# Real Numbers, Sequences and Series: part 8

Definition. A sequence is a function $f:N \rightarrow \Re$. It is usual to represent the sequence f as $(a_{n})_{n=1}^{\infty}$ where $f(n)=a_{n}$.

Definition. A sequence $(a_{n})_{n=1}^{\infty}$ is said to converge to a if for each $\varepsilon>0$ there is an $n_{0}$ such that

$|a_{n}-a|<\varepsilon$ for all $n > n_{0}$.

It can be shown that this number is unique and we write $\lim_{n \rightarrow \infty}a_{n}=a$.

Examples.

(a) The sequence $\{ 1, 1/2, 1/3, \ldots, 1/n \}$ converges to 0. For $\varepsilon>0$, let $n_{0}=[\frac{1}{\varepsilon}]+1$. This gives

$|\frac{1}{n}-0|=\frac{1}{n}<\varepsilon$ for all $n>n_{0}$.

(b) The sequence $\{ 1, 3/2, 7/4, 15/8, 31/16, \ldots\}$ converges to 2. The nth term of this sequence is $\frac{2^{n}-1}{2^{n-1}}=2-\frac{1}{2^{n-1}}$. So $|2-(2-\frac{1}{2^{n-1}})|=\frac{1}{2^{n-1}}$. But, $2^{n-1} \geq n$ for all $n \geq 1$. Thus, for a given $\varepsilon >0$, the choice of $n_{0}$ given in (a) above will do.

(c) The sequence $(a_{n})_{n=1}^{\infty}$ defined by $a_{n}=n^{\frac{1}{n}}$ converges to 1.

Let $n^{\frac{1}{n}}=1+\delta_{n}$ so that for $n >1$, $\delta_{n}>0$. Now, $n=(1+\delta_{n})^{n}=1+n\delta_{n}+\frac{n(n-1)}{2}(\delta_{n})^{2}+\ldots \geq 1+\frac{n(n-1)}{2}(\delta_{n})^{2}$, thus, for $n-1>0$, we have $\delta_{n} \leq \sqrt{\frac{2}{n}}$. For any $\varepsilon > 0$, we can find $n_{0}$ in N such that $n_{0}\frac{(\varepsilon)^{2}}{2}>1$. Thus, for any $n > n_{0}$, we have $0 \leq \delta_{n} \leq \sqrt{\frac{2}{n}} \leq \sqrt{\frac{2}{n_{0}}}<\varepsilon$. This is the same as writing

$|n^{\frac{1}{n}}-1|<\varepsilon$ for $n > n_{0}$ or equivalently, $\lim_{n \rightarrow \infty}n^{\frac{1}{n}}=1$

(d) Let $a_{n}=\frac{2^{n}}{n!}$. The sequence $(a_{n})_{n=1}^{\infty}$ converges to o. Note that

$a_{n}=\frac{2}{1}\frac{2}{2}\frac{2}{3}\ldots\frac{2}{n}<\frac{4}{n}$ for $n>3$.

For $\varepsilon>0$, choose $n_{1}$ such that $n_{1}\varepsilon>4$. Now, let $n_{0}=max \{ 3, n\}$ so that $|a_{n}|<\varepsilon$ for all $n > n_{0}$.

(e) For $a_{n}=\frac{n!}{n^{n}}$, the sequence $(a_{n})_{n=1}^{\infty}$ converges to 0. (Exercise!)

In all the above examples, we somehow guessed in advance what a sequence converges to. But, suppose we are not able to do that and one asks whether it is possible to decide if the sequence converges to some real number. This can be helped by the following analogy: Suppose there are many people coming to Delhi to attend a conference. They might be taking different routes. But, as soon as they come closer and closer to Delhi the distance between the participants is getting smaller.

This can be paraphrased in mathematical language as:

Theorem. If $(a_{n})_{n=1}^{\infty}$ is a convergent sequence, then for every $\varepsilon>0$, we can find an $n_{0}$ such that

$|a_{n}-a_{m}|<\varepsilon$ for all $m, n > n_{0}$.

Proof. Suppose $(a_{n})_{n=1}^{\infty}$ converges to a. Then, for every $\varepsilon$ we can find an $n_{0}$ such that

$|a_{n}-a|<\varepsilon/2$ for all $n > n_{0}$.

So, for $m, n > n_{0}$, we have

$|a_{m}-a_{n}|=|a_{n}-a+a-a_{m}|\leq |a_{n}-a|+|a-a_{m}|<\varepsilon$. QED.

The proof is rather simple. This very useful idea was conceived by Cauchy and the above theorem is called Cauchy criterion for convergence. What we have proved above tells us that the criterion is a necessary condition for convergence. Is it also sufficient? That is, given a sequence $(a_{n})_{n=1}^{\infty}$ which satisfies the Cauchy criterion, can we assert that there is a real number to which it converges? The answer is yes!

We call a sequence $(a_{n})_{n=1}^{\infty}$ monotonically non-decreasing if $a_{n+1} \geq a_{n}$ for all n.

Theorem. A monotonically non-decreasing sequence which is bounded above converges.

Proof. Suppose $(a_{n})_{n=1}^{\infty}$ is monotonic and non-decreasing. We have $a_{1} \leq a_{2} \leq \ldots \leq a_{n} \leq a_{n+1} \leq \ldots$. Since the sequence is bounded above, $\{ a_{k}: k=1, \ldots\}$ has a least upper bound. Let it be a. By definition, $a_{n} \leq a$ for all n, but for $\varepsilon>0$ there is at least one $n_{0}$ such that $a_{n_{0}}+\varepsilon>a$. Therefore, for $n > n_{0}$, $a_{n}+\varepsilon \geq a_{n_{0}}+\varepsilon>a\geq a_{n}$. This gives us $|a_{n}-a|<\varepsilon$ for all $n \geq n_{0}$. QED.

We can similarly prove that: A monotonically non-increasing sequence which is bounded below is convergent.

Suppose we did not have the condition of boundedness below or above for a monotonically non-increasing or non-decreasing sequence respectively, then what would happen? If a sequence is monotonically non-decreasing and is not bounded above, then given any real number $M>0$ there exists at least one $n_{0}$ such that $a_{n_{0}}>M$, and hence $a_{n}>M$ for all $n > n_{0}$. In such a case, we say that $(a_{n})_{n=1}^{\infty}$ diverges to $\infty$. We write $\lim_{n \rightarrow \infty}a_{n}=\infty$. More generally, (that is, even when the sequence is not monotone), the same criterion above allows us to say that $\{ a_{n}\}_{n=1}^{\infty}$ diverges to $\infty$ and we write $\lim_{n \rightarrow \infty}a_{n}=\infty$. We can similarly define divergence to $-\infty$.

We make a digression here: In case a set is bounded above, then we have the concept of least upper bound. For any set A or real numbers, we define supremum of A as

$sup \{ x: x \in A\}=sup A = least \hspace{0.1in} upper \hspace{0.1in}bound \hspace{0.1in} of \hspace{0.1in} A$, if A is bounded above and is equal to $\infty$ if A is not bounded above.

Similarly, we define infimum of a set A as

$inf \{ x: x \in A\}= inf A= greatest \hspace{0.1in} lower \hspace{0.1in} bound \hspace{0.1in} of \hspace{0.1in}A$, if A is bounded below, and is equal to $-\infty$, if A is not bounded below.

This would help us to look for other criteria for convergence. For any bounded sequence (a_{n})_{n=1}^{\infty} of real numbers, let

$b_{n} = \sup \{ a_{n}, a_{n+1}, a_{n+2}, \ldots\}=\sup \{ a_{k}: k \geq n\}$.

It is clear that $(b_{n})_{n=1}^{\infty}$ is now a non-increasing sequence. So, it converges. We set

$\lim_{n \rightarrow \infty}b_{n}=limit \hspace{0.1in} superior \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} sequence \hspace{0.1in} (a_{n})_{n=1}^{\infty}= \lim \sup a_{n}= \overline{\lim}_{n}a_{n}$,

Similarly, if we write

$a_{n}=\inf \{ a_{n}, a_{n+1}, \ldots\}=\inf \{ a_{k}: k \geq n\}$,

then $(a_{n})_{n=1}^{\infty}$ is a monotonically non-decreasing sequence. So we write

$\lim_{n \rightarrow \infty}c_{n}=limit \hspace{0.1in} inferior \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} sequence \hspace{0.1in} (a_{n})_{n=1}^{\infty}= \lim \inf a_{n}= \underline{\lim}_{n}a_{n}$,

We may not know a sequence to be convergent or divergent, yet we can find its limit superior and limit inferior.

We have in fact the following result:

Theorem.

For a sequence $(a_{n})_{n=1}^{\infty}$ of real numbers, $\underline{\lim}a_{n} \leq \overline{\lim}a_{n}$.

Further, if $l=\underline{\lim}a_{n}$, and $L=\overline{\lim}a_{n}$ are finite, then $l=L$ if and only if the sequence is convergent.

Proof.

It is easy to see that $l \leq L$. Now, suppose l, L are finite.If $l=L$ are finite. If $l=L$, then for all $\varepsilon >0$ there exists $n_{1}, n_{2}$ such that

$l - \varepsilon< \sup_{k \geq n} a_{k}< l + \varepsilon$ for all $n \geq n_{1}$ and $l - \varepsilon < \inf_{k \geq n} < l + \varepsilon$ for all $n > n_{2}$.

Thus, with $n_{0}=max{n_{1},n_{2}}$ we have

$l - \varepsilon for all $n > n_{0}$.

This proves that equality holds only when it is convergent. Conversely, suppose $(a_{n})_{n=1}^{\infty}$ converges to a. For every $\varepsilon > 0$, we have $n_{0}$ such that $a-\varepsilon for all $n > n_{0}$. Therefore, $\sup_{k \geq n}a_{k} \leq a+\varepsilon$ and

$a-\varepsilon \leq \inf a_{n}$ for all $n \geq n_{0}$. Hence, $a-\varepsilon \leq < L \leq a+\varepsilon$. Since this is true for every $\varepsilon > 0$, we have $L=l=a$.

QED.

# Real Numbers, Sequences and Series: Part 7

Exercise.

Discover (and justify) an essential difference between the decimal expansions of rational and irrational numbers.

Giving a decimal expansion of a real number means that given $n \in N$, we can find $a_{0} \in Z$ and $0 \leq a_{1}, \ldots, a_{n} \leq 9$ such that

$|x-\sum_{k=0}^{n}\frac{a_{k}}{10^{k}}|< \frac{1}{10^{n}}$

In other words if we write

$x_{n}=a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots +\frac{a_{n}}{10^{n}}$

then $x_{1}, x_{2}, x_{3}, \ldots, x_{n}, \ldots$ are approximate values of x correct up to the first, second, third, …, nth place of decimal respectively. So when we write a real number by a non-terminating decimal expansion, we mean that we have a scheme of approximation of the real numbers by terminating decimals in such a way that if we stop after the nth place of decimal expansion, then the maximum error committed by us is $10^{-n}$.

This brings us to the question of successive approximations of a number. It is obvious that when we have some approximation we ought to have some notion of the error committed. Often we try to reach a number through its approximate values, and the context determines the maximum error admissible. Now, if the error admissible is $\varepsilon >0$, and $x_{1}, x_{2}, x_{3}, \ldots$ is a scheme of successive is approximation of a number x, then we should be able to tell at which stage the desired accuracy is achieved. In fact, we should find an n such that $|x-x_{n}|<\varepsilon$. But this could be a chance event. If the error exceeds $\varepsilon$ at a later stage, then the scheme cannot be a good approximation as it is not “stable”. Instead, it would be desirable that accuracy is achieved at a certain stage and it should not get worse after that stage. This can be realized by demanding that there is a natural number $n_{0}$ such that $|x-x_{n}|<\varepsilon$ for all $n > n_{0}$. It is clear that $n_{0}$ will depend on $varepsilon$. This leads to the notion of convergence, which is the subject of a later blog.

More later,

Nalin Pithwa

# Real Numbers, Sequences and Series: Part 6

Theorem:

Given $x \in \Re_{+}$ and $n \in N$, we can find $y \in \Re$ such that $x=y^{n}$.

Proof:

Let $A={u \in \Re_{+}: u^{n}. If $x<1$, then $x^{n} and hence $x \in A$. On the other hand, if $x \geq 1$, then $1/2 \in A$. Thus, A is non-empty. Next, observe that if $v^{n}>x$, then v is an upper bound of A. In particular, $\{ 1+x\}$ is an upper bound of A.

By the least upper bound property, A admits a least upper bound. Let us denote it by y. We will rule out the possibilities $y^{n}>x$ and $y^{n} implying that $x=y^{n}$.

If $y^{n}, let $a=\frac{x-y^{n}}{nx}$ and $z=y(1+a)$. It can be checked that $z^{n} so that $z \in A$. But, $y , contradicting the fact that y is the least upper bound of A. (we have used the inequalities 7 and 8 in the previous blog).

On the other hand, if $y^{n}>x$, let $\frac{y^{n}-x}{ny^{n}}$ and $w=y(1-b)$. Again, it can be verified that $w^{n}>x$ and hence w is an upper bound of A. But, $w, contradicting the fact that y is the least upper bound of A. Hence, $y^{n}=x$QED.

In particular, we see that there is an element $\alpha \in \Re$ such that $\alpha^{2}=2$ and hence also $(-\alpha)^{2}=2$ which means that the equation $x^{2}=2$ has two solutions. The positive one of those two solutions is $\sqrt{2}$. In fact, the above theorem has guaranteed its extraction of the square root, cube root, even nth root of any positive number. You could ask at this stage, if this guarantees us extraction of square root of a negative number. The answer is clearly no. Indeed, we have

$x^{2} \geq 0$ for $x \in \Re$.

Remark.

We can further extend $\Re$ to include numbers whose squares are negative, which you know leads to the idea of complex numbers.

We have shown that Q is a subset of $\Re$. One can also show that between any two distinct real numbers there is a rational number. This naturally leads to the decimal representation of real number: Given any real number x and any $q \in N$, we can get a unique $a_{0} \in Z$ and unique $a_{1},a_{2}, \ldots a_{q} \in N$ such that $0 \leq a_{1} \leq a_{2} \ldots a_{q} \leq 9$ and

$|x-(a_{0}+a_{1}/10+a_{2}/100+\ldots + \frac{a_{q}}{10^{q}})|<\frac{1}{10^{q}}$

You are invited to try to  prove this familiar decimal representation.

If we have a terminating decimal representation of a real number, then surely it is rational.But, we know that rationals like 1/3, 1/7, 1/11, do not have a terminating decimal expansion.

It is clear that the decimal representation of $\sqrt{2}$ cannot terminate as it is not rational. There are many elements of $\Re$ which are not in Q. Any element of $\Re$ which is not in is called an irrational number, and irrational numbers cannot have terminating decimal representation.

More later,

Nalin Pithwa

# 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 limits: part IV

Representation of Q on the Number Line.

We have seen earlier that every element of can be represented on a straight line with a certain point representing 0. Positive integers are represented by points to the right of this marked point at equal lengths, and negative integers are similarly represented by points to the left of this.

We can use the same straight line to represent the newly constructed rational numbers as follows.

We know that we can divide a line segment, using a straight edge and compass, into q equal parts. So, the line segment between 0 to 1 can be divided into q equal parts for every positive integer q.

The points of division will now represent the numbers $\frac{1}{q}, \frac{2}{q}, \ldots , \frac{q-1}{q}$. It is now clear that any rational number of the form $\frac{p}{q}$ can be represented on the straight line. We have seen that between any two distinct rational numbers r and $r^{'}$, no matter how close they are, we can always find another rational number between them, example, $\frac{r+r^{"}}{2}$. Geometrically, this means that between any two distinct points on the line representing rational numbers, there is a point between them representing a rational number.

One can ask: does every point on the line correspond to a rational number? Note that we also have a point on the line representing length of the diagonal of a square of side 1. But, we shall show in the next blog that the length of the diagonal of unit square does not correspond to a rational number.

To prepare the ground for such discussions, let us start with a few definitions:

A subset $A \subset Q$ is said to be bounded below if there exists a rational number $\alpha$ such that

$\alpha \leq \beta$ for every $\beta \in A$

and $\alpha$ is called a lower bound of A. Similarly, $A \subset Q$ is said to be bounded above if there is a $\gamma \in Q$ such that $\beta \leq \gamma$, and $\gamma$ is called an upper bound of A. A set which is bounded above and also bounded below is called a bounded set. Observe that if a set has a lower bound $\alpha$ then it has many more like $\alpha -1$, $\alpha -2, \ldots$. Similarly, a set that is bounded above has many upper bounds. If there is a least element of all these upper bounds, then we call it the least upper bound of the set. But, it is not true that every set of rational numbers which is bounded above has a least upper bound. We can say similar things about sets which are bounded below. Some sets may have least upper bound. For example, the set of negative rational numbers is certainly bounded above and it is easy to see that 0 is the least upper bound.

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

# Real Numbers, Sequences and Series: Part II

Addition and multiplication of natural numbers.

For $m, n \in \textbf{N}$, we define inductively their sum $m+n$ inductively:

a) $m+1=m^{+}$ and (b) $m+n^{+}=(m+n)^{+}$

The above two are enough to define the sum $m+n$ for any m and $m \in \textbf{N}$. Indeed, if $A= \{ n: m+n \hspace{0.1in} is \hspace{0.1in} defined\}$, we have $1 \in A$ since $m+1$ is defined to be set $m^{+}$.

Now, if $n \in A$. We can now verify (Exercise):

(i) $m+n=n+m$ and (ii) $m. n^{+}=m.n+m$

It is easy to verify that (Exercise)

(iii) $m.n=n.m$ for all $m,n \in \textbf{N}$.

(iv) $m.(n.p)=(m.n).p$ for all $m,n,p \in \textbf{N}$.

(v) $m.(n+p)=m.n+m.p$, for all $m,n,p \in \textbf{N}$

For $m,n \in \textbf{N}$, we say that m is less than n, written as $m < n$, if there exists $k \in \textbf{N}$ such that $m+k=n$.

After we have defined natural numbers, we are in a position to define what is called a finite set. A non-empty set is called a finite set if there is a natural number n such that there is a bijection

$f: A \rightarrow \{ 1,2, \ldots, n \}$ —- Equation 2.1

The empty set is taken to be finite set by convention. A set which is non-finite set is called an infinite set. This means that for a set A to be infinite, one has to show that we cannot construct a bijection f as  in Equation 2.1 above. This may be incovenient to actually implement in practice. The following criterion, due to Dedekind, gives a more convenient way of deciding if a set is infinite or not.

Theorem.

A set is an infinite set if and only if it has a proper subset to which it is equivalent or equivalent.

(I am not presenting the formal proof here).

It is easily seen that is equinumerous to $\textbf{N} -\{ 1 \}$. Indeed, the bijection defined below shows this:

$f(n)=n+1$, that is, so 1 is mapped to 2, 2 is mapped to 3, is mapped to 4, and so on, so forth.

The bijection is interestingly put as follows: There was a hotel with a room corresponding to each natural number. That is to say, there are rooms with numbers $1,2,3, \ldots$ and so on. Once, when all the rooms were occupied by guests, a new customer arrived and wanted a room badly. How did the manager manage to accommodate the new customer without evicting any of the already existing guests from the hotel?

The manager knew mathematics. He simply requested the customer of room number 1 to move to room number 2, customer of room number 2 to move to room number 3 and so on. Now each customer has been accommodated without duplication where as room number 1 is vacant. He can now accommodate the new customer in that room. Thus, we have been able to show that is equivalent or equinumerous to a proper subset $\textbf{N}-1$ of itself. Hence, it must be an infinite set. A set is said to be countably infinite or just, countable, if it is equivalent to N.

Integers

We have already seen that we can add and multiply two elements of N. We can also “subtract” an element of from another provided the element from which we are subtracting is greater than or equal to the element we are subtracting. If $m,n \in \textbf{N}$, $m>n$, then there is a unique number $p \in \textbf{N}$ such that $m=n+p$, and we write $p=m-n$.

This operation is what is called subtraction as we had learnt in our elementary school. But, as we have seen, we can subtract a number only from larger numbers and not the other way round. This is the same situation as when I have Rs. 10000 in my bank account and I write a cheque for Rs. 9000 , the bank honours the cheque and pays the payee Rs. 9000 deducting it from my account leaving a balance of Rs. 1000. On the other hand, if I write a cheque  for Rs. 15000, the bank will usually dishonour the cheque. But, sometimes, depending on my creditworthiness, the bank obliges me by allowing an overdraft, that is, the bank remembers that I owe the bank Rs. 5000 and yet allows me to maintain my account. This is sometimes called an account with a “negative” balance.

Let us assume that the banker decides to designate deposits in natural numbers and withdrawals in bold face numbers. While bringing the account up-to-date, if the withdrawals total less than deposits, then the balance is entered in natural numbers. If the withdrawals total more than deposits, then the excess of the withdrawals over the deposits (what is called the overdraft) is entered in the balance column in bold face numbers, meaning that the account holder, meaning that the account holder owes the amount entered in bold face, to the bank. When he deposits some amount next time, the balance is entered after deducting the overdraft from the deposit. Thus, the banker introduces a new set of numbers which he writes in bold face. Similarly, he writes 0 when the balance is nil. But, the rules of addition and subtraction are more general than those for  the natural numbers. For example,

$5+\textbf{4}=1$

meaning that if one has a balance of Rs. 5/- and withdraws Rs. 4/- then the balance of Re. 1/- remains. Similarly, one writes

$100+\textbf{150}=\textbf{50}$

implying that if one withdraws Rs. 150/- out of a total deposit of Rs. 100/-, then one is left with an overdraft of Rs. 50/-, or

$\textbf{250} + 300=50$

meaning that if one has an overdraft of Rs. 250/- and deposits Rs. 300/-, then his net balance is Rs. 50/-. Also,

$180+0=180$ and $70+\textbf{70}=0$

meaning that if the balance is 180 and nothing is deposited nor withdrawn, then the balance is still Rs. 180 and if the entire balance (of Rs. 70) is withdrawn,  then the balance becomes nil.

Thus, we observe that the banker, by augmenting the natural numbers with a new set of numbers which are written in bold face, is able to add and subtract at will without any restriction. The banker’s numbers now consist of the natural numbers and a set of new numbers which he writes in bold face. So his  number system now is

$\{ \ldots, \textbf{3},\textbf{2}, \textbf{1},0, 1,2,3, \ldots\}$

It should be now be clear that the operations of subtraction which was earlier of restricted applicability can now be used in this new augmented number system by setting $0-n=\textbf{n}$, where $n \in \textbf{N}$.

We may use the notation $\textbf{1}=-1, \textbf{2}=-2, \textbf{3}=-3, \ldots$

where each bold face number is replaced by a number with a negative sign prefixed. Now our augmented number system is $\{ \ldots, -3,-2,-1,0,1,2,3 \ldots\}$.

We denote this changed set of numbers by $\textbf{Z}$, (which is the first of the German word Zahlen for integers).

Clearly, $\textbf{N} \subset \textbf{Z}$ and we would expect $\textbf{Z}$ to be amenable to at least the operations we could perform on $\textbf{N}$ Let us take the operation of addition first. For

$m, n \in \textbf{Z}$ there is a unique number written $m+n \in \textbf{Z}$ such that

a) $m+n=n+m$ for all $m, n \in \textbf{Z}$,

b) $m+(n+p)=(m+n)+p$ for all $m, n, p \in \textbf{Z}$

c) $m+0=m$ for all $m \in \textbf{Z}$

d) for every $m \in \textbf{Z}$ there is a number $n \in \textbf{Z}$ such that $m+n=0$.

One can show that this number is unique and we write $n=-m$. We have automatically

$m+n=m \Longrightarrow n=0$

This leads to an even more interesting property $-(-m)=m$ for all $m \in \textbf{Z}$.

Indeed, for every $m \in \textbf{Z}$, there is a unique $-m \in \textbf{Z}$ such that $m+(-m)=0$.

Again, by the same argument, there is $(-m) \in \textbf{Z}$ such that $-m+(-(-m))=0$, adding m to both sides, we get

$m+(-m+(-(-m)))=m+0=m$

But, by associativity, we have

$m+(-m+(-(-m)))=(m+(-m))+(-(-m))$

This gives $0+(-(-m))=m$ which is the same as $-(-m)=m$

Next, consider multiplication. For all $m, n \in \textbf{Z}$ we should have a unique $m.n \in \textbf{Z}$ having the following properties:

1) $m.n=n.m$ for all $m, n \in \textbf{Z}$

2) $m.(n.p)=(m.n).p$ for all $m, n, p \in \textbf{Z}$

Besides, where it involves addition and multiplication, the distributive rule:

3) $m.(n+p)=m.n+m.p$ for all $m, n, p \in \textbf{Z}$ should be satisfied.

We can now demonstrate:

i) $m.0=0.m=0$ for all $m \in \textbf{Z}$

ii) $(m+n).p=m.p+n.p$ for all $m, n, p \in \textbf{Z}$

iii) $m.1=1.m=m$ for all $m \in \textbf{Z}$

iv) $m.(-n)=(-m).n=-(m.n)$ for all $m,n \in \textbf{Z}$

v) $(-m).(-n)=m.n$ for all $m, n \in \textbf{Z}$

Part (i) is clear because we have the relation $n+0=n$. Multiplying by $m \in \textbf{Z}$, we get

$(n+0).m=n.m+0.m=n.m$ implying the required result.

ii) $(m+n).p=p.(m+n)=p.m+p.n=m.p+n.p$

Now, prove the rest and test your mettle for math!

The elements of $\textbf{Z}$ are called integers.

More later,

Nalin Pithwa

# Real Numbers, Sequences and Series: part I:

Natural Numbers.

Natural numbers are perhaps, the earliest mathematical notions of man. The difference between a pack of wolves and one wolf, a swarm of bees and one bee, a school of fish and a single fish, a heap of stones and a single stone, etc., may have led primitive man tbo systematize the idea of numbers. The fact that in Sanskrti (and in some other ancient Indo-European languages) there are usages where one object, two objects and many objects are distinguished, example, narah, nari, naraah — indicates man’s attempt to distinguish betweeh numbers. Heaps of stones found in caves, where early man lived, and the discovery of animal bones on which notches have been cut in regular series of five are indicative of devices by the human beings of yore to keep count of their livestock. That is how they understood plurality.

But, what do we understand by numbers? Is number the same thing as plurality ? We often use expressions like “there are five fingers on my hand”, “Pandavas were five brothers”, “there are five books on this table”, “there are five mangoes in the basket”, etc. Observe what we are talking about. In the first case, we were talking about a set of fingers on my hand, in the next about a set of brothers, next about a set of books  on a table, next about a set of mangoes. But to each of these sets we attached the common attribute “five”. Why was that? There must have been something common in all these sets to justify our attaching the common attribute. The commonality between them is that the elements of one set can be put in one-to-one correspondence with the elements of the other sets under discussion.

We say that two sets A and B are equinumerous (or of the same cardinality) if there is a map

$f: A \rightarrow B$

which is one-to-one and onto, that is, a bijection. One sees that if A and B are equinumerous and the sets B and C are also equinumerous, then so are A and C. One can easily observe that being equinumerous is an equivalence relation among sets.

This relation decomposes any class of sets into disjoint equivalence classes of sets. Two sets are in the same equivalence class if they are equinumerous. The equivalence class is characterized by a common property of its members, that is, any two of them are equinmerous. This characterizing property is what we would like to call the number of elements of the set belonging to the equivalence class. Since equivalence classes, into which we have classified the sets, are disjoint, the characterizing property associated with disjoint classes are distinct. “Commonness”, “common property” or ‘characteristics” is too imprecise an idea to be handled easily; some authors have defined the number of elements of a set as the equinumerous class to which it belongs. Thus, the number “two” would mean the class of all pairs, the number “three” the class of all triplets, …and so on. But, this too has some foundational problems which are hard to circumvent. We do not go into the details of such things here.

Instead, we adopt the axioms devised by Peano for the natural numbers N with the following properties:

1) There is a map (called the successor map) from N to N, sending an element n to $n^{+}$ (called the successor of n), which is injective, that is, $n \neq m \Longrightarrow n^{+} \neq m^{+}$

2) There is an element 1 that is not a successor of any element (that is, 1 is not in the image of any successor map, that is, $1 \neq n^{+}$ for all $n \in N$). Also, every other element of N is the successor of some element of N.

3) Suppose that $A \subseteq N$ and that (a) $1 \in A$ (b) $n^{+} \in A$ whenever $n \in A$. Then, $A=N$, that is, the only subset of N which contains 1 and the successor of each of its elements in N itself.

The third axiom is known as the Principle of Mathematical Induction.

More later,

Nalin Pithwa