10.14 Axioms of real numbers

Axioms of real numbers

\(\R\) has two basic operations: \(+\) and \(\cdot\)

Properties

Associativity for \(+\): \(\forall a,b,c\in\mathbb{R},\left(a+b\right)+c=a+\left(b+c\right)\)
Commutativity for \(+\): \(\forall a,b\in\mathbb{R},a+b=b+a\)
Neutral number: \(\exists sn\in R\) such that \(sn+a=a+sn=a\) (called zero 0) special number
Opposites: \(\forall a,\exists \tilde{a}:a+\tilde{a}=\tilde{a}+a=sn\) (denoted \(-a\))

If the set satisfies 1 2 3 4, then we call it Abelian Group

Theorem

There is only one \(sn\)

Proof

Assume we have \(2\) \(sn\) which is \(sn_1\) and \(sn_2\)

Then \(sn_1+sn_2=sn_2=sn_1\)

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Exercise: For each \(a\in R\), there is only one opposite of \(a\)

Proof

Assume we have \(2\) opposites of \(a\) which are \(a_1\) and \(a_2\)

Then \(a+a_1=0=a+a_2\)

Since \(a+a_1=a+a_2\), then \(a+a_1+a_1=a+a_2+a_1\Rightarrow (a+a_1)+a_1=(a+a_1)+a_2\Rightarrow 0+a_1=0+a_2\Rightarrow a_1=a_2\)

More Axioms

Associativity for \(\cdot\) : \((ab)c=a(bc)\)
Commutativity for \(\cdot\) : \(ab=ba\)
Identity: \(\exists1\neq0,~\forall a\in R:a\cdot1=1\cdot a=a\)
Inverse: \(\forall a\neq0,\exists a^{-1}:a\cdot a^{-1}=a^{-1}\cdot a=1\)

Theorem

\(1\) is unique

Proof

Assume we have \(2\) \(sn\) which is \(sn_1\) and \(sn_2\)

Then \(sn_1\cdot sn_2=sn_1=sn_2\)

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\(a^{-1}\) is unique

Assume we have \(2\) inverses of \(a\) which are \(a_1\) and \(a_2\)

Then \(a\cdot a_1=1=a_2\cdot a\)

Since \(a\cdot a_1=a_2\cdot a\), then \(a_1\cdot a\cdot a_1=a_1\cdot a_2\cdot a\Rightarrow\left(a_1\cdot a\right)\cdot a_1=a_2\cdot\left(a_1\cdot a\right)\Rightarrow1\cdot a_1=a_2\cdot1\Rightarrow a_1=a_2\)

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Distributive Law: \(\forall a,b,c\in\R,a(b+c)=ab+ac\)

If the set satisfies 1 2 3 4 5 6 7 9, then we call it Commutative Ring

If the set satisfies 1 2 3 4 5 6 7 8 9, then we call it Field

Example

Axioms 1,...9 for \(\R\): \(\R\) is a field

\(\mathbb{Q}\) is a field

\(\Z\) is a commutative ring

\(\N\) is none of this

\(\mathbb{C}\) is not a total ordered field but a field

Proposition

\(\forall a\in R,0\cdot a=a\cdot 0=0\)

Proof

Let \(x_0=0\cdot a\), we want to prove \(x_0=0\)

\(x_0+x_0=0\cdot a+0\cdot a=(0+0)\cdot a=0\cdot a=x_0\)

Then \(x_0+x_0=x_0\Rightarrow -x_0+(x_0+x_0)=-x_0+x_0=0\)

We have \(-x_0+(x_0+x_0)=0\)

Then \((-x_0+x_0)+x_0=0+x_0=x_0\)

Hence \(0=x_0\)

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Another part is totally same
If \(ax=ay\) and \(a\neq 0\), then \(x=y\) (Cancellation property)

Proof

We have \(ax=ay\) and \(a\neq 0\), then \(a^{-1}\left(ax\right)=a^{-1}\left(ay\right)\Rightarrow\left(a^{-1}a\right)x=\left(a^{-1}a\right)y\Rightarrow1\cdot x=1\cdot y\Rightarrow x=y\)

Subtraction and Division

We define subtraction as \(a-b=a+(-b)\) (the first \(-\) means a 运算, the second \(-\) means a 负数)

We define division as if \(b\neq 0\), \(\frac{a}{b}=a\cdot b^{-1}\)

Proposition

\(0-a=-a\)

Proof

\(0-a=0+(-a)=-a\)
\(\forall a,-(-a)=a\)

Proof

\((-a)+a=0\), and let \(b=-a\)

We have \(b+a=0\text{ and opposite is unique}\Rightarrow a\) is \(-b\Rightarrow a=-b=-(-a)\)
\((-a)\cdot b=-(a\cdot b)\), \((-a)(-b)=ab\)
1. Proof
  
  \(a\cdot b+(-(a\cdot b))=0=a\cdot b+(-1)\cdot(a\cdot b)=a\cdot b+((-1)\cdot a)\cdot b=a\cdot b+(-a)\cdot b\)
  
  Then we know \(a\cdot b\) is the opposite of \((-a)\cdot b\), thus \((-a)\cdot b=-(a\cdot b)\) 2. Proof
  
  \((-a)(-b)=(-1)\cdot a\cdot (-1)\cdot b=-(-1)\cdot a\cdot b=1\cdot a\cdot b=a\cdot b\)
\(-a=(-1)\cdot a\)

Proof

\((-1)\cdot a+a=(-1)\cdot a+1\cdot a=\left(\left(-1\right)\right.+1)\cdot a=0\cdot a=0\)

Then \(a\) and \((-1)\cdot a\) are opposite

Strategy

Use the additive inverse to prove the form like "\(a=-a\)" or "\(a=-(-a)\)"

More specifically, the additive inverse is useful to prove two things are equal or additive inverse

Order

\(\R\) has an order \(\leq\) with the following properties

Properties

Reflexivity: \(\forall a,a\leq a\)
Antisymmetry: If \(a\leq b\) and \(b\leq a\Rightarrow a=b\)
Transitivity: if \(a\leq b\) and \(b\leq c\Rightarrow a\leq c\)

Definitions

We say :

\(a<b\) if \(a\leq b\) and \(a\neq b\)

\(a\geq b\) if \(b\leq a\)

\(a>b\) if \(a\geq b\) and \(a\neq b\)

Positive numbers: all \(a>0\)

Intervals for \(a<b\) : \((a,b)=\left\lbrace x\in R:a<x<b\right\rbrace\)

Trichotomy

\(\forall a,b\) it happens only one of this \(a<b\), \(a=b\), \(a>b\) (in real number, and power set has not)

Compatibility

\(a\leq b\Rightarrow a+x\leq b+x, \forall a,b,x\in R\)
\(a\leq b\) and \(x\geq0\Rightarrow ax\leq bx,\forall a,b\in R\)

Proposition

\(1>0,\)\(2a>a,\frac12a<a\)(\(\frac12=1\cdot2^{-1}\)) \(a>0\)
1. Proof
  
  \(1=1\cdot1=1^2>0\) 2. Proof
  
  Since \(1>0\), then \(1+1>0+1\Rightarrow2>1\)
  
  Multiply \(a>0\) then \(2a>a\) 3. Proof
  
  Since \(2>1\), then multiply \(2^{-1} >0\) in both sides we have \(2\cdot2^{-1}>1\cdot2^{-1}\Rightarrow1>1\cdot2^{-1}\)
  
  Multiply \(a>0\) then \(a\cdot2^{-1}<a\Rightarrow\frac12a<a\)
\(a>0\Leftrightarrow\left(-a\right)<0~~~~a<0\Rightarrow-a>0\)
1. Proof
  
  Since \(a>0\), then \(a+(-a)>-a\Rightarrow0>-a\) 2. Proof
  
  Since \(a<0\), then \(a+(-a)<-a\Rightarrow0<-a\)
\(a\leq b~and~x<0\Rightarrow ax\geq bx\)

Proof

Since \(a\leq b\) and \(-x>0\), then \(a\cdot(-x)\leq b\cdot(-x)\Rightarrow-ax\leq-bx\)

Then \(ax+(-ax)\leq ax+\left(-bx)\Rightarrow0\leq ax+\left(-bx\right)\Rightarrow bx\leq ax+\left(-bx\right)\right.+bx\Rightarrow bx\leq ax\)
\(a\leq b\Leftrightarrow (-a)\geq (-b)\)

Proof

Just let \(x=-1\) and use 3

Or

Proof

\(\Rightarrow\)) Since \(a\leq b\), then \(-a+a\leq-a+b\Rightarrow0\leq-a+b\Rightarrow-b\leq-a+b+\left(-b\right)\Rightarrow-b\leq-a\)

Then, by definition of inequality, \(-b\leq -a\Rightarrow -a\geq -b\)

\(\Leftarrow\)) Since \(-a\geq -b\), \(a+(-a)\geq a+(-b)\Rightarrow0\geq a+\left(-b\right)\Rightarrow b\geq a+\left(-b\right)+b\Rightarrow b\geq a\)

Then, by definition of inequality, \(b\geq a\Rightarrow a\leq b\)
\(a^2>0,\forall a \neq 0\) (defined \(a\cdot a\))

Proof

Since \(\forall a \neq 0\), either \(a>0\) or \(a<0\)

If \(a>0\) then \(a\cdot a>0\cdot a\Rightarrow a^2>0\)

If \(a<0\) then \(-a>0,-a\cdot (-a)>0\cdot (-a)\Rightarrow a^2>0\)