# Real Analysis/Properties of Real Numbers

| ← Axioms of The Real Numbers | Real Analysis Properties of The Real Numbers | Exercises→ |

### Some simple results

At this point there are a large number of very simple results we can deduce about these operations from the axioms. Some of these follow, and some of them have proofs. The remaining proofs should be considered exercises in manipulating axioms. The aim of these results is to allow us to perform any manipulations which we think are "obviously true" due to our experience with working with numbers. Unless otherwise quantified, the following should hold for all $x,y\in \mathbb {R}$.
* $0$ is the only additive identity
    Proof. Suppose $x$ is an additive identity, s.t. $x\neq 0$. Then, let $y\in \mathbb {R}$. Then, $x+y=0+y$, and this leads to $x=0$. However, since we assumed that $x\neq 0$, this leads to a contradiction, therefore $0$ is the only additive identity. $\Box$
* $1$ is the only multiplicative identity
* Both additive and multiplicative inverses are unique. More formally: If both $x+y=0$ and $x+z=0$ then $y=z$; and if both $xy=1$ and $xz=1$ then $y=z$ (so that the notations $-x$ and $x^{-1}$ make sense).
    Proof: For the case of addition: We have $x+y=0$ and $x+z=0$, so adding $y$ to the latter equation, we get $(x+z)+y=0+y$, but then by commutativity and associativity of addition we deduce that $(x+y)+z=0+y$, and by our other assumption $0+z=0+y$, and then by identity of addition $z=y$. $\Box$
* $-(-x)=x$
* $\forall x\in \mathbb {R} \setminus \{0\}:(x^{-1})^{-1}=x$
* $0\times x=0$
* $0$ has no multiplicative inverse (so division by $0$ can not make sense)
* $\forall n,m\in \mathbb {Z} :x^{n}x^{m}=x^{n+m}$
* $\forall n,m\in \mathbb {Z} :(x^{n})^{m}=x^{nm}$
* $x>y\iff \neg x\leq y$ (Here $\neg$ is logical negation, so $\neg x\leq y$ means "it is not the case that $x\leq y$".)
    Proof: First we consider the implication $\implies$. Suppose $x>y$. By definition, this means that $x\not =y$ and $y<x$. If it were also true that $x\leq y$ then by anti-symmetry we have $x=y$, which is impossible. Thus $\neg x\leq y$.
    Conversely, suppose $\neg x\leq y$. First, if we had $x=y$ then by reflexivity $x\leq y$, which is impossible, so in fact $x\not =y$. Secondly, by totality we deduce that $y\leq x$. These two conditions are exactly those required for $x>y$. $\Box$
* $x<y\iff \neg x\geq y$
* $x$ is non-positive if and only if $x$ is not positive
* $x$ is non-negative if and only if $x$ is not negative
* If $x$ is both non-positive and non-negative then $x=0$
* $x$ is not both positive and negative
* $x\geq 0\iff -x\leq 0$
    Proof: Suppose $x\geq 0$. By one of the axioms we get $x+(-x)\geq 0+(-x)$. By additive inverse this gives $0\geq 0+(-x)$ and then by additive identity $0\geq -x$, as required.
    The converse implication follows similarly. $\Box$
* $(x\leq y{\mbox{ and }}z\leq 0)\implies xz\geq yz$
* $\forall x\in \mathbb {R} :x^{2}\geq 0$
    Proof: By totality of the order, we have either $x\geq 0$ or $x\leq 0$. In the first case we can apply the axiom linking the order to multiplication directly to $0\leq x$ and deduce $0\leq x^{2}$. In the latter case we apply the last result in this list to $0\leq x$ and obtain $x^{2}\geq 0$. $\Box$
* $1>0$ and $-1<0$

#### Applications

Although it might be said that the entirety of this book is devoted to studying the applications of completeness, there are in particular some simple applications we can give easily which provide an indication as to how completeness solves the problem with the rationals described above.

##### Theorem (Square roots)

Let $x\in \mathbb {R}$ be non-negative. Then $x$ has a unique non-negative square root, denoted ${\sqrt {x}}$, which satisfies $({\sqrt {x}})^{2}=x$.

##### Proof

We deal only with the case $x\geq 1$. The case $x\in [0,1)$ is left for the exercises.
First we note that when $y,z\in \mathbb {R}$ are non-negative, $y<z\implies y^{2}<z^{2}$ (In the terminology we will introduce later, this says that the function $y\mapsto y^{2}$ is strictly increasing). This makes it clear that there can be only one square root of $x$, and so it remains to find one.
Let $S=\{y\in \mathbb {R} :y^{2}\leq x\}$. We wish to apply the least upper bound axiom to $S$, so we must show that it is non-empty and bounded above.
That $S$ is non-empty is clear, since $1\in S$.
Furthermore, $x$ itself is an upper bound for $S$, since if $y>x\geq 1$, then $y^{2}>y$, so that $y^{2}>x$, and hence $y\not \in S$.
Putting these facts together, by the least upper bound axiom, we deduce that $S$ has a least upper bound, which we call $s$. We wish to show that $s$ is the square root of $x$ that we seek.
Certainly $s$ is positive, since $1\in S$ and so $s\geq 1$. In particular, we may divide by $s$.
To show that $s^{2}=x$, we eliminate the possibilities that $s^{2}>x$, and that $s^{2}<x$.
Suppose that $s^{2}>x$. Let $t=s-{\frac {s^{2}-x}{2s}}$. Then:
$t^{2}=s^{2}-(s^{2}-x)+{\frac {(s^{2}-x)^{2}}{4s^{2}}}=x+{\frac {(s^{2}-x)^{2}}{4s^{2}}}>x$
So $t$ is in fact an upper bound for $S$, but this is impossible, since $t<s$ and $s$ is the least upper bound for $S$.
Thus we have concluded that $s^{2}\leq x$.
Now suppose that $s^{2}<x$. Let $t=s+{\frac {x-s^{2}}{2s}}$. In a similar manner to the above, we deduce that $t^{2}<x$, so $t\in S$, but this is impossible since $t>s$ and $s$ is an upper bound for $S$.
Thus we have concluded that $s^{2}\geq x$, and so $s^{2}=x$ as required.$\Box$
This argument may appear excessively complex (especially since some details are left for the exercises), and indeed there is a sense in which it is, and we shall be able to present a much neater argument later. Nevertheless, it suffices to show that we can find a square root of 2, and so avoid the immediate problem with the rationals posed at the beginning of this section. To show that no more elaborate construction will give rise to the same problem will have to wait until we reach the study of continuity.

##### Theorem (Archimedes axiom)

(Note that despite the name, this theorem is not an axiom to us, but a theorem we deduce from the other axioms.)
a) $\forall x\in \mathbb {R} :\exists n\in \mathbb {N} :n>x$
b) $\forall x\in \mathbb {R} ^{*}+:\exists n\in \mathbb {N} :{\frac {1}{n}}<x$

##### Proof

a) Suppose the statement is not true, then we have the negation, which states:
$\exists x\in \mathbb {R} :\forall n\in \mathbb {N} :n\leq x$
but this is precisely the statement that $\mathbb {N}$ is bounded above. Certainly also it is non-empty, so we can apply the completeness axiom to get a least upper bound for $\mathbb {N}$. Call this least upper bound $l$.
Since $l$ is a least upper bound, we know that $l-1$ is not an upper bound, and thus $\exists n\in \mathbb {N} :n>l-1$. But then, $n+1>l$, and $n+1\in \mathbb {N}$ so we get the contradiction that $l$ is not an upper bound for $\mathbb {N}$ after all.
Thus, our supposition was false, and (a) holds.
b) Take $x\in \mathbb {R} ^{+}$. Certainly $x\not =0$, so that we can invert $x$ to get $x^{-1}\in \mathbb {R} ^{+}$. Applying part (a) to $x^{-1}$, we can find $n\in \mathbb {N}$ with $n>x^{-1}$, and then inverting this inequality, we deduce ${\frac {1}{n}}<x$ as required.$\Box$

##### Corollary (Density of rationals and irrationals)

If $x<y$ then $(x,y)$ contains both a rational number and an irrational number.

##### Proof

To find a rational in $(x,y)$, we apply Archimedes axiom (b) to $y-x$, getting $n\in \mathbb {N}$ with ${\frac {1}{n}}<y-x$. Thus $1<yn-xn$, so $xn<yn-1$.
We also apply Archimedes axiom (a) to $y+1$ to get $N\in \mathbb {N}$ satisfying $N>yn+2$.
Now choose the least $m\in \mathbb {N}$ satisfying $N-m<yn$. By the above, $m\geq 2$, and so, since $m$ is minimal, we know that:
$N-(m-1)\geq yn$
$N-m\geq yn-1$
Putting this together with the fact that $xn<yn-1$ deduced above, we get:
$N-m>xn$
So, in summary, we have $yn>N-m>xn$, so $y>{\frac {N-m}{n}}>x$, and we have found the rational number we want.
To find an irrational number, we use what we have just deduced to first find a rational $q\in (x+{\sqrt {2}},y+{\sqrt {2}})$, so that $q-{\sqrt {2}}\in (x,y)$. Furthermore, $q-{\sqrt {2}}$ must be irrational, for if it were rational then we would also have $q-(q-{\sqrt {2}})={\sqrt {2}}$ rational, and we know that it is not. $\Box$

## Properties of Least Upper Bounds

We'll be doing a lot of work with least upper bounds, so it will be important to know how to use them efficiently in proofs. Here are some definitions and properties that are helpful in this respect:

### Uniqueness of Least Upper Bounds

Every non-empty set that is bounded above has a unique least upper bound

#### Proof

Let $a$ and $b$ be two least upper bounds for a set $S$.
If $a>b$, then since $b$ is an upper bound for $S$, $a$ cannot be the least upper bound. Thus $a\leq b$. Similarly, $a\geq b$. Thus $a=b$, so $S$ can have only one least upper bound.

### Existence of Greatest Lower Bounds

Every non-empty set S that is bounded below has a unique greatest lower bound, or infimum (denoted $\inf S$).

#### Proof

Let S be non-empty and bounded below. Let $T:=\{-x:x\in S\}$.
Since S is non-empty, $\exists x\in S$. Thus $-x\in T$, so T is non-empty.
Since S is bounded below, $\exists M:\forall x\in S:x>M$.
Then $x\in T\implies -x\in S\implies -x>M\implies x<-M$.
Thus T is bounded above by M, and therefore T has a least upper bound, $\beta$.
Since $x\in S\implies -x\in T\implies -x<\beta \implies x>-\beta$, $-\beta$ is a lower bound for S.
Let $\alpha$ be a lower bound for S.
Then $x\in T\implies -x\in S\implies -x>\alpha \implies x<\alpha$, so $-\alpha$ is an upper bound for T.
Since $\beta$ is the least upper bound for T, $-\alpha >\beta$, and thus $\alpha <-\beta$.
Thus all lower bounds for S are less than $-\beta$
In other words, $-\beta$ is a greatest lower bound for S.
Uniqueness follows similarly to the uniqueness of least upper bounds.

### Theorem (Ordering of Sups and Infs)

If $S\subseteq T$, where S is non-empty and T is bounded above and below, then $\inf T\leq \inf S\leq \sup S\leq \sup T$

#### Proof

Since S is non-empty, it contains an element x. By definition, $\inf S\leq x$ and $x\leq \sup S$, so $\inf S\leq \sup S$.
Since T is bounded above, it has a least upper bound, $\sup T$.
Since t is in particular an upper bound for T, $\forall x\in T:x\leq \sup T$. Since $S\subseteq T$, $x\in S\implies x\in T\implies x\leq \sup T$.
Thus $\sup T$ is an upper bound for S, so $\sup S$ exists and by definition $\sup S\leq \sup T$.
Similarly, $\inf S\geq \inf T$.

## Sum and Product Notation

We often need to take a sum or product of several real numbers at a time. Since "..." is given no meaning by our axioms, we can't just write "$a_{1}+a_{2}+\dots +a_{n}$". Thus we use the symbols $\sum _{k=1}^{n}a_{k}$ and $\prod _{k=1}^{n}a_{k}$ to denote the sum and product, respectively, over an arbitrary finite number of real numbers. We do this inductively, as follows:
* $\sum _{k=1}^{1}a_{k}=a_{1}$ and $\prod _{k=1}^{1}a_{k}=a_{1}$
* $\sum _{k=1}^{n}a_{k}=a_{n}+\sum _{k=1}^{n-1}a_{k}$ and $\prod _{k=1}^{n}a_{k}=a_{n}\prod _{k=1}^{n-1}a_{k}$
Now we can prove some properties of sums and products:

### Properties

* The order of summation can be changed arbitrarily. That is, if $\{a_{k}:1\leq k\leq n\}=\{b_{k}:1\leq k\leq n\}$, then $\sum _{k=1}^{n}a_{k}=\sum _{k=1}^{n}b_{k}$ and $\prod _{k=1}^{n}a_{k}=\prod _{k=1}^{n}b_{k}$
Proof: This follows from commutativity and a rather nasty induction.
* $\sum _{k=1}^{n}a_{k}+\sum _{k=1}^{n}b_{k}=\sum _{k=1}^{n}(a_{k}+b_{k})$ and $\prod _{k=1}^{n}a_{k}\prod _{k=1}^{n}b_{k}=\prod _{k=1}^{n}(a_{k}b_{k})$
Proof: We proceed by induction. First, note that $\sum _{k=1}^{1}a_{k}+\sum _{k=1}^{1}b_{k}=a_{k}+b_{k}=\sum _{k=1}^{1}(a_{k}+b_{k})$.
Now assume that $\sum _{k=1}^{n-1}a_{k}+\sum _{k=1}^{n-1}b_{k}=\sum _{k=1}^{n-1}(a_{k}+b_{k})$. Then $\sum _{k=1}^{n}a_{k}+\sum _{k=1}^{n}b_{k}$$=\sum _{k=1}^{n-1}a_{k}+a_{n}+\sum _{k=1}^{n-1}b_{k}+b_{n}$$=\sum _{k=1}^{n-1}a_{k}+\sum _{k=1}^{n-1}b_{k}+a_{n}+b_{n}$$=\sum _{k=1}^{n-1}(a_{k}+b_{k})+(a_{n}+b_{n})$$=\sum _{k=1}^{n}(a_{k}+b_{k})$.
The statement for products follows similarly.
* $c\sum _{k=1}^{n}a_{k}=\sum _{k=1}^{n}ca_{k}$
Proof: Another induction. For $n=1$, $c\sum _{k=1}^{1}a_{k}=ca_{1}=\sum _{k=1}^{1}ca_{k}$. Now assume the statement is true for n-1. Then $c\sum _{k=1}^{n}a_{k}=c(\sum _{k=1}^{n-1}a_{k}+a_{n})=\sum _{k=1}^{n-1}ca_{k}+ca_{n}=\sum _{k=1}^{n}ca_{k}$.
* $\sum _{k=1}^{n}(a_{k})\sum _{l=1}^{m}(b_{l})=\sum _{k=1}^{n}\sum _{l=1}^{m}a_{k}b_{l}$
Proof: We induct on n. The previous property takes care of the case n=1. Assume the statement is true for n-1. Then $\sum _{k=1}^{n}(a_{k})\sum _{l=1}^{m}(b_{k})$$=(\sum _{k=1}^{n-1}(a_{k})+a_{n})\sum _{l=1}^{m}(b_{k})$$=\sum _{k=1}^{n-1}(a_{k})\sum _{l=1}^{m}(b_{k})+a_{n}\sum _{l=1}^{m}(b_{k})$$=\sum _{k=1}^{n-1}\sum _{l=1}^{m}(a_{k}b_{l})+\sum _{l=1}^{m}(a_{n}b_{k})$$=\sum _{k=1}^{n}\sum _{l=1}^{m}(a_{k}b_{l})$
Most familiar properties of sums and products can be deduced by similar methods.