# Real numbers, sequences and series: part V

Real Numbers.

Our scheme of real numbers has so far been extended to rational numbers so that we could not only add and subtract without any hindrances, but could also multiply and divide without any restriction, save division by zero. But, when it comes to extraction of square roots, it seems “incomplete”. For example, there is no rational number whose square gives 2. This question crops up when we try to relate numbers to geometry.

We have seen before that we can always represent a rational number on a line. Now suppose we have a square with sides of unit length. By the theorem of Pythagoras, the square of the length of diagonal is equal to $1^{2}+1^{2}=2$. We can also see that the length of the diagonal can be geometrically represented on the line. Now, we ask here, does there exist a rational number whose square is 2, or equivalently, does the equation $x^{2}=2$ have a solution in rational numbers? This amounts to asking if we can find two non-zero integers p and q such that

$p^{2}=2q^{2}$Equation I.

We can demonstrate that there are no such integers. Suppose that there were two such integers p and q satisfying equation I. We may assume without loss of generality, that p and q have no common factors. If there were any, we could cancel them from both the sides of equation I till there were no common factors. Now as there are no common factors between p and q, for equation I to hold, 2 must be a factor of $p^{2}$, and hence, of p. So we may write

$p=2m$ Equation 2.

Substituting this in Equation 2, we get $2m^{2}=q^{2}$Equation 3.

As before, we conclude that for equation 3 to hold, 2 must divide q giving $q=2n$ for some integer n. This contradicts our assumption that p and q have no common factors. This only proves that equation I has no solution in integers (except $p=0, q=0$). Thus, the length of the diagonal of the unit square, though it has a point representing it on the line, does not correspond to a rational number. This shows that is not large enough to accommodate a number such as the length of the diagonal of a unit square. So we must extend Q. We could simply include the new number which we write as $\sqrt{2}$ in the scheme along with Q. But, then this ad hoc extension may not stand up to all our demands to include newer and newer numbers which arise out of algebraic equations, e.g.,$\sqrt[3]{6}$, etc. Even if we somehow include these numbers in our scheme, we must know how to perform arithmetic operations like addition, subtraction, multiplication and division in it. Also, how does one decide which of a given pair of new numbers is smaller? There are many ways in extending to accommodate all these new numbers. We illustrate one of the simplest ways of doing this.

Let $E=\{ a \in Q: a^{2}<2\}$. It is clear that E has an upper bound, for example, 2. Next, we note that if $y \in E$ is an upper bound of E, then there exists $n \in N$ with $n(2-y^{2})>1+2y$ (we can always choose an n because of the Archimedean property of Q). Then, one has $2-(y+\frac{1}{n})^{2}=(2-y^{2})-\frac{1}{n}(2y+\frac{1}{n}) \geq (2-y^{2})-\frac{2y+1}{n}>0$, showing that $y^{2}<(y+\frac{1}{n})^{2}<2$ and thus no upper bound of E can be in E. Let $x \in Q$ be one such upper bound of E, then so is $\frac{x}{2}+\frac{1}{x}$ since $(\frac{x}{2}+\frac{1}{x})^{2}-2=(\frac{x}{2}-\frac{1}{x})^{2}>0$. On the other hand, $x-(\frac{x}{2}+\frac{1}{x})=\frac{x^{2}-2}{2x}>0$ since $x \not\in E$. Therefore, the E has no least upper bound in Q.

This tells us what to do. We now extend the field of rationals to a larger field containing it which has all properties of along with an additional property, called the least upper bound property and abbreviated as lub property, namely, every set that is bounded above has a least upper bound.

More precisely, we postulate that there is a field $(\Re, +, .)$, $\Re$ containing Q and satisfying all the properties of Q listed above and an additional one:

The least upper bound property (lub): If $A \subseteq \Re$ is bounded above, then A admits a least upper bound, that is, (1) there is a $\gamma \in \Re$ such that $\alpha \leq \gamma$ for every $\alpha \in A$ and (2) for every $\gamma^{'}< \gamma$, there is a $\beta \in A$ such that $\gamma^{'} < \beta$.

Remark:

As we constructed out of Z, there are ways of constructing $\Re$ out of Q. However, an explicit construction of $\Re$ from is beyond the scope of the present blog. Nevertheless, we are going to use the listed properties of $\Re$ in all subsequent discussions and deduce many interesting consequences.

Since $\Re$ has all the properties of including the order property, we write (as for Q),

$\Re_{+}=\{ \alpha \in \Re|\alpha >0\}$

and this has the same properties as $Q_{+}$. Now, we use the same notation like $\leq, <, \geq, >$ for elements of $\Re_{+}$, exactly as we did for the elements of Q. As before, we write $\beta > \alpha$ or $\beta \geq \alpha$ according as $\alpha < \beta$ or $\alpha \leq \beta$.

Clearly, 1, $1+1, 1+1+1, \ldots$ belong to $\Re$ and represent the natural numbers. So, one sets up a one to one correspondence between $n \in N$ and $1+1+1+\ldots+1 (n times)$ to conclude that $N \subset \Re$. Writing $-n$ as the additive inverse of $n \in N$, of course $0 \in \Re$, we see that $Z \subseteq \Re$. Since $\Re$ is a field, for $m,n \in Z$, $n \neq 0$, we must have

$m.n^{-1} \in \Re$

where $n^{-1}$ is the multiplicative inverse of n. We agree to write $m.n^{-1}=\frac{m}{n}$. This way we have $Q \subseteq \Re$. We have thus effectively extended to $\Re$.

More later,

Nalin Pithwa

# Real Numbers, Sequences and Series: Part III: Rational Numbers

On Z, we now have unlimited subtraction. Can we divide in Z? In a limited sense, yes. Division is essentially an inverse problem. That is to say, $m \div n$ would stand for a number which when multiplied by n gives m. As long as we are in Z, the question does not always have an answer. For example, we know that 5 multiplied by 2 gives 10 as the product, hence 10 divided by 5 gives 2 as the quotient.  What if we asked which multiplied by 2 gives us 5?  We know that there is no integer which when multiplied by 2 gives 5. Such questions frequently arise when we have to distribute m things equally among n persons. If m is not an integral multiple of n, then becomes inadequate for such distribution.

If 10 cakes are to be shared equally among 5 children, then each child gets 2 cakes. But how do we proceed if the same 10 cakes are to be shared equally among 4 children? One possibility is to give 2 cakes to each of the 4 children and then divide each of the remaining 2 cakes into two equal halves so that there are 4 halves. Each child can now be given half a cake.

What we are doing is introducing a new kind of numbers called “fractions”, meaning a “part” of the whole number. Each half of the cake is represented by $1/2$. Suppose we had 3 cakes to be shared equally by 5 children. What we do is to divide each cake into 5 equal parts. So we have in all 15 similar pieces of cake. Now we can distribute these 15 pieces among 5 children, each getting a share of 3 pieces. Each child’s share is represented by $3/5$. Suppose we divided each of the cakes into 10 equal parts instead, then we would have thirty equal pieces of cake. When shared equally by 5 children, each child gets a share of 6 pieces. Each child’s share can be represented by $6/10$. The share of each child in both ways of division ought to be the same, as in both cases each child has an equal share and nothing of the original three cakes remains. This is what would amount to saying that $3/5$ and $6/10$ represent the same number. Geometrically, they will look identical.

Unless this “equivalence” of fractions is allowed, it would be hard to add fractions. For example, how does one add $1/2$ to $1/3$, though we have no difficulty in adding $1/5$ to $3/5$ giving us $frac{1}{5}+\frac{3}{5}=\frac{4}{5}$? We argue that since $1/2$ represents the same number as $3/6$ and $1/3$, the same as $2/6$, we have

$\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$.

This means that if somebody has three parts out of six parts, and next two parts out of six parts, then he has in all five parts out of six. We can perform subtraction similarly. Multiplication and division can be defined as we did in our elementary school. This presupposes that a fraction represents measurement of some physical quantity like cake, stick, pole, etc. But we should have a definition which should capture the essence of the process of “dividing a certain physical object like a cake into 5 equal parts”. To formalize our construction of fractions, we proceed as follows:

Consider the Cartesian product

$\textbf{Z} \times (\textbf{Z}-\{ 0 \})= \{ (m,n): m,n \in \textbf{Z}, n \neq 0\}$

and define the relation $\sim$ in $latex \textbf{Z} \times (\textbf{Z}-\{ 0 \})$ by $(m,n) \sim (m^{'}, n^{'})$ if $mn^{'}=m^{'}n$. It is easy to see that $\sim$ is an equivalence relation in $\textbf{Z} \times (\textbf{Z}-\{ 0 \})$.:

Now, we can decompose the set $\textbf{Z} \times (\textbf{Z}-\{ 0 \})$ into disjoint equivalence classes. Let us denote the equivalence class containing $(m,n)$ by $\frac{m}{n}$.

Suppose $(m^{'},n^{'})$ is another element of the same equivalence class. Then, we ought to denote it also by

$\frac{m^{'}}{n^{'}}$. But, since $(m,n)=(m^{'},n^{'})$ we have $mn^{'}=m^{'}n$. So, we can say, $\frac{m}{n}=\frac{m^{'}}{n^{'}}$ if $mn^{'}=m^{'}n$. We can now define addition, subtraction, multiplication and division among the new numbers denoted by $\frac{m}{n}$ in the following way:

(i) $\frac{m}{n}+\frac{m^{'}}{n^{'}}=\frac{mn^{'}+m^{'}n}{nn^{'}}$ for all $m,n,m^{'},n^{'} \in \textbf{Z}$ and $n \neq 0, n^{'} \neq 0$.

(ii) $\frac{m}{n}.\frac{m^{'}}{n^{'}}=\frac{mm^{'}}{nn^{'}}$ for all $m,n,m^{'},n^{'} \in \textbf{Z}$ and $n \neq 0, n^{'} \neq 0$.

(iii) $\frac{m}{n} \div \frac{m^{'}}{n^{'}}=\frac{mn^{'}}{nm^{'}}$ for $m \in \textbf{Z}$ and

$n,n^{'},m^{'} \in \textbf{Z}-\{ 0 \}$.

With these rules of addition, we have

$\frac{m}{n} +\frac{0}{n^{'}}=\frac{mn^{'}+0.n}{nn^{'}}=\frac{mn^{'}}{nn^{'}}=\frac{m}{n}$

for all $m \in \textbf{Z}$ and $n, n^{'} \in \textbf{Z}-\{0\}$. This is to say that $\frac{0}{n^{'}}$ for every element $n^{'} \in \textbf{Z}-\{ 0 \}$ behaves like the “zero element” which when added to any of the numbers of the form $\frac{m}{n}$ gives us the same number $\frac{m}{n}$. Such an element is sometimes called additive identity. We have also

$\frac{m}{n}+\frac{-m}{n}=\frac{0}{n}$.

This tells us that for every element $\frac{m}{n}$, we have an element of the same type, which when added to it gives the zero element. This is what we usually understand as the “negative” of the number $\frac{m}{n}$ which is sometimes called the additive inverse of $\frac{m}{n}$. Again, the rules of multiplication tell us that

$\frac{m}{n}.\frac{n^{'}}{n^{'}}=\frac{mn^{'}}{nn^{'}}=\frac{m}{n}$ for all $m,n,n^{'} \in \textbf{Z}$ and $n \neq 0, n^{'} \neq 0$.

This means that $\frac{n^{'}}{n^{'}}$ acts the way the multiplicative identity like 1 behaves in $\textbf{Z}$ (which when multiplied by any number gives us back the same number). For simplicity of notation, we denote $\frac{0}{n}$ by 0 (note that $\frac{0}{n} \sim \frac{0}{n^{'}}$) and $\frac{n}{n}$ by 1  so that we may identify $\frac{mn}{n}$ with $m \in \textbf{Z}$. Since $\frac{m}{n}.\frac{n}{m}=\frac{m.n}{m.n}=1$, when $m, n \in \textbf{Z}$ and $m \neq 0, n \neq 0$, we conclude that every non-zero element of the form $\frac{m}{n}$ has a multiplicative inverse.

We call the new numbers $\frac{m}{n}$ rational numbers which is in fact, an extension of Z. We denote the set of rational numbers by

$\textbf{Q}= \{ \frac{}{}: m, n \in \textbf{Z}, n \neq 0\}$

In $\textbf{Q}$, we can add, subtract and multiply at will. We can also verify easily (Exercise):

(i) $\alpha+\beta=\beta+\alpha$ for all $\alpha, \beta \in \textbf{Q}$

(ii) $\alpha + (\beta +\gamma)=(\alpha+\beta)+\gamma$ for all $\alpha, \beta, \gamma \in \textbf{Q}$

(iii) $\alpha + 0=\alpha$ for all $\alpha \in \textbf{Q}$

(iv) for every $\alpha \in \textbf{Q}$, there is a unique $\beta \in \textbf{Q}$ such that $\alpha+\beta=0$.

We call this $\beta$ the negative of $\alpha$, denote it by $-\alpha$, and write $\alpha + (-\alpha)=0$.

(v) $\alpha.\beta=\beta.\alpha$ for all $\alpha, \beta \in \textbf{Q}$

(vi) $\alpha.(\beta. \gamma)=(\alpha.\beta).\gamma$ for all $\alpha, \beta, \gamma \in \textbf{Q}$.

(vii) $\alpha.1=\alpha$ for all $\alpha \in Q$

(viii) $\alpha.(\beta +\gamma)=\alpha.\beta+\alpha.\gamma$ for all $\alpha, \beta, \gamma \in \textbf{Q}$

(ix) For every $\alpha \in \textbf{Q}$, $\alpha \neq 0$, there is a unique $\beta \in \textbf{Q}$ such that $\alpha.\beta=1$. We denote $\beta=\frac{1}{\alpha}$ or $\alpha^{-1}$.

Given $\alpha \in \textbf{Q}$ and $\beta \in \textbf{Q}-0$, we define $\alpha$ divided by $\beta$, written as $\frac{\alpha}{\beta}$ by $\alpha.\beta^{-1}$. Note that for $\beta \neq 0$, $\frac{\alpha}{\beta}=\gamma \Longleftrightarrow \alpha=\beta\gamma$

Remark.

Note that if $\alpha \neq 0$ and $\beta=0$, then there does not exist a $\gamma$ such that $\alpha = \beta \gamma$. Thus, in this case, $\frac{\alpha}{\beta}$ cannot be defined. Also, when $\alpha=0$ and $\beta=0$, any choice of $\gamma$ will satisfy $\alpha=\beta \gamma$. Hence, again $\frac{\alpha}{\beta}$ cannot be defined. So, division by zero is not defined.

Remark.

Any set $\mathcal{F}$ with two operation $(+,.)$ satisfying all the nine properties listed above is called a field. This means that $\textbf{Q}$ is a field. We will be doing a detailed discussion of fields later in these blogs.

The set $\textbf{Q}$ is called the set of rational numbers. No  doubt $\textbf{Q}$ is a field, where usual arithmetic operations can be performed, but it has an additional feature that is the order relation. That is to say, given two rational numbers we can decide, if they are not equal, which is the larger of the two. To describe the modus operandi of this, we proceed as follows:

Let $\textbf{Q}_{+}=\{ \frac{m}{n}:m,n >0\}$, the subset of positive rational numbers, then $\textbf{Q}_{+}$, is a subset of $\textbf{Q}$ having the properties:

i) for $\alpha \in \textbf{Q}$ either $\alpha \in \textbf{Q}_{+}$, or $\alpha=0$, or $-\alpha \in \textbf{Q}_{+}$. In other words, $\textbf{Q}_{+}\bigcup \{0\}\bigcup \{ -\alpha: \alpha \in \textbf{Q}_{+}\}$.

ii) $\alpha, \beta \in \textbf{Q}_{+}$ implies that $\alpha + \beta \in \textbf{Q}_{+}$, $\alpha \beta \in \textbf{Q}_{+}$.

This is sometimes paraphrased by saying that $\textbf{Q}_{+}$ is closed under addition and multiplication. We define an order relation in $\textbf{Q}$ with the help of $\textbf{Q}_{+}$ by saying that

for $\alpha, \beta \in \textbf{Q}$ and $\alpha \neq \beta$

$\alpha < \beta$ if $\beta - \alpha \in \textbf{Q}_{+}$, or $\beta < \alpha$ if $\alpha - \beta \in \textbf{Q}_{+}$.

This is meaningful because either $\beta - \alpha \in \textbf{Q}_{+}$, or $-(\beta -\alpha) \in \textbf{Q}_{+}$, or $\beta - \alpha=0$.

i) Given $\alpha , \beta \in \textbf{Q}$ and $\gamma \in \textbf{Q}_{+}$, we have $\alpha < \beta$ implies $\alpha\gamma < \beta\gamma$; this is obvious because we have $\beta - \alpha \in \textbf{Q}_{+}$, and hence, $(\beta-\alpha)\gamma \in \textbf{Q}_{+}$.

ii) $0 < \alpha , \beta$ and $\alpha < \beta \Longrightarrow \frac{1}{\beta} < \frac{1}{\alpha}$

iii) $\alpha < \beta \Longrightarrow -\beta < -\alpha$

iv) $\alpha < \beta$, $\gamma < 0 \Longrightarrow \beta\gamma < \alpha\gamma$

v) Given $\alpha, \beta \in \textbf{Q}$ and $0 < \alpha$, there is an $n \in \textbf{N}$ such that $\beta < n \alpha$.

For a proof of this, first note that if $\beta < \alpha$, then there is nothing to prove (we might take $n=1$). If, on the other hand, $\alpha < \beta$, set $\alpha=\frac{p}{q}$ and $\beta = \frac{p^{'}}{q^{'}}$, $p,q,p^{'},q^{'} \in \textbf{N}-\{0\}$. This is true because $0 < \alpha < \beta$.

Now, $\alpha q=p$. Let $\frac{p^{'}}{q^{'}}=n_{1}+\frac{r_{1}}{q_{1}}$, where $n_{1} \textbf{N} \bigcup \{0\}$ (since $0 \leq \frac{r^{'}}{q^{'}}<1$). Finally, we set $qm=n$ so that we have

$n\alpha=mq\alpha=mp>\frac{p^{'}}{q^{'}} = \beta$.

This last property is  called the Archimedean property of $\textbf{Q}$. (This was used by Archimedes in his works, but he later credited Eudoxus for this.).

vi) Given $\alpha, \beta \in \textbf{Q}$, $\alpha < \beta$, there is a $\gamma \in \textbf{Q}$ such that $\alpha < \gamma < \beta$. For example, take $\frac{\alpha + \beta}{2}$.

vii) $\alpha \leq \beta \Longrightarrow \alpha + \gamma \leq \beta + \gamma$ for all $\gamma \in \textbf{Q}$.

It can be verified that $0.\alpha=0$ and $(-\alpha).(-\beta)=\alpha.\beta$ for all $\alpha , \beta \in \textbf{Q}$.

More later,

Nalin Pithwa