8.29 Real numbers tutorial
Remark
We have an equality between a finite decimal representation and infinite decimal representation
Suppose that we denote the infinite decimal representation \(0.9999....\) by \(x\) that is
\(x=0.999...(E1)\)
\(10x=9.9999....=9+0.9999....(E2)\)
and from \((E2)-(E1)\): \(9x=9\) then \(x+1\)
therefore \(1=0.99999....(E3)\)
Now upon division by \(10,10^2,10^3,10^4,...\) the (E3) we obtain
\(0.1=0.0999....\)
\(0.01=0.00999.....\)
\(0.001=0.000999....\)
...........
(E4)
We can use these result to convert any finite decimal representation into an infinite decimal representation
For example \(0.758=0.757+0.001=0.757+0.000999...=0.757999...\)
Conversely, (E3) and (E4) can be used to convert any infinite decimal representation that has infinite nines into a finite decimal representation
-
Prove that given \(x,y\in Q\), \(x<y\) if and only if there exist \(p,r\in Z\) and \(q\in N\) such that \(x=\frac{p}{q},y=\frac{r}{q}\) with \(p<r\)
\(\Rightarrow\))Suppose that \(x<y\).
As \(x,y\in Q\) there exist \(a,b\in Z\) and \(c,d\in N\) such that \(x=\frac{a}{c},y=\frac{b}{d}\), hence \(\frac{a}{c}<\frac{b}{d}\), then \(ad<bc\).
Now let \(p=ad,r=bc\) and \(q=cd\) then \(x=\frac{p}{q},y=\frac{r}{q}\) with \(p<r\)
\(\Leftarrow\))Suppose that there exist \(p,r\in \Z\) and \(q\in\N\) such that \(x=\frac{p}{q},y=\frac{r}{q}\) with \(p<r\)
Hence \(pq<rq\) and thus \(x<y\)
Properties of inequality
Let \(a,b\in R\)
-
If \(a<b\) and \(c<0\) then \(ac>bc\)
-
If \(a<b\) and \(c<d\) then $a+c<b+d $
-
If \(a<b\) then \(-a>-b\)
-
If \(a<0\) and \(b<0\) then \(ab<0\)
-
\(a<b\) if and only if \(b-a>0\)
Let \(x,y\in R\) and \(n\in N\). Prove that if \(x,y>0\), \(x<y\) if and only if \(x^n<y^n\)
\(y^n-x^n=(y-x)(y^{n-1}+y^{n-2}x+...+yx^{n-2}+x^{n-1})\)