1.3 Real numbers are just the decimals

In most of the real analysis books, the real numbers are defined a more complicated way.

First of all, we should realize that for each Cauchy-sequence we have already associated its “limitvalue” decimal and Lemma 1.2.1 states that for two Cauchy-sequences we associate the same “limitvalue” if and only if they are equivalent. So, the second definition is not so scary anymore. We will not even use the second definition, only as a guiding light. You will see that it is not a bad guiding light. Just for fun, let us show that $\equiv$ is an equivalence relation (we recall the definition from MATH111).

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  $p\equiv p$. That is, $(p_n-p_n)\to 0$. Well, this is obvious.

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  If $p\equiv q$ then $q\equiv p$. That is, if $|p_n-q_n|\to 0$, then $|q_n-p_n|\to 0$. The two sequences are the same…

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  If $p\equiv q$ and $q\equiv r$ then $p\equiv r$. That is, if $(p_n-q_n)\to 0$ and $(q_n-r_n)\to 0$ then $|p_n-r_n|=|(p_n-q_n)-(q_n-r_n)|\to 0$. This is also obvious.

Of course the equivalence classes are formed by the guys whose “limitvalue”’s are the same.

Proof:  By definition, $r+s$ is the limitvalue of ${\{r_n+s_n\}}_{n=1}^{\mathrm{\infty }}$. Also, it is the limitvalue of ${\{(r+s)\}}_{n=1}^{\mathrm{\infty }}$. Hence the proposition follows from Lemma 1.2.1. $\mathrm{\square }$

(Commutativity) $a+b=b+a$ and $ab=ba$.
(Associativity) $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$.
(Distributivity) $a(b+c)$ = $ab+ac$.
(Zero Rule) $a+0=a=0+a$.
(Unity) $aÂ\mathrm{·}1=a$.
(Subtraction) The equation $a+x=0$ has a unique solution $x=-a$. Similarly, the equation $a+x=b$ has a unique solution $x=b-a$.
(Division) If $a\ne 0$, then the equation $ax=b$ has a unique solution $x=b/a$.

Let us show how to prove the Distributivity Law.

Proof:  The statement clearly holds for rational numbers. The point is that we can transfer the proofs for the reals using Cauchy-sequences. We need to prove that the two Cauchy-sequences ${\{{((r+s)t)}_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{(rt+st)}_n\}}_{n=1}^{\mathrm{\infty }}$ are equivalent (well they are exactly the same).

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  ${\{{((r+s)t)}_n\}}_{n=1}^{\mathrm{\infty }}\equiv {\{{(r+s)}_n{t}_n\}}_{n=1}^{\mathrm{\infty }}$ (by Proposition 1.3.1)

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  ${\{{(r+s)}_n\}}_{n=1}^{\mathrm{\infty }}\equiv {\{r_n+s_n\}}_{n=1}^{\mathrm{\infty }}$ (similarly)

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  So, ${\{{((r+s)t)}_n\}}_{n=1}^{\mathrm{\infty }}\equiv {\{(r_n+s_n)t_n\}}_{n=1}^{\mathrm{\infty }}={\{r_n{t}_n+s_n{t}_n\}}_{n=1}^{\mathrm{\infty }}$ .

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  ${\{r_n{t}_n+s_n{t}_n\}}_{n=1}^{\mathrm{\infty }}\equiv {\{{(rt)}_n+{(st)}_n\}}_{n=1}^{\mathrm{\infty }}$ (again by Proposition 1.3.1)

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  and finally ${\{{(rt)}_n+{(st)}_n\}}_{n=1}^{\mathrm{\infty }}\equiv {\{{(rt+st)}_n\}}_{n=1}^{\mathrm{\infty }}$.

$\mathrm{\square }$

It is not magic. One can explain all these things using just the digits of the decimals, only it will take much more time. In Section 1.1, we defined an ordering of the rationals. We can compare real numbers the following way.

Note that this is exactly the way you learn to compare decimals in high school. The following lemma is trivial.

Also, it is clear that if $q$ and $r$ are rational numbers and $q>r$ according to this definition, then $q$ is larger than $r$ according to the definition in Section 1.1. Using Cauchy-sequences again, we can prove the following laws for real numbers (you learned all of them in high school).

(Trichotomy) Either or .
(Addition Law)  if and only if .
(Multiplication Law) If $c>0$, then if and only if . If , then if and only if .
(Transitivity) If and , then .

Yes, the triangle inequality also holds for real numbers:

Of course, $|a|=a$ if $a>0$ and $|a|=-a$ if

Exponentiation Rules If $x>0$ and $y\in ℝ$, then $x^y$ is a well-defined real number satisfying the following rules.

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  $x^n=x\times x\times \mathrm{\dots }x$ (exactly $n$-times)

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  $x^{\frac{p}{q}}=a$, then $a$ is the number for which $a^q=x^p$.

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  If , and $x>1$, then .

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  ${(xy)}^t=x^t{y}^t$.

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  $x^y{x}^z=x^{(y+z)}$.

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  ${(x^y)}^z=x^{yz}$.

(hopefully, at the end of the course you will completely convince yourself that the exponentiation is well-defined, using only simple principles, nothing complicated)

We end the chapter with a little theorem.

Proof:  By Lemma 1.3.1, there exists a rational number $\frac{a}{b}>0$ and an integer $N>0$ such that if $n\ge N$ then $r_n\ge \frac{a}{b}$. Hence, $br\ge b{r}_n\ge b\frac{a}{b}=a\ge 1.$ $\mathrm{\square }$