3.3 The irrationals and the reals
We first show that there exists irrational numbers (i.e., numbers that are not rational) and show that this means that there exists subsets of \(\mathbb{Q}\) whose lower upper bound is not rational (and hence we need a bigger number system).
There does not exists \(x\in \mathbb{Q}\) such that \(x^2=2\).
For the sake of a contradiction, suppose there exists \(x =\frac{a}{b} \in \mathbb{Q}\) such that \(x^2 =2\). Without loss of generality, since \(x^2=(-x)^2\) and \(0^2=0\), we can assume \(x>0\), i.e. \(a,b \in \mathbb{Z}_+\). Furthermore, if \(0<x\leq 1\), then \(x^2\leq 1<2\), and if \(x\geq 2\) then \(x^2\geq 4>2\). So we assume that \(1<x<2\).
Let \(A = \{r \in \mathbb{Z}_+:rx \in \mathbb{Z}\} \subseteq \mathbb{Z}\). Note that \(A\) is non-empty since, \(bx = a \in \mathbb{Z}_+\), so \(b \in A\). We have that \(A\) is bounded below by \(0\), so by the Well Ordering Principle, \(A\) contains a minimal element, call if \(m\). We will prove that \(m\) is not minimal by finding \(0<m_1<m\) with \(m\in A\). This will be a contradiction to the Well Ordering Principle.
Define \(m_1=m(x-1)= mx-m \in \mathbb{Z}\). Since \(1<x\), we have \(x-1>0\) so \(m_1=m(x-1)>0\). Similarly, since \(x<2\) we have \(x-1<1\) so \(m_1=m(x-1)<m\). Hence \(0<m_1<m\). Now \[m_1x=m(x-1)x=mx^2-mx = 2m-mx \in \mathbb{Z}.\] Hence \(m_1 \in A\) and \(0<m_1<m\) which is a contradiction.□
Since irrational numbers exists if we restrict ourselves to only using rational number then there are many unanswerable questions. From simple geometrical problems (what is the length of the diagonal of a square with length side 1), to the fact that there are some bounded set of rationals which do not have a rational least upper bower.
Consider the set \[A = \{x\in \mathbb{Q}:x^2<2\}.\] We have \(A\) is bounded above (for example, by \(2\) or \(10\)), but we show it does not have a least upper bound in the rational.
Let \(\alpha =\frac{a}{b} \in \mathbb{Q}\) be the least upper bound of \(A\) and note we can assume \(\alpha>0\). We either have \(\alpha^2<2\), \(\alpha^2=2\) or \(\alpha^2>2\). We will show that all three of these cases leads to a contradiction.
Case 1: \(\alpha^2<2\). We show that in this case \(\alpha\) is not an upper bound by finding \(x \in A\) such that \(\alpha<x\).
[Scratch work: We look for \(c\in \mathbb{Z}\) such that \(\left(\alpha+\frac{1}{bc}\right)^2=\left(\frac{ac+1}{bc}\right)^2<2\). Rearranging, we get \(a^2c^2+2ac+1<2b^2c^2\), then \(2ac+1<c^2(2b^2-a^2)\). Since \(\alpha^2<2\), we know \(a^2<2b^2\), so \(2b^2-a^2>0\). Since \(2b^2-a^2 \in \mathbb{Z}\), we have \(2b^2-a^2\geq 1\), so \(c^2(2b^2-a^2)\geq c^2\). So to find \(c\) such that \(2ac+1<c^2\), i.e. \(0<c^2-2ac-1\), i.e. by completing the square \(0<(c-a)^2-(a^2+1)\). To simplify our life, let us take \(c\) to be a multiple of \(a\), say \(ka\), then we are looking for \(0<(k-1)^2a^2-a^2-1 = ((k-1)^2-1)a^2-1\), i.e. \((k-1)^2-1)>2\), so \(k=3\) should work, i.e. \(c=3a\).]
Let \(x = \alpha+\frac{1}{3ab}>\alpha\). We prove that \(x\in A\) (and hence \(\alpha\) is not an upper bound) by showing \(x^2<2\). We have \[\begin{align*} x^2 &= \left(\frac{3a^2+1}{3ab}\right)^2 &\\ &= \frac{9a^4+6a^2+1}{9a^2b^2} &\\ &< \frac{9a^4+9a^2}{9b^2a^2} & \text{ (since $6a^2+1<9a^2$ )}\\ &=\frac{a^2+1}{b^2}& \\ &< \frac{a^2+(2b^2-a^2)}{b^2}& \text{ (since $a^2<2b^2$)}\\ &<2.& \end{align*}\]
Case 2: \(\alpha^2=2\). This is a contradiction to Theorem 3.8
Case 3: \(\alpha^2>2\). We leave this as an exercise. [Hint: Find appropriate \(c\) so that \(x = \alpha - \frac{1}{bc} \in \mathbb{Q}\) is such that \(x^2>2\). Argue that \(x\) is an upper bound for \(A\) and \(x<\alpha\) to conclude \(\alpha\) is not the least upper bound. ]
Notice that in the above we made sure to have \(x>\alpha\) by setting \(x = \alpha+\epsilon\) where \(\epsilon>0\). By doing so, we reduced the numbers of properties we needed \(x\) to have.
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We use this as a motivation to introduce the real numbers.
The set of real numbers, denoted \(\mathbb{R}\), equipped with addition \(+\), multiplication \(\cdot\) and the order relation \(<\) satisfies axioms (A1) to (A11), (O1) to (O4) and the Completeness Axiom
Completeness Axiom: Every non-empty subset \(A\) of \(\mathbb{R}\) which is bounded above has a least upper bound.
It can be shown that there is exactly one quadruple \((\mathbb{R};+;\cdot;<)\) which satisfies these properties - up to isomorphism. We do not discuss the notion of isomorphism in this context here (although we will later look at it in the context of groups) save to remark that any two real number systems are in bijection (we will define this later) and preserves certain properties. This allows us to speak of the real numbers.
There are several ways of constructing the real number system from the rational numbers. One option is to use Dedekind cuts. Another is to define the real numbers as equivalence classes of Cauchy sequences of rational numbers (you will explore Cauchy sequences in Analysis).You may continue to imagine the real numbers as a number line as you did pre-university.
The definition of the absolute value is the same for real numbers as for the rational numbers, as is the notion of bounded sets (and therefore all the results in the previous section still holds for \(\mathbb{R}\)).
We can use the absolute value to define a distance or metric on \(\mathbb{R}\). To do so we define \[d(x,y) = |x-y|\] for any two points \(x,y\in \mathbb{R}\). This distance has the following properties, for any \(x,y,z \in \mathbb{R}\):
\(d(x,y)\geq 0\) and \(d(x,y)=0\) if and only if \(x=y\);
\(d(x,y)=d(y,x)\);
\(d(x,y)\leq d(x,z)+d(z,y)\).
We deduce two important result about \(\mathbb{R}\).
Every non-empty subset \(A\) of \(\mathbb{R}\) bounded below has a greatest lower bound.
Let \(A\subset \mathbb{R}\) be non-empty and bounded below. Let \(c\in \mathbb{R}\) be a lower bound. Define the set \[B=\{-x:x\in A\}.\] Let \(x\in B\), so \(-x\in A\). Since \(c\) is a lower bound, \(-x\geq c\), i.e. \(x\leq -c\). So \(-c\) is an upper bound for \(B\), and \(B\) is non-empty [since \(A\) is non-empty]. By the Completeness Axiom \(B\) has a least upper bound, \(u\in \mathbb{R}\). Let \(\ell = -u\). We prove that \(\ell\) is the greatest lower bound for \(A\).
We first show \(\ell\) is a lower bound. Let \(x\in A\), then \(-x\in B\) so \(-x\leq u\). Hence \(x\geq -u=\ell\).
We now show \(\ell\) is the greatest lower bound by showing any real number bigger than \(\ell\) is not a lower bound. Let \(y\in \mathbb{R}\) be such that \(y>\ell\), so \(-y<-\ell=u\). Now \(-y\) is not an upper bound for \(B\) since \(u\) is the least upper bound of \(B\). So by definition, there exists \(b\in B\) such that \(b>-y\), i.e. \(-b<y\). Since \(b\in B\), we have \(-b\in A\). Hence \(y\) is not a lower bound for \(a\).
So \(\ell\) is the greatest lower bound for \(A\).□
For any \(x\in \mathbb{R}\) there exists \(n\in \mathbb{Z}_+\) such that \(x\leq n\).
We prove this by contradiction.
\(\big[\neg(\forall x \in \mathbb{R}, \exists n\in \mathbb{Z}_+\) such that \(x\leq n\)) \(\iff \exists x \in \mathbb{R}\) such that \(\forall n \in \mathbb{Z}_+, x>n \big]\).
Suppose there exists \(x\in \mathbb{R}\) such that \(n<x\) for all \(n\in \mathbb{Z}_+\). In particular, this means \(\mathbb{Z}_+\subseteq \mathbb{R}\) is bounded above. So by the completeness axiom, \(\mathbb{Z}_+\) has a least upper bound \(\alpha\) [in \(\mathbb{R}\)]. Since \(\alpha\) is the least upper bound, \(\alpha-\frac{1}{2}\) is not an upper bound, i.e. there exists \(b \in \mathbb{Z}_+\) such that \(\alpha-\frac{1}{2}<b<\alpha\). But then, \(b+1\in \mathbb{Z}\) and \(\alpha+\frac{1}{2}<b+1<\alpha+1\) so \(\alpha<b\). This contradicts the fact \(\alpha\) is an upper bound for \(\mathbb{Z}_+\).□
The above property is named after Archimedes of Syracuse (Sicilian/Italian mathematician, 287BC - 212 BC) although when Archimedes wrote down this theorem, he credited Eudoxus of Cnidus (Turkish mathematician and astronomer, 408BC - 355BC). It was Otto Stolz (Austrian mathematician, 1842 - 1905) who coined this property - partly because he studied fields where this property is not true and therefore needed to coin a term to distinguish between what is now know as Archimedean fields and non-Archimedean fields.
We finish this section by introducing notation for some common subsets of \(\mathbb{R}\).
Let \(a,b \in \mathbb{R}\) are such that \(a\leq b\), we denote:
the open interval of \(a,b\) by \((a,b)=\{x\in \mathbb{R}:a<x<b\}\);
the closed interval of \(a,b\) by \([a,b] = \{x \in \mathbb{R}:a\leq x\leq b\}\).
\([a,b) = \{x \in \mathbb{R}: a\leq x<b\}\); \((a,b] = \{x\in \mathbb{R}:a<x \leq b\}\);
\((a,\infty) = \{x \in \mathbb{R}:a<x\}\); \([a,\infty) = \{x\in \mathbb{R}:a\leq x\}\);
\((-\infty,b) = \{x \in \mathbb{R}: x<b\}\); \((-\infty,b] = \{x \in \mathbb{R}: x\leq b\}\).
By convention we have \((a,a)=[a,a)=(a,a]=\emptyset\), while \([a,a]=\{a\}\).