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Section 1.1 Sets of Numbers

In this course we are mainly interested in the counting numbers, which we will call the natural numbers, \(\mathbb{N};\) more specifically,

\begin{equation*} \mathbb{N}=\{1,2,3,4,5,6,7,8,9,10,...\}. \end{equation*}

If we add \(0\) to the naturals we will denote the set by \(\mathbb{N}_0;\) that is,

\begin{equation*} \mathbb{N}_0=\mathbb{N}\cup\{0\}. \end{equation*}

If we then add the negatives, we come up with the integers 1 ,

\begin{equation*} \mathbb{Z}=\{..., -3, -2, -1, 0, 1, 2, 3,...\}. \end{equation*}

If we divide integers by non-zero integers, we get the rational numbers, or quotients,

\begin{equation*} \mathbb{Q}=\left\{\frac{a}{b}:a,b\in\mathbb{Z}, b\neq 0\right\}. \end{equation*}

It is easily seen from the definitions above that

\begin{equation*} \mathbb{N}\subset\mathbb{N}_0\subset\mathbb{Z}\subset\mathbb{Q}, \end{equation*}

where "\(\subset\)" is used to mean "is a subset of"; symbolically

\begin{equation*} A\subset B \Leftrightarrow (\forall x\in A, a\in B). \end{equation*}

Unfortunately, not all numbers are rational. For example, in number theory we are sometimes interested in solutions to polynomial equations, such as

\begin{equation} x^2-2=0\text{.}\label{Eq-1_1}\tag{1.1.1} \end{equation}

The solution to (1.1.1) is \(x=\sqrt{2},\) which most of us know is an irrational number. But to be sure we must prove it. First we introduce some notation.

\(\mathbb{Z}\) comes from the German word for numbers, Zahlen

Convention 1.1.1.

We say that \(a\) divides \(b,\) and write \(a\vert b,\) if there is some integer \(c\) such that \(b=ac.\)

By contradiction. Suppose that \(\sqrt{2}\in\mathbb{Q}.\) Then we may write \(\sqrt{2}=\frac{m}{n},\) where \(\frac{m}{n}\) is a completely reduced fraction (in lowest terms). Since \(\sqrt{2}=\frac{m}{n}\) we have

\begin{equation*} 2n^2=m^2 \end{equation*}

and so \(2\vert m^2,\) which implies that \(2\vert m.\) That there is an integer \(k\) such that \(m=2k.\) Now we have

\begin{equation*} 2n^2=m^2=(2k)^2=4k^2, \end{equation*}

which reduces to

\begin{equation*} n^2=2k^2. \end{equation*}

This is similar to the earlier case, and so it must be the case that \(2\vert n.\) Since \(2\vert m\) and \(2\vert n,\) the rational number \(\frac{m}{n}\) is not in lowest terms, a contradiction. Thus \(\sqrt{2}\) is irrational.

Similar to a calculus class, we denote the set of rationals and irrationals as \(\mathbb{R},\) and call this the real numbers.