Whole numbers

For natural numbers subtraction doesn’t always makes sense. In this chapter we will investigate how to solve this problem and why it is worthwhile to do so.

As we have demonstrated in the previous chapter, for natural numbers we can always change the order of subtractions. For example:

(5+13)-10=8

however, in:

(5-10)+13

5-10 is a nonsense: we cannot eat more apples than we have. How can we fix that? Let’s simply accept 5-10 as a new type of number and assume that 5 extra apples were borrowed and we owe them.

Table of Contents

Negative numbers

Let’s start with a concrete example: 5-10. We can simplify this into:

5-10=5-5-5=0-5

In general, if natural b>a, then there must exist natural c such that b=a+c>a thus:

a-b=a-(a+c)=a-a-c=0-c

To shorten the notation, let’s write: (0-c)=-c. We will call numbers like that negative.

Subtraction as addition of negative numbers

To start with, it would be useful to find a scenario in which negative numbers do make sense. Imagine walking forward and backward along a straight line from some starting point. Steps taken forward are expressed by positive numbers and steps taken backward are negative.

With this in mind we can put together a list of n objectives: s₁, s₂, s₃, … ,s_n, each representing a number of steps to be taken either forward or backward. To keep all steps equally long, you simply make sure that the heel of the foot in front always touches the toes of the back foot. Each next task from the list is always performed from the place you currently are. Taking negative number of steps “-b” is equivalent to subtraction:

a+(-b)=a+(0-b)=(a+0)-b=a-b

where “a” denotes your current location i.e. the total number of steps from the starting point before moving “b” steps backward.

It doesn’t matter in what order we go after the objectives. Eventually we always arrive at the same destination which is either in front (positive) or behind the starting point (negative). Why? Because irrespective of the order, the total effective number of steps taken forward “f” and the total effective number of steps taken backward “b” do not change. You always arrive at f-b steps from the starting point.

Conclusion: we can add positive and negative numbers in any order, and the result is always the same, thus it can be shown that commutative and associative properties of addition must be preserved for negative numbers too. It can be demonstrated in exactly the same way as for natural numbers so I won’t repeat the reasoning here again.

Note that after we supplement natural numbers with their negatives, we fix the problems we had with subtraction:

a-b always makes sense i.e. the result is either positive, 0 or negative
subtraction doesn’t commute i.e. a-b≠b-a, yet thanks to negative numbers we can replace it with addition which does commute:
a-b=a+(-b)=(-b)+a

Set ℕ (the one containing all natural numbers) supplemented with all negative numbers and 0 is denoted with the symbol ℤ. Members of ℤ are called whole numbers. ℕ is a subset of ℤ.

Negative of a negative

In the previous section we were adding positive and negative numbers. What happens if we subtract a negative number:

a-(-b)=?

Number negative to “c” (variable “c” represents any whole number) has the following property:

(-c)+c=c+(-c)=c-c=0

Note that for every positive number there is a unique negative number and for every negative number we have a unique positive number!

“c” has its negative -c, what is negative of -c ? The sum of a number with its negative is zero so the sum of -c and its negative –(-c) must be zero:

(-c)+(-(-c))=(-c)-(-c)=0

If you compare the above equation with the previous one, it’s clear that a negative of a negative must be positive:

-(-c)=c

thus

a-(-b)=a+(-(-b))=a+b

Can you figure out the result of: -(a-b) and why?

As we know from this section, for positive “a” and “b” we have:

-(a+b)=-a-b

That is: we can subtract a+b or “a” first and then “b”. By the same token we can write:

Eq.1: -(a-b)=-(a+(-b))=

=-a-(-b)=-a+(-(-b))=

=-a+b=b+(-a)=b-a

Order

Which is greater -10 or -5? To see it clearly, we will use the equivalence that is the consequence of the properties of natural numbers discussed in this chapter:

$b > a$ $\boldsymbol{\Leftrightarrow}$ $b - a > 0$

Using this as a template, let’s accept that whole number “b” is greater than whole “a” only if b-a>0. With this in mind for b=-5 and a=-10 we have:

b-a=-5-(-10)=-5+10=10-5=5>0

so -5 is greater than -10:

-5 > -10.

By the same token, -c is smaller than 0 for c>0:

0-(-c)=0+c=c>0

thus:

0>-c

Of course, any two whole “a” and “b” are either equal or one of them must be greater. That means we can put all whole numbers in a descending or ascending order.

Equalities, inequalities and addition

Let’s find out whether adding the same whole number on both sides of equalities and inequalities preserves them.

For whole “a”, “b” and “c”, let’s add/subtract “c” to/from “a” and “b” to see if it affects the distance between them:

(b+c)-(a+c)=b+c-a-c=b-a

(b-c)-(a-c)=b-c-a+c=b-a

If b=a then b-a=0 and using the formulas above we conclude that:

b+c=a+c

and

b-c=a-c

If on the other hand b>a, then:

(b+c)-(a+c)=b-a>0

(b-c)-(a-c)=b-a>0

and from the equivalence adopted in the previous section:

$b > a$ $\boldsymbol{\Leftrightarrow}$ $b - a > 0$

it follows that:

b+c>a+c

and

b-c>a-c

By the same token if b≥a, then:

b+c≥a+c

b-c≥a-c

All these properties can be recapped as:

$b = a$ $\boldsymbol{\Rightarrow}$ $b \pm c = a \pm c$

$b > a$ $\boldsymbol{\Rightarrow}$ $b \pm c > a \pm c$

$b \ge a$ $\boldsymbol{\Rightarrow}$ $b \pm c \ge a \pm c$

where ± means + or -.

Negative numbers and multiplication

Are negative numbers compatible with multiplication? According to our definition of multiplication a∙(-b) means repeating “a” times the action of moving “b” steps backward.

If we move a∙b steps forward and then a∙(-b), we arrive where we started so:
a∙b+a∙(-b)=0

We can also write:
a∙b+(-(a∙b))=a∙b-a∙b=0

Comparison of the two equalities above, leads to the following conclusion:
a∙(-b)=-(a∙b)

What about: (-a)∙b =? Here our definition stops working because taking a negative number of steps makes no sense. We can either surrender or get creative. What is needed is an equality that would produce the answer.

What is the result of:
a∙b+(-a)∙b?

We know that:
a∙(b-b)=a∙(b+(-b))=0

For the sake of symmetry, let’s simply demand the following to be true as well:
a∙b+(-a)∙b=(a+(-a))∙b

Of course:
(a+(-a))∙b=(a-a)∙b=0

so:
a∙b+(-a)∙b=0

For natural a and b we have:
a∙b+(-(a∙b))=a∙b-a∙b=0

Comparison of these two equalities leads to the conclusion:
(-a)∙b=-(a∙b)

Altogether we have the following relations:
(-a)∙b=-(a∙b)=a∙(-b)

so the brackets can be dropped:
Eq.2: (-a)∙b=-(a∙b)=a∙(-b)=-a∙b

No matter how we interpret -a∙b, the result is the same.

So far we have examined all the cases apart from two negative numbers multiplied by each other. Can you figure out what will be the result?

Two negative numbers can be expressed with variables: -a and -b where a,b>0. After applying Eq.2 twice, we get:
(-a)∙(-b)=-(a∙(-b))=-(-(a∙b))=a∙b

Done!

Now that we know how to handle multiplication of negative numbers, let’s get back to our “walking along a path” thought experiment to see what are the consequences. Taking a negative number of steps doesn’t make sense. On the other hand, we have:
(-a)∙b=a∙(-b)
(-a)∙(-b)=a∙b

so to fix our old definition of multiplication, we simply have to remember that “-” in front of “a” reverses the direction of steps.

Multiplication of natural numbers is commutative:
Eq.3: a∙b=b∙a

and associative:
Eq.4: (a∙b)∙c=a∙(b∙c)

Let’s check whether these properties are preserved for whole: a, b and c.

If we assume “b” to be negative, then we can write b=-b’ where b’>0. After inserting this into Eq.3 we get:

left hand side of the commutative property: a∙b=a∙(-b’)=-(a∙b’)
right hand side of the commutative property: b∙a=(-b’)∙a=-(b’∙a)=-(a∙b’)

Both sides are equal, so the commutative property holds for negative b=-b’.

Next, let’s insert b=-b’ into Eq.4:

left hand side of the associative property:
(a∙b)∙c=(a∙(-b’))∙c=(-(a∙b’))∙c=-((a∙b’)∙c)
right hand side of the associative property:
a∙(b∙c)=a∙((-b’)∙c)=a(-(b’∙c))=-(a∙(b’∙c))=-((a∙b’)∙c)

The sides are equal, so the associative property holds for negative b=-b’.

In the same fashion we can prove that Eq.3 and 4 hold for negative “a” and “c” as well. Conclusion: whole numbers turned out to be consistent with the commutative and associative properties of multiplication.

Distributive property of multiplication over addition

Is the equality:
(a+b)∙c=a∙c+b∙c
valid for whole numbers?

If b is negative, then b=-b’ where b’>0 (b’ is simply a new variable: b’=-b) so:
(a+(-b’))∙c=(a-b’)∙c

We already know that If a≥b’≥0 then:
(a-b’)∙c=a∙c-b’∙c

that with Eq.2 can be transformed into:
a∙c-b’∙c=a∙c+(-b’)∙c

so:
(a+(-b’))∙c=a∙c+(-b’)∙c

thus:
(a+b)∙c=a∙c+b∙c
must be valid for negative b as well as long as a≥b’≥0.

What if b’>a≥0? We have:
(a+(-b’))∙c=(a-b’)∙c

that using Eq.1 can be transformed into:
(a-b’)∙c=(-(b’-a))∙c

and thanks to Eq.2 rewritten into:
(-(b’-a))∙c=-((b’-a)∙c)

We have assumed: b’>a≥0 so for the term (b’-a)∙c we have:
-((b’-a)∙c)=-(b’∙c-a∙c)=

and finally:
=-(b’∙c+(-a∙c))=-b’∙c-(-a∙c)=-b’∙c+a∙c=(-b’)∙c+a∙c=a∙c+(-b’)∙c

Our starting point was (a+(-b’))∙c so ultimately for b’>a≥0 we can write:
(a+(-b’))∙c=a∙c+(-b’)∙c

In other words, we have proved that:
(a+b)∙c=a∙c+b∙c
holds for every negative “b” as well. The situation for negative “a” is analogous, so we don’t have to check it separately.

What if both “a” and “b” are negative: a=-a’, b=-b’, a’,b’>0? We can write:
((-a’)+(-b’))∙c=(-a’-b’)∙c=(-(a’+b’))∙c=-((a’+b’)∙c)

Now all terms in the brackets on the right are positive, so we can transform it further as follows:
-((a’+b’)∙c)=-(a’∙c+b’∙c)=-(a’∙c)-(b’∙c)=(-a’)∙c+(-(b’∙c))=(-a’)∙c+(-b’)∙c

Altogether:
((-a’)+(-b’))∙c=(-a’)∙c+(-b’)∙c

that is:
(a+b)∙c=a∙c+b∙c
must hold for when both “a” and “b” are negative as well.

The last case to check involves negative c: 0>c=-c’, c’>0. In this can we can write:
(a+b)∙(-c’)=-((a+b)∙c’)

We know that for whole “a” and “b” and positive c’ we can transform it further into:
-((a+b)∙c’)=-(a∙c’+b∙c’)=

and finally into:
=-a∙c’-b∙c’=a∙(-c’)+(-(b∙c’))=a∙(-c’)+b∙(-c’)

so:
(a+b)∙(-c’)=a∙(-c’)+b∙(-c’)

Conclusion:
(a+b)∙c=a∙c+b∙c
holds for whole a, b and c.

Equalities, inequalities and multiplication

It is easy to see that for whole numbers we must have:
$b = a$ $\boldsymbol{\Rightarrow}$ $b\cdot c = a\cdot c$

yet the situation for inequalities is not that straightforward.

Consider: -3<-2.

Let’s multiply both sides times -1:
3>2

so the multiplication flipped “<”.

Let’s take a look at a general case. We have the equivalence:
$b > a$ $\boldsymbol{\Leftrightarrow}$ $b-a>0$

Let’s assume b>a and multiply (b-a)>0 times c:

if c>0 then (b-a)∙c=b∙c-a∙c>0 and from the equivalence above it follows that: b∙c>a∙c
if c=0 then (b-a)∙0=0 and the inequality is no longer true
if c<0 then (b-a)∙c=b∙c-a∙c<0 and the inequality is flipped: b∙c<a∙c

b≥a behaves similarly, yet this time b∙c≥a∙c is true for c=0 as well.

Division

Just like for natural numbers, division of a whole number usually doesn’t produce a whole number. Additionally, the result is not unique. For example for 10/4 we have:

10=2∙4+2

10=3∙4-2.

The remainder of division is either 2 or -2.

Another problem is with division by a negative number. What does it mean to divide something into a negative number of parts? We will get back to this problem later.

Negative numbers and positional systems

A negative number in the positional system is simply obtained with a minus put in front of the positional representation of this number:

$-\sum_{i=0}^{n}d_i\cdot \Gamma^{i}$

Addition or subtraction of whole numbers discussed in this chapter can be transformed into operations on natural numbers. Let’s take a look at a few problematic cases to illustrate the point:

5-10=-(10-5)=-5
-3-4=-(3+4)=-7
-3-(-4)=-3+4=4-3=1

The same can be said for multiplications. For example:
5∙(-2)=-(5∙2)
(-5)∙2=-(5∙2)
(-5)∙(-2)=5∙2

In other words, after some minor transformations, the procedures for addition, subtraction and multiplication of natural numbers can be used in the realm of whole numbers.

Summary

The set of whole numbers ℤ is ordered and all the rules of addition and multiplication hold for it just like they do for ℕ (the set of natural numbers).

Using whole numbers we can evaluate every complicated expression involving subtraction without having to worry whether it makes sense or not. Let’s take a look at the procedure for multiplying on fingers:

((a-5)+(b-5))∙10+(5-(a-5))∙(5-(b-5))=

Now that we know the rules for operations on whole numbers, we can easily transform this into:
=(a+b-10)∙10+(10-a)∙(10-b)=

=10∙a+10∙b-100+(10-a)∙10-(10-a)∙b=

=10∙a+10∙b-100+100-10∙a–(10∙b-a∙b)=

=(100-100)+(10∙a-10∙a)+(10∙b-10∙b)+a∙b=a∙b

We have just proved that this procedure is valid for every pair of whole numbers and yields positive results as long as “a” and “b” are both positive or negative.

Whole numbers are useful for counting non-divisible things. Numbers adequate for handling divisible objects are the subject of the next chapter.