Proposition B.1.1.
The additive identity and multiplicative identity are unique. For each \(a\in\R\text{,}\) the additive inverse \(-a\) is unique; if \(a\ne0\text{,}\) then the multiplicative inverse \(a^{-1}\) is unique.
Section B.1 Foundations and Consequences
Subsection B.1.1 Ordered-Field Axioms
We start with a set \(\R\) with two operations \(+\) and \(\cdot\text{,}\) and a subset \(\R_{+}\subseteq \R\text{.}\) We call this an ordered field when the following hold.
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\(+\) and \(\cdot\) are commutative and associative, and \(a(b+c)=ab+ac\) for all \(a,b,c\in\R\text{.}\)
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There are distinct elements \(0,1\in\R\) such that \(a+0=a\) and \(a\cdot 1=a\) for all \(a\in\R\text{.}\)
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Every \(a\in\R\) has an additive inverse \(-a\) with \(a+(-a)=0\text{,}\) and every nonzero \(a\in\R\) has a multiplicative inverse \(a^{-1}\) with \(a\cdot a^{-1}=1\text{.}\)
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The set \(\R_{+}\) is closed under addition and multiplication.
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For each \(a\in\R\text{,}\) exactly one of \(a=0\text{,}\) \(a\in\R_{+}\text{,}\) \(-a\in\R_{+}\) holds.
We define \(a<b\) if and only if \(b-a\in\R_{+}\text{,}\) and \(a\le b\) if and only if \(a<b\) or \(a=b\text{.}\)
To characterize the real numbers, we add the least upper bound property: every nonempty subset of \(\R\) that is bounded above has a least upper bound in \(\R\text{.}\) An ordered field with this property is a complete ordered field.
Subsection B.1.2 Uniqueness Up to Isomorphism
A standard theorem says that any two complete ordered fields are isomorphic (as ordered fields). This is the precise sense in which there is only one real number system up to isomorphism.
Reference: [1], Chapter II, end of Section 4 (after the proof of existence of square roots), where he states that Properties I-VII determine the real number system up to a unique one-to-one correspondence preserving sums and products.
Subsection B.1.3 Immediate Consequences of the Axioms
The next results are typical short derivations from the axioms (compare [1], Chapter II, Sections 1-2).
Proof.
If \(0,0'\) are additive identities, then \(0=0+0'=0'\text{.}\) The same argument gives uniqueness of \(1\text{.}\) If \(b,c\) both satisfy \(a+b=0\) and \(a+c=0\text{,}\) then \(b=b+0=b+(a+c)=(b+a)+c=0+c=c\text{.}\) The multiplicative case is identical for \(a\ne0\text{.}\)
Proposition B.1.2.
For every \(a,b\in\R\text{:}\) \(a\cdot0=0\text{,}\) \((-1)a=-a\text{,}\) and \((-a)(-b)=ab\text{.}\)
Proof.
Since \(0+0=0\text{,}\) distributivity gives \(a\cdot0=a(0+0)=a\cdot0+a\cdot0\text{.}\) Adding \(-(a\cdot0)\) to both sides yields \(a\cdot0=0\text{.}\)
Next, \(((-1)+1)a=(-1)a+a=0\cdot a=0\text{,}\) so \((-1)a\) is an additive inverse of \(a\text{;}\) by uniqueness, \((-1)a=-a\text{.}\)
Then \((-a)(-b)=((-1)a)((-1)b)=((-1)(-1))ab\text{.}\) Also \(((-1)+1)(-1)=0\) gives \((-1)(-1)+(-1)=0\text{,}\) hence \((-1)(-1)=1\text{.}\) Therefore \((-a)(-b)=ab\text{.}\)
Proposition B.1.3.
If \(a<b\text{,}\) then \(a+c<b+c\) for every \(c\in\R\text{.}\) If \(a<b\) and \(0<c\text{,}\) then \(ac<bc\text{.}\) Also, \(a^2\ge0\) for every \(a\in\R\text{.}\)
Proof.
If \(a<b\text{,}\) then \(b-a\in\R_{+}\text{.}\) But \((b+c)-(a+c)=b-a\in\R_{+}\text{,}\) so \(a+c<b+c\text{.}\)
If additionally \(0<c\text{,}\) then \(c\in\R_{+}\) and \(b-a\in\R_{+}\text{.}\) Closure of \(\R_{+}\) under multiplication gives \(c(b-a)=bc-ac\in\R_{+}\text{,}\) hence \(ac<bc\text{.}\)
For \(a\in\R\text{,}\) exactly one of \(a=0\text{,}\) \(a\in\R_{+}\text{,}\) \(-a\in\R_{+}\) holds. If \(a=0\text{,}\) then \(a^2=0\text{.}\) Otherwise either \(a^2\in\R_{+}\) or \((-a)^2\in\R_{+}\text{,}\) and \((-a)^2=a^2\text{.}\) Thus \(a^2\) is never negative, i.e. \(a^2\ge0\text{.}\)
