Rational numbers

Number types described in previous chapters are useful for describing the world of indivisible things. Here we will extend our system to be able to handle divisible objects as well.

We can divide a piece of string into as many parts as we wish, so the remainder of division is not applicable in this case. How can we measure how much string we have to begin with though, or to be more general, how can we specify the length?

Measurement of length

Measuring length is quite straightforward. We grab a string or tape and cover it with evenly spaced marks. Then we simply extend the tape along the chosen object and count the number of marks between chosen points. Accuracy of the procedure depends on the density of marks.

Of course, it is important that all measuring tapes are identical. To do that we simply have to agree on a unit, that is: a distance between the marks.

There is no such thing as a unit good for everything. However, instead of using a multitude of incompatible units, it’s better to agree on a single one, and multiply or divide it by a natural number to get the adequate resolution. The most common unit of length is meter and its derivatives: millimeter=m/1000, centimeter=m/100, kilometer=1000m etc.

Multiplying a unit times natural numbers is straightforward, but what about division? Dividing by 2 is easy: we simply fold the unit on itself. Dividing by 4 is easy too: we divide by 2 twice.

Is it possible to divide a piece of string into 3 or 5 equally long parts? For now, we are going to assume that to be possible. The procedure for this will be demonstrated in the chapter on geometry.

Ratios

Let’s divide 11 meters of string among 4 people. 11 is not divisible by 4 (11=2∙4+3), but the meter is, so perhaps instead of dividing 11∙m by four, we could add “m/4” eleven times:
11∙(m/4)=(2∙4+3)∙(m/4)=(2∙4)∙m/4+3∙(m/4)=2m+3∙(m/4)

Everyone gets a piece of string having the same length of 2 and three quarters of a meter. How do we know though that it is equivalent to dividing 11∙m by 4:
11∙(m/4)=(11∙m)/4

We can’t just claim that these two are equal, it has to be demonstrated. For this purpose we will use the following easy-to-prove formula:
Eq.1:     (a∙b)∙(L/n)=a∙(b∙(L/n))

To make it general, we use variables a, b, n instead on concrete numbers, and L representing any unit instead of just meter. Note that irrespective whether we add L/n “a∙b” times (left-hand side) or add L/n in blocks of “b” terms in each and then add all the blocks together (right-hand side), the result must be the same.

We also know that adding L/n n times results in L:
n∙(L/n)=L

so we can write:
a∙L=a∙(n∙(L/n))=

and using Eq.1 (with b=n), we can move the brackets (…) in the following way:
=(a∙n)∙(L/n)=

then replace the order of multiplication inside (…):
=(n∙a)∙(L/n)=

and using Eq.1, move the brackets (…) again:
=n∙(a∙(L/n))

The resulting formula reads:
a∙L=n∙(a∙(L/n))

After dividing both sides into “n” equal parts, we get:
(a∙L)/n=a∙(L/n)

One n-th of the distance a∙L is equal to the distance L/n added “a” times so:
(11∙m)/4=11∙(m/4)

and we have proved what we wanted.

To be able to measure distances in terms of meters, we need to be able to tell how many times meter fits into any other arbitrary distance. In other words, we have to perform a division:
“some distance”/meter=”how many meters”

that by definition is equivalent to:
“some distance”=”how many meters” ∙ meter

Let’s take a look at some examples:
(n∙L)/L=n

No problems here, n∙L is n times bigger than L.

What about:
(L/n)/L=?

L/n is n times shorter than L irrespective of how long L is. How can we express this relation? We are going to use “1/n” to represent this and accept “1/n” as a new type of a number.

Note that instead of 1 we can insert some other arbitrary natural number “a” that is either divisible or not divisible by n. “a/n” is called a ratio and all numbers that can be represented like this are called rational numbers.

From now on, a multiplication of a rational number “a/n” times some divisible object L will be by definition equal to:
Eq.2:     (a/n)∙L:=(a∙L)/n=a∙(L/n)

in agreement with examples presented above. (L/n)/L=1/n is equivalent to:
L/n=(1/n)∙L

Negative ratios

What about a negative distance (-11m)/4? Does it make sense? In one of the previous chapters we have played with a thought experiment in which steps were counted using whole numbers. Now, thanks to ratios, every step can have a different length e.g. L, L/2, (3/2)L etc.

Just like before, minus will be used to denote steps taken backwards. Keep this experiment in mind. It will help you interpret the meaning of the operations described below.

If we agree that L represents a foot’s length and “x” the number of steps taken, we can write:
(-x)∙L=x∙(-L)=-(x∙L)
(-x)∙(-L)=x∙L

because we have discovered earlier that minus simply flips the direction of moving. Note that thanks to (-x)∙L=-(x∙L) we can drop the brackets -x∙L because no matter where we place them, the result is the same.

In the previous section we have established a few facts. Do they still hold in the negative direction?

Let’s start with Eq.1: (a∙b)∙(L/n)=a∙(b∙(L/n)).

It is easy to see that it must be true if we replace L with its negative –L because we can simply perform the same reasoning as for L just in the opposite direction.

The next obvious question is whether Eq.1 holds for negative “a” and/or “b”. To find out, let’s insert for example b=-b’ (‘ in b’ is simply used to distinguish it from b), b’>0 i.e. b must be negative. Thanks to the properties of whole numbers we can pull the minus out and place it in front of a∙b’:
(a∙(–b’))∙(L/n)=–(a∙b’)∙(L/n)

In the equality -x∙L=x∙(-L), x and L can be anything so we can insert a∙b’ in place of x, and L/n in lieu of L:
-(a∙b’)∙(L/n)=(a∙b’)∙(-(L/n))=

In the next step we move the brackets ( ) in agreement with Eq.1 (we already know Eq.1 does hold in the negative direction):
=a∙(b’∙(-(L/n)))=

Finally, we can place the minus in front of b’ instead of L/n (thanks to -x∙L=x∙(-L)):
=a∙((-b’)∙(L/n))

Altogether we have proved:
(a∙(-b’))∙(L/n)=a∙((-b’)∙(L/n))

that is: Eq.1 is true for negative b=-b’, b’>0. For negative “a” the reasoning is the same so Eq.1 must hold for when a and b are whole numbers.

In the previous section, equation Eq.2:
(a/n)L:=(a∙L)/n=a∙(L/n)

was established with the help of Eq.1, so Eq.2 must make sense for whole “a” and negative L too.

Note that after inserting n=1, we get:
(a/1)L:=aL/1=aL

thus a/1=a, i.e. whole numbers are also ratios. Moving forward we will denote a set of all rational numbers with the symbol ℚ.

The last thing we need to establish is: what is the negative of a rational number a/n? Following the convention introduced for whole numbers, we will denote it with -(a/n). By definition positive and negative cancel each other out so:
(a/n)L+(-(a/n))L=0

We also know that:
(a/n)L+((-a)/n)L=(a∙L)/n+(-a∙L)/n=0

irrespective of what divisible object we insert in lieu of L, so (-a)/n must be equal to the negative of a/n:
-(a/n)=(-a)/n

To summarize, a ratio x/y makes sense for negative x=-a. What about negative y=-n? Dividing a segment into negative number of parts sounds bonkers but maybe we could extend our system somehow to include these situations as well. Let’s take a look.

Obviously adding -a∙L/n n times is equal to -a∙L:
n∙(-a∙L/n)=-a∙L

From this it follows that:
(-n)∙(-a∙L/n)=n∙(a∙L/n)=a∙L

because taking negative number of steps -n simply flips the direction we are walking: (-x)∙(-L)=x∙L. Next, if we divide both sides of the equality above by -n, it makes sense that on the left-hand side we end up with -a∙L/n because we simply undo the multiplication by -n. After doing that, the formula reads:
(-a∙L)/n=(a∙L)/(-n)

As we know, L can be pulled to the right:
((-a)/n)L=(a/(-n))L

so:
(-a)/n=a/(-n)

In other words, a rational number is a ratio of any two whole numbers as long as the one at the bottom is not zero. After all, it’s not clear how to divide a segment into zero parts. Can we pull another hat trick to extend our system to accommodate this scenario as well? We will get back to this problem later.

Fraction of a fraction

Let’s take a fraction of some segment L: (a/n)L and then take a fraction of a fraction: (b/k)∙((a/n)L). Surely the result must be equal to some fraction of L itself?

Using Eq.2:
(a/n)L:=(a∙L)/n=a∙(L/n)

we can transform (b/k)∙((a/n)L) into:

After dividing aL into n parts and then dividing each of them into k parts, we end up with n∙k equally long pieces so the following must be true:

Subsequent application of Eq.2 and Eq.1: (a∙b)∙(L/n)=a∙(b∙(L/n)) leads to the following formula:

so altogether we have:

Eq.3:

for whole a, b, k, n where k and n cannot be 0. In other words, fraction of a fraction (left-hand side) is also a fraction (right-hand side).

Addition of segments of differing length

Let’s add (a/n)L and (b/n)L:

Easy!

What about adding aL/n and bL/k for n≠k? Well, we have to make them compatible first:

After that addition is straightforward:
Eq.4:

Addition of rational numbers

Can we add rational numbers without referring to distance in a similar way as natural numbers are independent from the objects we count?

For whole m and p we have:
(m+p)L= mL+pL

Let’s use it as a template for a definition (:=) of rational numbers:

After combining the above with Eq.4 we get:

Now we can drop L on the left and right because the equation is valid irrespective of what L is:
Eq.5:

Note that from the commutative and associative properties of addition we proved for whole numbers, and from Eq.5 it follows that addition of rational numbers must be also commutative:

and associative:

Subtraction of rational numbers

Using whole numbers as a template, we will adopt the following definition of subtraction:
a/b-c/d:=a/b+(-(c/d))

where ratio -(c/d) is a negative of c/d:
(c/d)+(-(c/d))=0=-(c/d)+(c/d)

Multiplication of rational numbers

So far we managed to define addition and subtraction of rational numbers without a direct reference to any concrete divisible object L. Can we do the same for multiplication?

If we insert n=1 in the Eq.1:
(a∙b)∙(L/n)=a∙(b∙(L/n))

we get:
(a∙b)L=a∙(bL)

that can be used as a template for defining (:=) a multiplication of ratios. To do that we will replace “a” with “c/k” and “b” with “d/h”:

According to Eq.3 we have:

so altogether:

and L can be anything divisible so we can drop it:
Eq.6:

Note that thanks to the properties of whole numbers, the multiplication defined in this way is commutative:

and associative:

Addition mixed with multiplication

As we have seen the commutative and associative properties of addition and multiplication for rational numbers follow straight from the properties of whole numbers. The same can be said about the distributive property of multiplication over addition:

Preservation of equalities and inequalities

We will examine equalities and inequalities using the thought experiment from the beginning of this chapter where fractional steps were introduced.

Adding some rational number s on both sides of the equality:
p₁=p₂

obviously preserves it:
p₁+s=p₂+s

because if two people start in the same position (p₁=p₂), move in the same direction and cover the same distance (s), they must arrive at the same destination.

It’s also quite obvious that if step s₁ of person 1 is equal to the step s₂ of person 2 (s₁=s₂), then:
w∙s₁=w∙s₂
where w represents number of steps.

What about inequalities? If we denote a starting point with p, step taken with s, and arrival point with n, we can write:
p+s=n

After adding -p on both sides we get:
p+s+(-p)=n+(-p)
so
s=n–p

If s corresponds to moving forward then the step is positive s>0, and if we move backwards, then the step is negative: s<0. From this it follows that we can define n>p as:

n>p    n–p>0

that is:

if n is greater than p: n>p, then to get from p to n we must move forward, and if moving forward gets us from p to n, then n must be greater than p.

Using this definition, rational numbers can be sorted into ascending or descending order.

Note that using the equivalence above, we can prove all the inequalities listed below in exactly the same fashion as for whole numbers (here and here).

n>p n+w>p+w
n≥p n+w≥p+w

For w>0:
n>p w∙n>w∙p

For w<0:
n>p w∙n<w∙p

For w≥0:
n≥p w∙n≥w∙p

For w≤0:
n≥p w∙n≤w∙p

Greatest common divisor (GCD)

A negative rational number is equal to the corresponding positive one with minus in front, so let’s focus on positive numbers. Certainly a ratio expressed with two natural numbers is positive. This representation is not unique, though. Many different pairs represent the same rational number e.g.:
2/4=(2/2)∙(1/2)=1/2
4/8=(4/4)∙(1/2)=1/2 etc.

Note that we can’t simplify the above beyond 1/2. How can we reduce a ratio of any two natural numbers A and B to its most basic representation?

If m is the greatest common divisor (GCD) of A and B, then there exist such a and b that:

and a/b is what we are after. What’s left to do is finding the greatest common divisor (GCD). Let’s take a look at a few examples.

Example 1

What is the GCD for 1517 and 1073?

Let’s divide the greater by the smaller:
1517=1∙1073+444

If 1517 and 1073 are both divisible by some number m, then 444 must be also. Why? Because if 1073 is divisible by m and 444 isn’t, then 1517 cannot be either (see the equality above).

Next, let’s assume that m is the GCD of 1517 and 1073. In such case, it must also be the GCD of 1073 and 444 because if 1073 and 444 had GCD>m, then it would have to be a divisor of 1517 and we know that m is the greatest.

Conclusion: if we find the GCD of 1073 and 444, it must also be the GCD of 1517 and 1073.

In the next step, we break apart 1073 in terms of 444:
1073=2∙444+185

and using the same reasoning as before, we conclude that the GCD of 444 and 185 is also the greatest for 1073 and 444, and it must also be the GCD for 1517 and 1073.

The procedure described above generates the following sequence of equations:
1517=1∙1073+444
1073=2∙444+185
444=2∙185+74
185=2∙74+37
74=2∙37

37 is the GCD of 74 and 37, so it must also be the GCD of (185,74), (444,185), … and (1517, 1073).

Let’s insert 74=2∙37 into 185=2∙74+37 and this one into the one above etc. all the way up to 1517:
1517=41∙37
1073=29∙37

The following two ratios must be equal then:
1517/1073=(37∙41)/(37∙29)=41/29

Example 2

What is the GCD of 44 and 105?

105=2∙44+17
44=2∙17+10
17=1∙10+7
10=1∙7+3
7=2∙3+1
3=3∙1

The GCD in this case is 1, so the ratio 44/105 cannot be simplified.

Example 3

In example 1 it was shown that the GCD for 1517 and 1073 is 37:
1517/1073=(37∙41)/(37∙29)=41/29

and 41/29 cannot be simplified any further. In other words, the GCD of 41 and 29 must be 1:
41=1∙29+12
29=2∙12+5
12=2∙5+2
5=2∙2+1
2=2∙1

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Let’s up our game a little bit by looking at the problem from an abstract point of view, i.e. instead of using concrete numbers we will describe the procedure presented above using variables.

For any two numbers such that A>B>0, we can break apart A in terms of B:
A=c∙B+R₀

where R₀<B is a rest of division A/B (why variable R has an index 0 will become clear very soon).

Next, let’s break apart R₀ in terms of m (R₀/m):
R₀=r∙m+d
d<m
where m is the GCD of A and B: A=a∙m, B=b∙m.

Combining the above two equations produces:
A=c∙B+R₀=c∙B+(r∙m+d)
A=a∙m=c∙(b∙m)+(r∙m+d)

and it is apparent that d must be zero because otherwise A wouldn’t be divisible by m, and from this it follows that R₀ must be divisible by m too.

Also, if m is the GCD of A and B then it must be the GCD of B and R₀. That’s because if B and R₀ had a greater common divisor, it would be a divisor of A as well.

We have A>B>R₀, so finding the GCD for B and R₀ will be easier than for A>B because it is easier to deal with smaller numbers.

In the next step, we will break apart B in terms of R₀ (B/R₀). w₀ denotes how many times R₀ fits into B:
B=w₀∙R₀+R₁
R₀>R₁

B and R₀ are divisible by m, so R₁ must be as well. We also know that m is the GCD of B and R₀, so it must be the GCD of R₀ and R₁.

For R₀>R₁ we can break apart R₀ in terms of R₁:
R₀=w₁∙R₁+R₂
R₁>R₂

and the GCD of R₁ and R₂ is also the GCD of R₀ and R₁. The continuation of this process can be described with a formula:
R_i=w_i+1∙R_i+1+R_i+2
R_i>R_i+1>R_i+2

where “i” denotes how many times we have repeated the procedure; e.g. for i=0 the above equation is identical to the previous one: R₀=w₁∙R₁+R₂.

Note that with each next iteration, we are moving towards smaller and smaller numbers, so at some point, after iteration i=n, we must have:

R_n=w_n+1∙R_n+1+0

that is: R_n is divisible by R_n+1 without the rest and R_n+1 must be the GCD of every pair R_i, R_i+1 including A and B.

Division of rational numbers

A ratio of whole numbers is simply a rational number. But what about a ratio of two rational numbers?

By definition, division is the inverse of multiplication. This fact is expressed by the following equivalence:

After inserting z=a/b and y=d/c it turns into:

Now we have to figure out what ratio x multiplied by d/c equals to a/b. Let’s guess x=(a/b)∙(c/d):

and thanks to the associative property of multiplication we can move the brackets:

thus indeed:

The equation on the left of the above equivalence has some interesting consequences. After inserting a=1 and b=1, we get:

so the following must be true:

In other words, the division by d/c (on the right) is equivalent to the multiplication by the inverse of d/c (on the left) that is equal to c/d (in the middle).

Note that while division is not commutative:

multiplication is:

Of course, multiplication of a/b times its inverse must be equal 1:

In closing, let’s take a look at the properties of inverse numbers.

Every rational number p has a negative: -p:
p+(-p)=(-p)+p=0

It seems that every number p has an inverse 1/p:
p∙(1/p)=(1/p)∙p=p/p=1

However, 0 doesn’t have such an inverse:
0∙(1/0)=(1/0)∙0=0
No matter what ratio we insert for 1/0, the result will never be 1.

Infinity

For every number G>0, we can find r so small that 1/r is greater than G, e.g. r=1/(G+1). In other words, when r is getting closer to 0, 1/r shoots past every limit. This fact is expressed in mathematics as:

where ∞ represents infinity.

Is ∞ a number? In the following statement:
2/r-1/r=1/r
if r approaches 0, both 2/r and 1/r shoot to infinity, yet their difference explodes to infinity too:
∞-∞=∞

so ∞ cannot be a number because it behaves differently. Despite of this, ∞ is a very handy concept as we will soon discover.

Negative powers

In one of the earlier chapters, we have introduced powers as a shorthand for multiplying number x by itself n times. This meant that powers had to be natural. Do negative powers make sense?

Obviously, we have:
xⁿ ∙ 1/xⁿ = (1/xⁿ) ∙ xⁿ = 1

Power for natural numbers have the following property:
x^a+b = x^a ∙ x^b

Let’s demand this to be true for negative powers too:
xⁿ ∙ x^-n = x^-n ∙ xⁿ = x⁰ = 1

Comparison of the above equation with the first one in this section leads to the conclusion that a negative power can be defined as:
Eq.7:     x^-n := 1/xⁿ

With this definition in mind, can x be equal to anything? For x=0 we have:
0^-n:=1/0ⁿ=1/0

and 1/0 makes no sense, so x in Eq.7 cannot be equal to 0.

Other than that, it is easy to check that for any whole a, b and c, the following must be true:

x^a+b = x^a ∙ x^b = x^b ∙ x^a = x^b+a

x^(a+b)+c = (x^a ∙ x^b) ∙ x^c = x^a ∙ (x^b ∙ x^c) = x^a+(b+c)

Our definition of power (Eq.7) is consistent with commutative and associative properties of addition and multiplication.

Positional representation of rational numbers

Let’s take a look at a rational number a/k for natural a>k>0. If we break apart “a” in terms of k:

a=w∙k+u
u<k

then the number a/k can be rewritten as:

a/k=w+u/k
0<u/k<1

If instead of positive “a” or “k” we had a negative number, we could simply pull the minus sign out and place it in front:

-(a/k)=-(w+u/k)

to make the part in the brackets positive: w>0, u>0, k>0. In other words, we don’t have to analyse negative numbers separately.

As we know, every natural number “w” can be expressed in terms of powers of 10:

where d_i is a digit i.e. d_i<10. The smallest of these powers is equal to 10⁰=1, so it is too great to express a fraction u/k, u<k.

In the previous section we have learnt about negative powers 10^-n=1/10ⁿ. They can be as small as we wish so we could use them to express fractions.

The fractional part of a rational number

Let’s try to express fractions in terms of digits multiplied by negative powers of 10:

That is: we want to find such digits d_-1, d_-2, d_-3 etc. that the sum on the right is equal to u/k.

Why bother? Because as we will see, in this form the procedure for adding rational numbers is exactly the same as for whole numbers.

Example 1

In this example we will use the quinary system i.e. the system where every number is coded with five digits: 0, 1, 2, 3, 4, and 10 represents five. How would we go about expressing 1 divided by 4: 1/4?

There is no 1/10 in 1/4 so let’s introduce it:
1/4 = 10/4 ∙ 1/10

Now we have 10 twice. Let’s get rid of the one in red.

In the quinary system we can express 10 in terms of available digits as:
10 = five = 1 ∙ 4 + 1

so:
1/4 = (1 ∙ 4 + 1)/4 ∙ 1/10 = 1 ∙ 1/10 + 1/4 ∙ 1/10

The first term on the right: 1 ∙ 1/10 looks exactly like we want: the digit 1 multiplied by 1/10. The next term however contains a fraction 1/4 and we would like to have a digit instead.

Let’s replace 1/4 it with: 1/4=1 ∙ 1/10 + 1/4 ∙ 1/10:
1/4 = 1 ∙ 1/10 + (1 ∙ 1/10 + 1/4 ∙ 1/10) ∙ 1/10 = 1 ∙ 1/10 + 1 ∙ 1/100 + 1/4 ∙ 1/100

Now the first two terms on the right are in the form we want but the third one contains a fraction: 1/4 again. Let’s get rid of it in the same way as before. Repetition of this procedure generates a series:
1/4 = 1/10 + 1/100 + 1/1000 + 1/10000 + 1/100000 + …

We can continue this forever, so 1/4 can be written as a sum:

and we have found what we were looking for: 1/4 expressed in terms of digits multiplied by negative powers of 10: 1/10ⁱ=10^-i.

Example 2

For 1/4 in the decimal system we can write:
1/4 = 10/4 ∙ 1/10

to introduce the missing 1/10.

To get rid of the excess 10, we replace it with its equivalent broken apart in terms of a divisor i.e. 4: 10=2∙4+2. After that we have:

1/4=(2∙4+2)/4 ∙ 1/10=(2 + 2/4) ∙ 1/10=2 ∙ 1/10 + 2/4 ∙ 1/10

and we need to transform 2/4 =1/2 into digit times 1/10. The divisor (2) multiplied by 5 is equal to 10 so let’s multiply the dividend by 5 as well:

2/4 = 1/2 = 5/5 ∙ 1/2 = 5 ∙ 1/10

After inserting this into the term for 1/4, we get:

1/4 = 2 ∙ 1/10 + 5 ∙ 1/10² = 2 ∙ 10^-1 + 5 ∙ 10^-2

and that’s it, we are done!

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To practice reasoning independent of concrete numbers, let’s demonstrate how this procedure would work for an arbitrary positional system with the base Γ (memory refresher).

As we have seen in the example 1, the series of digits corresponding to the fraction u/k may be infinite:

We need to figure out a procedure generating digits d_-1, d_-2, d_-3 …

As in our examples, the missing 1/Γ can be introduced in the following way:

Next, let’s break apart u∙Γ in terms of k:

u∙Γ=d_-1 ∙ k + r_-1

where variables with indices _-1 tell us respectively:
d_-1 – how many times k fits into u∙Γ
r_-1 – what is the leftover, r_-1<k

Why the index was chosen to be -1 will become clear shortly.

After dividing u∙Γ by k, we get:

u∙Γ/k must be smaller than Γ because u/k<1, so d_-1 must be a digit i.e. d_-1<Γ.

Let’s insert the equation above into u/k:

The index of d was chosen to be -1 because d_-1 is multiplied by 1/Γ=Γ^-1.

Next, we need to express the fraction r_-1/k in terms of digits in a similar way as we did for u/k:

In the next step we break apart r_-1 ∙ Γ in terms of k:

r_-1 ∙ Γ = d_-2 ∙ k + r_-2

so

where d_-2 is a digit and r_-2<k. After inserting this into u/k we get:

and again, the choice of index -2 is apparent: d_-2 is multiplied by 1/Γ².

We continue this process until in some step n, r_-n=0 or alternatively the series is infinite. Whatever the case may be the result can always be expressed as:

where n may be infinity.

Notation for the fractional part of a number

Every positive rational number can be expressed as:
w+u/k

where w,u and k are natural. From the chapter on natural numbers and the previous section we know how to express the whole and fractional part in the positional system:

The terms of the first and the second summation look similar, so we can rewrite this into:

where index “i” runs from some number m to minus infinity.

To make the notation more compact, we will use the following shorthand:

The dot between d₀ and d_-1 in the series on the left separates the whole from the fractional part. For negative numbers we simply put a minus in front of the series.

Thanks to the shorthand, we can write 1/4 in the quinary system (that’s our example 1 from the previous section) as:

1/4=0.1111111111111111111111111111111…

In the decimal system the series is finite (as seen in the example 2):

1/4=0.25

Multiplication times n-th power of 10

In the chapter about natural numbers it was demonstrated that multiplications by 10ⁿ simply result in adding zeros at the end of the number. Does this work the same for rational numbers? Let’s take a look at a few examples.

Example 1

123.456 ∙ 10² = (1 ∙ 10² + 2 ∙ 10¹ + 3 ∙ 10⁰ + 4 ∙ 10^-1 + 5 ∙ 10^-2 + 6 ∙ 10^-3) ∙ 10² =

= 1 ∙ 10⁴ + 2 ∙ 10³ + 3 ∙ 10² + 4 ∙ 10¹ + 5 ∙ 10⁰ + 6 ∙ 10^-1 =

= 12345.6

Example 2

123.456 ∙ 10^-2 = (1 ∙ 10² + 2 ∙ 10¹ + 3 ∙ 10⁰ + 4 ∙ 10^-1 + 5 ∙ 10^-2 + 6 ∙ 10^-3) ∙ 10^-2 =

= 1 ∙ 10⁰ + 2 ∙ 10^-1 + 3 ∙ 10^-2 + 4 ∙ 10^-3 + 5 ∙ 10^-4 + 6 ∙ 10^-5 =

= 1.23456

Conclusion: multiplying a rational number times 10ⁿ moves the dot separating the whole and fractional part of a number. If “n” is positive, we move the dot “n” times right, if negative then we move it left.

Addition and subtraction of rational numbers

It is always possible to transform addition/subtraction into a summation of positive numbers or subtraction of the smaller one from the greater, we just have to remember to put the correct sign in front of the result. We can also turn the operation on rational numbers into the operation on whole numbers e.g.

11.23 – 123.1211 = -(123.1211 – 11.23)

123.1211-11.23=(123.1211-11.23) ∙ 10000/10000=
=(1231211-112300)/10000=1118911/10000=111.8911

so

11.23 – 123.1211 = -111.8911

Multiplication of rational numbers

If we deal with negative numbers, we simply extract the sign and place it in front of the result of multiplication performed on two positive numbers.

Just like for addition, we can move the dot before the multiplication, so it can be performed in the same way as for natural numbers e.g.

123.121 ∙ (-11.23) = -(123.121 ∙ 11.23)

123.121 ∙ 11.23 = (123.121 ∙ 1000/1000) ∙ (11.23 ∙ 100/100) =
= (123121 ∙ 1/1000) ∙ (1123 ∙ 1/100) = (123121 ∙ 1123) ∙ 1/100000 =
= 1382.64883

so

123.121 ∙ (-11.23) = -1382.64883

Division of rational numbers

Division of rational numbers can be transformed into division of natural numbers with the correct sign placed in front. It is accomplished in a similar way as for multiplication. Let’s take a look at an example:

123.121/(-11.23) = -(123.121/11.23)

123.121/11.23 = 123.121 ∙ 1/11.23 = (123121 ∙ 1/1000) ∙ (100/1123) =
= (123121/1123) ∙ 100/1000 = (123121/1123) ∙ 1/10

In the decimal system:
123121 = 100 ∙ 1123 + 821

so:
123121/1123 = 100 + 821/1123

and the greatest common divisor of 821 and 1123 is 1, so 821/1123 cannot be simplified. What’s left to do then, is finding a positional representation of 821/1123. The series corresponding to 821/1123 may never end. Is it possible to tell under what conditions the series is finite? We will get back to this problem in the next chapter.

Conversion of a fraction from one positional system to another

It may sound easy at first. We could move the dot like in the previous sections to obtain a natural number multiplied by 10 to some power and then convert. The problem is that 10ⁿ in one system usually doesn’t have the form 10^x in another. To illustrate the point, let’s convert decimal 0.375 into binary:

0.375=375/1000

375 in the binary system is represented by 101110111 and 1000 by 1111101000. So, 0.375 in the binary system is 101110111/1111101000. In the next step, we need to find a series representing this ratio. It looks like a daunting task. Is there a simpler way?

We know that:
0.375<1

so after multiplying both sides by the base of the binary system (2), the result is:
2 ∙ 0.375 = 0.75 = 0 + 0.75 < 2

On the right side, the result is split into the natural part: 0 and a fraction: 0.75. Note that the natural part must be a digit in the binary system because it is smaller than 2 no matter what fractional number we insert instead of 0.375.

After dividing the above equation by 2, we get:
0.375 = 0/2 + 0.75/2

If you don’t see where it is going yet, bear with me, it will become apparent soon.

Let’s give 0.75 the same treatment as we gave to 0.375:
2 ∙ 0.75 = 1.5 = 1 + 0.5

0.75 = 1/2 + 0.5/2

After inserting this into the formula for 0.375, we end up with:
0.375 = 0/2 + 1/2² + 0.5/2²

Next, for 0.5 we have:
2 ∙ 0.5 = 1

0.5 = 1/2

so after inserting this into the equation for 0.375, we get:

0.375 = 0 ∙ 2⁰ + 0 ∙ 2^-1 + 1 ∙ 2^-2 + 1 ∙ 2^-3

Voila! That’s exactly what we want, decimal 2 is binary 10:

0 ∙ 10⁰ + 0 ∙ 10^-1 + 1 ∙ 10^-2 + 1 ∙ 10^-3 = 0.011

so 0.375 must be equivalent to the binary 0.011

Let’s convert 0.011 back to decimal using the same method as above. Ten in the binary system is 1010, so we will be multiplying times 1010 instead of times two:

0.011 ∙ 1010 = 0.011 ∙ (1000+10) = 11+0.11

0.11 ∙ 1010 = 110 + 1.1 = 111+0.1

0.1 ∙ 1010 = 100 + 1 = 101

From the above it follows:

0.011 = 11/1010+0.11/1010

0.11 = 111/1010+0.1/1010

0.1 = 101/1010

After inserting the last two equalities into the one for for 0.011, we get:

0.011 = 11/1010 + 111/1010² + 101/1010³

11, 111 and 101 in the decimal system are: 3, 7 and 5 respectively, so binary 0.011 is decimal: 3/10+7/100+5/1000.

The reasoning described above can also be presented using variables. Operating on variables is useful because it is applicable to all cases all at once instead of a single, concrete situation. Let’s take a look.

Let r be some rational number 0<r<1 and Γ the base of the system we want r to convert into. Because r<1 we have:
r ∙ Γ = d_-1 + f_-1 < Γ

where we split r ∙ Γ into natural d_-1<Γ (i.e. a digit in the system with the base Γ) and fractional component f_-1.

f_-1<1 so
f_-1 ∙ Γ = d_-2 + f_-2 < Γ
d_-2< Γ and f_-2 represent natural and fractional components of f_-1 ∙ Γ respectively.

If f_-2>0, we repeat the whole procedure again:
f_-2 ∙ Γ = d_-3 + f_-3 < Γ

Obviously r = r ∙ Γ/Γ. After inserting all the terms described above into r = r ∙ Γ/Γ, we get:
r = d_-1/Γ + d_-2/Γ² + d_-3/Γ³ + f_-3/Γ³

and if f_-3>0, we continue the process until for some n, f_-n=0 and we are done. It is also possible that this process never ends as we have seen for 1/4 expressed in the binary system: 0.1111111111111111…

Rational powers

If a power can be whole, perhaps it can be rational too. According to the definition of power we have:

x^a := x multiplied by itself “a” times

(x^a)^b := x^a multiplied by itself “b” times

so the following must be also true:

(x^a)^b = x^a∙b

Let’s demand the above equation to be valid for a=1/k and b=k:

(x^1/k)^k = x¹

and we’ve just found a definition of a fractional power: x^1/k is such a number that if we multiply it by itself k times, the result is x.

Let’s take a look at a few examples (in the decimal system):

(4^1/2)² = 4 so 4^1/2 must be equal to 2

(27^1/3)³ = 27 so 27^1/3 = 3

How about 2^1/2=?

Rational numbers have an infinite granularity (n in 1/n can be as big as we wish), so we expect that every number should be rational. If that’s the case, we ought to be able to find such natural “a” and “b” that:
2^1/2 = a/b

If that’s the case, then:
(2^1/2)² = 2 = a²/b²

so:
2 ∙ b² = a²

that is: a² must be even i.e. divisible by 2 without a remainder (a²/2=b²).

“a” itself can be either even or odd so there must exist such number a₁ that:
a = 2 ∙ a₁
or
a = 2 ∙ a₁ + 1

If “a” was odd, a² would be odd too: a² = 4 ∙ a₁²+4 ∙ a₁+1, but we know that a² is even (a² =2 ∙ b²) so a must be even as well:
a = 2 ∙ a₁

After inserting the above into 2 ∙ b² = a² we get:
2 ∙ b² = 4 ∙ a₁²
so
b² = 2 ∙ a₁²

and using the same reasoning as for “a”, we can show that b must be even because b² is even. In other words, there must exist such b₁ that b = 2 ∙ b₁. After inserting this in the above equation we obtain:
4 ∙ b₁² = 2 ∙ a₁²
so
2 ∙ b₁² = a₁²

thus a₁ must be also even i.e. there exists such a₂ that a₁ = 2 ∙ a₂. It’s obvious that we can continue this game indefinitely. In other words, we have proved that:

if 2^1/2 is rational i.e. 2^1/2=a/b
then “a” and “b” must both be divisible by two an infinite number of times.

That is not possible so 2^1/2 cannot be a rational number!

Conclusion: rational numbers are not up to the task to express fractional powers!

Summary

We have introduced rational numbers to be able to handle divisible objects.

Studying their properties led us to the discovery that division can be converted into multiplication by number inverted to given. The only exception in this context is zero because it cannot be inverted (1/0 is a nonsense).

It also turned out that for rational numbers, both addition and multiplication have the same properties (associative, commutative etc.) as their counterparts for whole numbers.

Regarding the positional system, we have demonstrated that it can be extended to represent fractions. However, for some of them, an infinite series of digits is required. On a positive note, the procedure of performing mathematical operations for fractions in a positional representation doesn’t differ from the procedure for whole numbers.

In the last chapter we have introduced rational powers. That led to quite an unexpected discovery: even though the rational numbers have an infinite granularity (1/n can be as small as we wish), some numbers cannot be expressed as a ratio a/b of whole a and b. Can we extend our number system to cover this situation too? We will find out in the next chapter.