8.27 Real numbers and completeness
1.Rational numbers and decimal expansions=9=
Proposition (Criterion for a rational number to have a finite decimal expansion)
Let \(x\in\mathbb{Q}\), \(x=\frac{p}{q},(p:q)=1\)
Then \(x\) has a finite decimal expansion \(\Leftrightarrow\) \(q=2^j\times 5^k\) with \(j,k\in\N_0\) (Equivalently \((p ~prime\wedge p|q)\Rightarrow (p=2\wedge q=5)\) )
\(\Leftarrow\)) Let \(x=\frac{p}{q}=\frac{p}{2^j\times 5^k}=\frac{p\times 5^{j-k}}{2^j\times 5^j}=\frac{m}{10^j}\) with \(m\in\N\)
\(\Rightarrow\)) By hypothesis: \(\frac{p}{q}=a_0+\frac{a_1}{10}+\frac{a_2}{100}+...+\frac{a_m}{10^m}\)
\(10^m\times \frac{p}{q}=10^m\times \sum_{j=0}^{m}a_j.10^{-j}\)
\(10^m\times \frac{p}{q}=\sum_{j=0}^{m}a_j.10^{m-j}\in\N\)
Then \(10^{m}\cdot p=q\cdot m~~~~with~m=\sum_{j=0}^{n}a_{j}10^{n-j}.\)
\(\Rightarrow q\mid10^{m}\cdot p~~~~~ Since~(p:q)=1,we~conclude~that~q\mid10^{m}\)
Now if r is a prime number such that \(r|q\), then \(r|10^m\Rightarrow r|10\Rightarrow r|2\times 5\Rightarrow r=2~or~r=5\)
2.Order and density in Q
Cross-multiplication method (order)
Let \(x,y\in Q,x=\frac{p}{q},y=\frac{r}{s}\)
We say that \(x<y\Leftrightarrow ps<qr\)
Density in Q
Let \(x,y\in Q,x<y\), then there exists \(z\in Q|x<z<y\)
We call this property Density of Q in Q
Consider \(z=\frac{x+y}{2}\)
\(\begin{aligned}z&=\frac{x+y}{2}<\frac{2\times y}{2}=y\\z&=\frac{x+y}{2}>\frac{2\times x}{2}=x\end{aligned}\Rightarrow x<z<y,z\in Q.\)
Proposition: Q has holes=9=
Recall that \(\nexists x\in Q|x^2=2\)
Such number is indeed a distant
Consider the sets \(A=\{x\in Q^+:x^2<2\}, B=\{x\in Q^+:x^2>2\}\)
Note that \(Q^+=\{x\in Q:x\geq0\}=A\cup B\),\(A\cap B=\emptyset\)
Also \(x\in A\wedge y\in B\Rightarrow x<y\)
Claim: \(\nexists z\in Q:x<z~~\forall x\in A~and~z<y~~\forall y\in B\)
We will prove that \(\begin{aligned}&1)~z>a~~~\forall a\in A\Rightarrow z^{2}\geq2\\&2)~z<b~~~\forall b\in B\Rightarrow z^{2}\leq2\end{aligned}\). Then \(\Rightarrow z^{2}=2\Rightarrow z\notin Q.\)
Assume \(z>a~~~\forall a\in A\). Suppose that \(z^2<2\). Now consider a positive but very small \(h\in (0,1)\)
Consider:
We conclude that: \((z+h)^2<2\Rightarrow z+h\in A\Rightarrow z<z+h<z\) Contradiction!
Conclusion: \(z>a~~~\forall a\Rightarrow z^2\geq 2\)
Similarly: \(z<b~~~\forall b\in B\Rightarrow z^2\leq 2\)
Remember that we are under the hyp: \(z^2<2\)
Choose \(h\) so small that \(h(2z+1)<2-z^2\)
\(0<h<\frac{2-z^{2}}{2z+1}\)
3.Real numbers
The set \(R\) is a set containing \(\mathbb{Q}\) extending the operations \(+,\cdot\) with the same order relation.
The Axiom of continuity
If \(A,B\subseteq R,A\cup B=R\), \(\left(a\in A\wedge b\in B\right)\Rightarrow a<b\)
Then there exists \(\alpha\in R/a\leq\alpha\leq b~~~\forall a\in A~~~\forall b\in B\)
4.Absolute value function
\(|\cdot |:R\rightarrow R_{\geq 0}\)
\(\left.|x|=\begin{cases}x&if~x\geq0\\-x&if~x<0.\end{cases}\right.\)
\(\left|7\right|=7\\\left|-3\right|=-\left(-3\right)=3\)
Properties
\(|a|\geq0~~~\forall a\in R\\|a|=0\Rightarrow a=0\\|a\cdot b|=|a|\cdot|b|\)
Triangular inequality: \(a,b\in\mathbb{R}\)
\(|a+b|\leq|a|+|b|\)
Interpretation: \(|a|~is~the~size~of~a\in R\)
Also, we may define a notion of distance in R: \(d\left(a,b\right)=\left|b-a\right|=\left|a-b\right|\)
Reversed triangular inequality: \(a,b\in R\)
\(|a|=|a-b+b|\leq|a-b|+|b|\Rightarrow|a|-|b|\leq|a-b|\\|b|=|b-a+a|\leq|b-a|+|a|\Rightarrow|b|-|a|\leq|a-b|\)
Then we can get \(\vert\vert a\vert-\vert b\vert\vert\leq\vert a-b\vert\)
Bernoulli's inequality
Let \(x\in R,x>-1\). Then for any \(n\in\N\)
\((1+x)^{n}\geq1+nx\)
\(P(n):(1+x)^{n}\geq1+nx\)
We will use induction:
\(P(1):1+x\geq1+x~~~~true!!\)
Assumer \(P(n)\) true and prove \(P(n+1)\) also true.
We assume that \((1+x)^n\geq 1+nx\)
Now consider \(\left(1+x\right)^{n+1}=\left(1+x\right)^{n}\cdot\left(1+x\right)\geq\left(1+nx\right)\cdot\left(1+x\right)\)
\(=1+x+nx+nx^{2}=1+\left(n+1\right)\cdot x+nx^{2}\)
\((1+x)^{n+1}\geq1+(n+1)\cdot x~~~Since~nx^{2}\geq0\)
Towards a different characterization of completeness in R=9=
\(A=\{1-\frac{1}{m}:m\in N\}=\{0,\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{m-1}{m},...\}\) \(\frac{1}{m}\) hints you to use 3.Archimedean property of N in R
\(S=1\) is an upper bound
\(1-\frac{1}{m}\leq1.~Hence~0\leq\frac{1}{m}\)
Is there any upper bound \(S\prime <1\)?
Suppose \(S\prime <1\) is upper bound
\(1-\frac{1}{m}\leq S\prime<1\)
Consider \(m\) such that \(0<\frac{1}{m}<1-s\prime \\s\prime-1<-\frac{1}{m}\) (It is possible since \(1-s\prime >0\) by assumption)
Then \(1-\frac{1}{m}>1+S\prime -1=S\prime\Rightarrow S\prime\) is not an upper bound
We conclude that \(S=1\) is the lowest upper bound for A
Supremum and infimum
\(A\subset R\), we say that \(S\in R\) is the supremum of A if
\(1)a\leq S~~~\forall a\in A\left(S~is~an~upper~bound\right)\\2)If~S^{\prime}\geq a~~~\forall a\in A\Rightarrow S\leq S^{\prime}\)
This means that S is the lowest upper bound(l.u.b)
Similarly:
We say that \(i\in R\) is the infimum of \(A\) if:
\(1)i\leq a~~~\forall a\in A\left(i~is~an~lower~bound~for~A\right)\\2)If~i^{\prime}\in R~|~i^{\prime}\leq a~~~~\forall a\in A\Rightarrow i^{\prime}\leq i\)
\(A=[-1,3)\cup\{4\} ~~~Sup(A)=4~~~~~Inf(A)=-1\)