Definition 1.1.1. Let \(\mathcal L_{\mathrm{field}}\) be the first-order language \(\mathcal L_{\mathrm{field}} = \lbrace 0, 1, +, -, \cdot\rbrace,\) where \(0\) and \(1\) are constant symbols, \(+\) and \(\cdot\) are binary function symbols, and \(-\) is a unary function symbol.
Definition 1.1.2. Let \(T_{\mathrm{field}}\) be the \(\mathcal L_{\mathrm{field}}\)-theory which consists of the following axioms:
- \(\forall x,y,z \ ((x + y) + z = x + (y + z)),\)
- \(\forall x,y \ (x + y = y + x),\)
- \(\forall x \ (x + 0 = x),\)
- \(\forall x \ (x + (-x) = 0),\)
- \(\forall x,y,z \ ((x \cdot y) \cdot z = x \cdot (y \cdot z)),\)
- \(\forall x,y \ (x \cdot y = y \cdot x),\)
- \(\forall x \ (x \cdot 1 = x),\)
- \(\forall x,y,z \ ((x \cdot y) + (x \cdot z) = x\cdot (y+z) ),\)
- \(0 \neq 1,\)
- \(\forall x \ (x \neq 0 \rightarrow \exists y \ (x \cdot y = 1)).\)
Proposition 1.1.3. For all \(x,y,z\)
- \(x+y=x+z\Rightarrow y=z,\)
- \(x+y=x\Rightarrow y=0,\)
- \(x+y=0\Rightarrow y=-x,\)
- \(-(-x)=x,\)
Proof.
(1) Since \(x+y=x+z,\) \((x+y)+(-x) = (x+z) + (-x).\) Hence \(y=z.\)
(2) Since \(x+0=x,\) we have \(x+y=x+0.\) Hence \(y=0,\) by (1).
(3) Since \(x+(-x)=0,\) we have \(x+y=x+(-x).\) Hence \(y=-x,\) by (1).
(4) Since \(x+(-x)=0,\) \(-(-x)=x,\) by (3).\(\square\)
Proposition 1.1.4. For all \(x,y,z\)
- \((x\ne 0 \land x\cdot y=x\cdot z)\Rightarrow y=z,\)
- \((x\ne 0 \land x\cdot y=x)\Rightarrow y=1,\)
- \(x\cdot y=(-x)\cdot (-y),\)
- \((-1)\cdot x = -x.\)
Proof.
(1) Let \(x\cdot w=1.\) Since \((x\cdot y)\cdot w = (x\cdot z)\cdot w,\) \((x\cdot w)\cdot y = (x\cdot w)\cdot z.\) Hence \(y=z.\)
(2) Since \(x\cdot 1 = x,\) \(x\cdot y = x\cdot 1.\) Since \(x\ne 0,\) by (1), \(y=1.\)
(3) Since \(x\cdot y+(-x)\cdot y=0,\) we have \((x\cdot y) = - ((-x)\cdot y).\) Similarly, that \((-x)\cdot y+(-x)\cdot (-y)=0\) implies \((-x)\cdot (-y) = - ((-x)\cdot y).\) Since the additive inverse of \((-x)\cdot y\) is unique, \(x\cdot y=(-x)\cdot (-y).\)
(4) Since \(x+((-1)\cdot x)=(1+(-1))\cdot x = 0,\) by (3) in Proposition 1.1.3, \((-1)\cdot x = -x.\)\(\square\)
Definition 1.1.5. \(0\) and \(1\) are called the additive identity and multiplicative identity, respectively. \(-x\) is called the additive inverse of \(x.\)
We write \(y=1/x,\) if \(x\ne 0\) and \(x\cdot y = 1.\)
Further, we often write
in place of
\[x+(-y),x\cdot \left(\frac 1 y \right),x+x,x+x+x,x\cdot x,x\cdot x\cdot x,\dots .\] An \(\mathcal L_{\mathrm{field}}\)-structure \({\mathcal F}=(\mathbb F, 0, 1, +, -, \cdot)\) is called a field if \({\mathcal F}\vDash T_{\mathrm{field}},\) and \(x\) is called a scalar if \(x\in {\mathbb F}.\)
Definition 1.1.6. Let \(\mathcal{L}_{\mathrm{of}} = \mathcal{L}_{\mathrm{field}} \cup \{<\},\) where \(<\) is a binary relation symbol, and let \(T_{\mathrm{of}}\) be the \(\mathcal{L}_{\mathrm{of}}\)-theory that consists of the field axioms together with the following five axioms:
- \(\forall x \ \neg(x < x),\)
- \(\forall x,y,z\ ((x < y \land y < z) \rightarrow x < z),\)
- \(\forall x,y\ (x\ne y \rightarrow (x < y \lor y < x)),\)
- \(\forall x,y,z\ (x < y \rightarrow x + z < y + z),\)
- \(\forall x,y,z\ ((0 < x \land 0 < y) \rightarrow 0 < x \cdot y).\)
An \(\mathcal{L}_{\mathrm{of}}\)-structure \(\mathcal{F} = (\mathbb{F}, 0, 1, +, -, \cdot, <)\) is called an ordered field if \(\mathcal{F} \vDash T_{\mathrm{of}}.\)
Proposition 1.1.7. For all \(w,x,y,z,\)
- \(0<x \Rightarrow -x<0,\)
- \((x<0\land y<0) \Rightarrow 0<x\cdot y,\)
- \((x<y\land z<w) \Rightarrow x+z<y+w,\)
- \(0\le x^2.\)
Proof.
(1) From (OF14) in Definition 1.1.6, for all \(x,y,u,v,\) if \(x<y\) then \(u<v\) where \(v-u = y-x.\) Thus \(0<x\) implies \(-x<0.\)
(2) Since \(0<-x\) and \(0<-y,\) \(0<(-x)\cdot (-y).\) Thus, by (3) of Proposition 1.1.4, \(0<x\cdot y.\)
(3) Since \(x<y,\) it follows that \(x+z<y+z.\) Similarly, since \(z<w,\) we have \(z+y<w+y.\) Since the order relation is transitive, \(x+z<w+y.\)
(4) Since the order is comparable if \(0\ne x\) then \(0<x\) or \(x<0.\) If \(0<x\) then \(0\cdot x<x\cdot x,\) so that \(0<x^2.\) If \(x<0\) then \(0<x^2.\) The only case left is that \(x=0.\) If \(x=0\) then \(x^2=0\cdot x=0.\) Therefore \(0\le x^2.\) \(\square\)
Theorem 1.1.8. The real number system \(\mathcal R=(\mathbb R,0,1,+,-,\cdot,<)\) has the least-upper-bound property. Also, \(\mathcal R\) is an ordered field. Equivalently,
\[\mathcal R \vDash T_{\mathrm{of}}.\]We henceforth identify \(\mathbb N\), \(\mathbb Z\), and \(\mathbb Q\) with their respective images \(\mathbb N^{\ast}\), \(\mathbb Z^{\ast}\), and \(\mathbb Q^{\ast}\) in \(\mathbb R.\) Under this identification, we have
\[\mathbb N \subseteq \mathbb Z \subseteq \mathbb Q \subseteq \mathbb R.\]In particular, the rational field \(\mathcal Q=(\mathbb Q,0,1,+,-,\cdot,<)\) is a subfield of \(\mathcal R.\)
Theorem 1.1.9. For all \(x\in \mathbb R\) and all \(y\in \mathbb R_{>0},\) there exists \(n\in \mathbb N\) such that
\[x<ny.\]Proof. Let \(x\in \mathbb R\) and \(y\in\mathbb R_{>0}\) be given, and let \(S\) be the set of all \(ny,\) where \(n\) runs through the positive integers. That is,
\[S = \lbrace ny : n\in \mathbb N\rbrace.\]We suppose, for contradiction, that \(S\) is bounded above. Then \(S\) has the supremum in \(\mathbb R,\) by Theorem 1.1.8. Put \(\alpha = \sup S.\) There is an element \(my\) such that \(\alpha - y < my.\) But then we have \(\alpha < (m+1)y\in S,\) which contradicts that \(\alpha\) is an upper bound for \(S.\) Therefore it is concluded that the Theorem 1.1.9 is in fact true. \(\square\)
Theorem 1.1.10. For all \(x,y\in \mathbb R,\) if \(x<y\) then there exists \(p\in \mathbb Q\) such that
\[x<p<y.\]Proof. Let \(x\) and \(y\) be reals such that \(x<y.\) There exists \(n\in \mathbb N\) such that \(1<n(y-x),\) by Theorem 1.1.9. Hence we can choose such a number and denote it by \(n.\) Then we consider the set
\[S = \lbrace k \in \mathbb Z : nx < k\rbrace.\]Theorem 1.1.9 implies that \(S\) is nonempty. It is also bounded below, so \(S\) has the least element; denote it by \(m.\) Then we have
\[nx < m \le nx + 1<ny.\]Since \(0<n\) it follows that
\[x < \frac m n < y.\]This proves Theorem 1.1.10. \(\square\)
Theorem 1.1.11. For all \(y\in \mathbb R_{>0}\) and all \(n\in \mathbb N,\) there exists the unique \(x\in \mathbb R_{>0}\) such that
\[x^n = y.\]Proof. It is clear that there is at most one such \(x,\) since \(0<x_1<x_2\) implies \(x_1^n<x_2^n.\)
We consider the set
If \(x=y/(1+y)\) then \(0<x<1.\) Since \(x^n<x<y,\) \(x\in S.\) Thus \(S\) is nonempty. If \(x>1+y\) then \(y<x<x^n,\) so that \(x\notin S.\) Thus \(1+y\) is an upper bound for \(S.\) Therefore there exists \(\alpha\) such that \(\alpha = \sup S.\)
We suppose, for contradiction, that \(\alpha^n<y\) or \(y<\alpha^n.\)
We first assume \(\alpha^n<y.\) Choose \(h\) so that
Then
\[(\alpha + h)^n - {\alpha}^n < h\cdot n(\alpha + h)^{n-1}<h\cdot n(\alpha + 1)^{n-1} < y-{\alpha}^n.\]Thus \((\alpha + h)^n < y,\) so that \(\alpha + h \in S.\) Since \(\alpha<\alpha+h,\) this result contradicts the fact that \(\alpha\) is an upper bound for \(S.\)
We now assume \(y<\alpha^n.\) Choose
Then \(0<h<\alpha.\) If \(\alpha - h \le x,\) we conclude that
\[\alpha^n - x^n \le \alpha^n - (\alpha - h)^n<h\cdot n\alpha^{n-1} = \alpha^n - y.\]Thus \(y<x^n\) so that \(x\notin S.\) Since \(\alpha -h<\alpha,\) this results contradicts the fact that \(\alpha\) is the supremum for \(S.\)
Therefore \(\alpha^n=y,\) and this prove Theorem 1.1.11.\(\square\)
Definition 1.1.12. Let \(x\in \mathbb R.\) We define \(n\)-th root of \(x\) as follows:
\[x^{1/n} = \begin{cases} \text{the unique }y>0\text{ such that }y^n=x, & \text{if } x>0. \\ 0, & \text{if } x=0,\\ \end{cases}\]Proposition 1.1.13. For all \(x,y\in \mathbb R_{>0}\) and \(n\in \mathbb N,\)
\[(x\cdot y)^{1/n} = x^{1/n}\cdot y^{1/n}.\]Proof. Let \(x,y\in \mathbb R_{>0}\) be given. We let \(a=(x\cdot y)^{1/n}, b=x^{1/n},\) and \(c=y^{1/n}.\) Then we have \(a^n=x\cdot y, b^n=x,\) and \(c^n=y.\) Hence \(a^n=b^n \cdot c^n=(b\cdot c)^n,\) so that
\[\left(a^n\right)^{1/n} = \left((b \cdot c)^n\right)^{1/n}\]by the uniqueness assertion of Theorem 1.1.11. Since
\[\left(a^n\right)^{1/n} = a\quad \text{and} \quad \left((b\cdot c)^n\right)^{1/n}=b\cdot c,\]it follows that \(a=b\cdot c.\) By definition of \(a,b,\) and \(c,\) therefore,
\[(x\cdot y)^{1/n} = x^{1/n}\cdot y^{1/n}.\tag*{\(\square\)}\]Exercises
Exercise 1. Prove that for all \(x\in \mathbb R\) and \(n\in \mathbb N,\)
\[nx = n\cdot x.\]Proof. By Definition 1.1.2 and Definition 1.1.5,
\[\begin{align} nx &= \underbrace{x + \cdots + x}_{n\ \text{times}} = 1\cdot x + \cdots + 1\cdot x \\ &= (\underbrace{1 + \cdots + 1}_{n\ \text{times}})\cdot x\\ &= n\cdot x.\tag*{\(\square\)} \end{align}\]Exercise 2. Prove that for all \(x,y\in \mathbb R,\)
- \(0<1,\)
- \(x<y \Rightarrow -y<-x,\)
- \((0<x \land 0<xy) \Rightarrow 0<y,\)
- \(0<x<y \Rightarrow 1/y<1/x,\)
- \(x<y \Rightarrow x<(x+y)/2<y.\)
Proof.
(1) We have \(0\le 1^2.\) Since \(1^2 = 1\cdot 1= 1\) and \(0\ne 1,\) \(0<1.\)
(2) Suppose, for contradiction, that \(-x\le -y.\) Since \(x<y,\) it follows that \(x+(-x)<y+(-y),\) which contradicts \(x+(-x)=y+(-y).\) Hence \(-y<-x.\)
(3) Suppose, for contradiction, that \(y\le 0.\) If \(y=0,\) then \(x\cdot y = 0,\) which contradicts \(0<xy.\) If \(y<0,\) then \(0<-y,\) by (1). Thus \(0<x\cdot (-y).\) It yields \(0<-(x\cdot y),\) since \(x\cdot(-y)= -(x\cdot y).\) Therefore \(xy<0,\) which contradicts \(0<xy.\) Consequently, it is concluded that \(0<y.\)
(4) If \(1/x\le 1/y\) then \(x\cdot(1/x)<y\cdot(1/x)\le y\cdot(1/y).\) If follows that \(x\cdot(1/x)<y(1/y),\) which contradicts \(x\cdot(1/x)=1=y\cdot(1/y).\) Therefore it is concluded that \(1/y< 1/x.\)
(5) Since \((1/2)\cdot (1+1) = 1,\) we have \(((x+y)/2)\cdot(1+1) = x+y.\) We also have \(x\cdot (1+1)= x + x\) and \(y\cdot (1+1) =y+y.\) Since \(x<y\) implies \(x+x<x+y<y+y,\) we finally get \(x<(x+y)/2<y.\) \(\square\)
Exercise 3. Let \(r\in \mathbb Q,r\ne 0,\) and \(x\notin \mathbb Q.\) Prove that \(r+x\notin \mathbb Q\) and \(rx\notin \mathbb Q.\)
References
- Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.
- Munkres, J. R. (2000). Topology (2nd ed.). Pearson.