# 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