2}{b^2},$ and $nb^2=a^2.$ By the Fundamental Theorem of Arithmetic, we can write $a=p_i^sc$ and $b=p_i^td,$ where $s$ and $t$ are non-negative integers and $c$ and $d$ are integers that are not divisible by $p_i.$ That is, $p_i$ occurs $s$ times as a factor of $a,$ and it occurs $t$ times as a factor of $b.$ (We have simply factored all possible factors of $p_i$ out of $a$ and $b.$) But then $p_i$ occurs $2s$ times as a factor of $a^2,$ and it occurs $2t$ times as a factor of $b^2.$ Since $p_i$ occurs an odd number of times as a factor of $n,$ and it occurs an even number of times as a factor in $a^2$ and in $b^2,$ it follows that $p_i$ occurs an odd number of times as a factor of $nb^2$ and an even number of times as a factor in $a^2.$ But, by the Fundamental Theorem of Arithmetic, this means that $nb^2$ cannot equal $a^2,$ since they have different prime factorizations. This contradicts the fact that $n=\frac{a^2}{b^2},$ and the contradiction shows that $n$ cannot be rational. $\qed$
But taking square roots of integers does not produce all irrational numbers. For example, $\pi$ and $e$ are irrational, but $\pi^2$ and $e^2$ are not integers. The number $\pi$ can be defined geometrically as the circumference of a circle that has diameter equal to 1. The number $e$ is the base of the natural logarithm, and its definition requires calculus.
For lots more irrational numbers, we can turn to non-repeating decimals. (See Exercises 17 through 21 in Section 1.1.) Real numbers are often represented in decimal form, $a.d_1d_2d_3\dots,$ where $a$ is an integer and $d_1,d_2,d_3,\dots$ are digits, that is, integers in the range 0 to 9. (This really represents the number as the sum of an infinite series, $a+\sum_{n=1}^{\infty}\frac{d_n}{10^n}.$) A number in decimal form is rational if and only if it is repeating. In a repeating decimal, after some point, there is a finite block of one or more digits that is just repeated over and over forever. For example, $\frac 13=0.3333333\dots=0.\overline{3}$ or $\frac{223}{165}=1.351515151\dots=1.3\overline{51}$ or $\frac12=0.500000\dots = 0.5\overline{0}.$ The line over 3 or 51 or 0 indicates that that digit or block of digits repeats forever.
But this means that any non-repeating decimal represents an irrational number. For example:
0.101001000100001000001000000100000001000000001000000000...
In this decimal, the number of zeros between the ones increases by one from each group of zeros to the next, making it impossible for the decimal to terminate in a finite, repeating block. The ones could be replaced with other digits to give other irrational numbers. For example, we could replace some of the ones with twos:
0.102002000100002000001000000100000002000000001000000000...
We can choose any subset of the ones to replace with twos. Since there is a countable number of ones, there is an uncountable number of subsets of the ones, so we can get an uncountable number of different irrational numbers from this one example. (This assumes that you know about countable versus uncountable infinities, and the fact that the set of subsets of an infinite set is uncountable. If so, you probably know that the set of rational numbers is countable — that is, it can be put into one-to-one correspondence with the set of natural numbers — while the set of real numbers is uncountable. But it's nice to have a concrete example of uncountably many irrational numbers.)
The question remains, is there some way that we can produce all of the irrational numbers? That is the topic of the next reading.