Walking the path from counting numbers to complex numbers.
Contents
Preface
Many years ago I read that Richard Feynman
gave a talk to a room full of scientists
in which he rederived basic abstract algebra on real numbers
in under an hour.
Since then I found that Feynman gave this derivation in a discussion on Algebra
in his Lectures on Physics, for which I give a link a few paragraphs below.
I'm not going to compete with Feynman,
but doing this derivation seemed like a fun challenge to undertake.
Below I present my explanation of how one gets to complex
numbers based on a few simple concepts: repetition,
inverse and
closure.
Along the way I try to throw in a few comments about
abstract algebra.
By the end, we will look at
Euler's Identity,
eiπ+1=0,
and maybe make it a little less mystical than it might appear.
It is not necessary for you to understand all of the references
to math terms, so you don't need to follow those links unless you
want to learn about that concept.
Similarly, it is not necessary for you to follow and understand in
detail every proof.
Hopefully you can simply ignore any parts you don't
immediately understand and yet still get something out
of the overall presentation.
I walked this path mostly for my own entertainment, but I thought
perhaps others might get something out of it.
It is quite long and likely contains some errors,
so
caveat lector.
Here are a couple of other documents that discuss Algebra that you might find interesting:
Introduction
Imagine that none of this stuff exists, so we are making it all up
as we go. We are going to define our numbering system from the ground up,
gradually building up a structure of definitions and operations
that all manage to work together nicely.
It's not just by random chance that things work nicely:
we are defining our numbers and operations precisely to make
them work together nicely.
In the code blocks below, I label each assumption (or definition)
with a name such as A1 enclosed in square brackets,
like this: [A1].
Lemmas (things which can be proved from the assumptions and are
used in later proofs) are labeled
similarly but with L rather than A.
Other intermediate steps in a proof which are not referenced outside
of that proof are labeled similarly but with I.
These names may be referenced later to build up additional lemmas.
The references look the same,
but appear in the text or in comments after an equation rather than before.
Concepts
There are three basic ways we will be extending our system:
- Repetition: performing the same operation many times.
For example, multiplication is repeated addition.
- Inverse: an operation that has the opposite effect of some
other operation.
For example, subtraction is the inverse of addition.
- Closure: the results of an operation are in the same set
as the operands.
For example, the natural numbers (or positive integers) are closed under
addition, because you can add any two natural numbers
and get another natural number;
but they are not closed under
subtraction, because there are some expressions on natural numbers
using subtraction whose results are not natural numbers,
such as (3 - 5).
Preview
Here is the quick preview of how we will move from counting to complex:
- start with zero and the successor function
- repeated successors yields counting and the natural numbers
- repeated counting yields addition
- inverse of addition yields subtraction
- closure on subtraction yields negative numbers
- repeated addition yields multiplication
- inverse of multiplication yields division
- closure on division yields rational numbers
- repeated multiplication yields exponentiation
- inverse of exponentiation yields logarithms
- closure on exponentiation with positive rational
numbers yields real numbers
- closure on exponentiation with negative rational
numbers yields complex numbers
- all of our operations on complex numbers are already closed, so we are done
If you enjoy playing with math you might want to try doing
all of these derivations yourself before reading my derivations.
Counting
At the most basic level, we start with some simple assumptions,
which happen to be a subset of the Peano axioms.
We define a starting point for counting.
Historically, people typically started with one,
but for later simplicity in this exercise we start with zero.
We define a successor function s(x) that takes a number x
and produces the next number, which by definition is distinct
from x.
[A1] zero exists
[A2] given x, s(x) generates another number, where s(x) is not the same as x
Equals
We define an equals operator (=) so that the statement a=a is true,
and the statement a=b
means that, for any true statement containing a, we can replace any or all
instance of
a by b and the resulting statement will also be true.
We further assume that if a=b is false, then the same replacements as
described above will generally (but not always) yield a false statement.
[A3] a=a is true for all a
[A4] a=b is a replacement rule (described above)
The equals operator is:
- Reflexive: a=a (by definition)
- Symmetric: if a=b then b=a. Starting with the true statement a=a
and the predicate a=b,
by our definition of equals we can replace any instance of a by b
in a=a and still have a true statement; we chose to replace
the first a by b, yielding b=a.
- Transitive: if a=b and b=c, then a=c.
Taking the assumed true statement a=b, and applying our equals rule
using the second statement b=c, we replace b by c in the first
statement, yielding a=c.
[L5.1] if a=b then b=a (demonstrated above)
[L5.2] if a=b and b=c then a=c (demonstrated above)
For convenience, we define the not-equals operator != to be false
whenever equals on the same values is true, and vice=versa.
The above definition also leads almost directly to one of the
common ways of solving algebraic equations: performing the same
operation to both sides of an equation, such as adding the same
number to both sides of an equation, or multiplying both sides
by the same number.
Here's an example of adding the same amount to both sides of an equation.
a = b Assume this is our starting equation we are working with
a + c = a + c True by definition [A3]
a + c = b + c From [A4]
Note that this works for any function:
[I6.1] a = b Assume this is our starting equation we are working with
[I6.2] f(a) = f(a) True by definition [A3]
[I6.3] f(a) = f(b) From [A4] using [I6.2] as a starting equation
and [I6.1] as our replacement rule
[L6.4] if a = b then f(a) = f(b) for any f defined for a
f(x) might be 2*x, x+3, sin(x), or anything else we desire.
Thus we can start with any true equation, perform the same valid
operation on both sides, and still have a true equation.
Natural Numbers
Given our previously defined starting point of zero,
we now define the natural numbers:
[A7.0] 0=zero
[A7.1] 1=s(0)
[A7.2] 2=s(1)
[A7.3] 3=s(2)
etc. to infinity.
By definition, s(x)!=x, so 1!=0, 2!=1, etc.
Note that we did not assume that repeated application of s(x) would not
eventually give us the same number.
Without that assumption it is possible that, for example, s(s(s(x)))=x,
or in other words, 3=0.
This yields a "modulo" system, which can be useful.
But for this particular exposition, I want to use the "normal" numbers,
so we will add the assumption that s(x) is never equal to any previous
value in the sequence.
More precisely, we assume:
[A8] For any x, repeated application of the successor function
any number of times will never generate x.
We have now defined an unending stream of distinct numbers, each of which is
a successor to one other number.
Greater Than
We next define the relational operators less than (<)
and greater than (>) with the
following statements:
[A9] s(a) > a
[A10] if (a > b) and (b > c) then (a > c)
[A11] (b < a) always has the same truth value as (a > b)
We are now at the point where we can count and know
(by definition) that each time we
count we get a number that is greater than all of the previous numbers.
We can start with any number and count up from there by repeated
application of the successor function.
For example, if we start with 4 (which is s(s(s(s(zero))))) we can
count up from there by three by repeated application of
the successor function three
times to get s(s(s(4))), which we can calculate is 7.
This gets unwieldy pretty fast.
To make this simpler, let's define an "addition" operator + that gives us
the same results as repeated counting.
Addition
We define the addition operator (
+) as follows:
[A21] a + 0 = a
[A22] a + s(b) = s(a + b)
Some quick examples:
[L23.1] a + 1 = a + s(0) = s(a + 0) = s(a)
[L23.2] a + 2 = a + s(1) = s(a + 1) = s(s(a))
Since s(a) = a+1, we also have
[L23.3] a + s(b) = a + (b+1)
[L23.4] s(a + b) = (a + b) + 1
For some of what we want to do below,
we are going to need to use the rule of induction:
[A24] If an equation is true for a known value of n,
and it can be demonstrated to be true for n+1 for any n when true for n,
then it is true for all natural numbers x where x > n.
Associative
We now show that our addition operator is associative.
We want to prove that (a+b)+n = a+(b+n) for all n.
We start by showing this is true for n=1,
then use induction:
[L25.1] a + (b + 1) = (a + b) + 1 From [A22], [L23.3] and [L23.4]
[I25.2] a + (b + n) = (a + b) + n Inductive assumption, true for n=1
a + (b + (n + 1))
= a + ((b + n) + 1) From [L25.1] on (b+(n+1))
= (a + (b + n)) + 1 From [L25.1] with (b+n) for b
= ((a + b) + n)+ 1 From [I25.2] applied to (a+(b+n))
= (a + b) + (n + 1) From [L25.1] in reverse with (a+b) for a and n for b
[L26] a + (b + c) = (a + b) + c Above lines summarized, with c for n+1
Thus by induction we have our proof of associativity.
Commutative
We use a similar approach to show that addition is commutative,
such that a+b=b+a.
We start by showing that 0 commutes with a for any a.
[I27.1] 0 + 0 = 0 From [A21] with 0 for a
0 + 1 = 0 + s(0) From [L23.1]
= s(0 + 0) From [A22] with 0 for a and b
= s(0) From [L27]
= 1
[L27.2] 0 + 1 = 1 Summary of the above few lines
[I27.3] 0 + n = n Inductive assumption, true for n=1 from [L27.2]
[I27.4] 0 + (n + 1) = (0 + n) + 1 From [L26]
[I27.5] 0 + (n + 1) = n + 1 By induction from [I27.3] and [I27.4]
[L27.6] 0 + a = a From [I27.5] with a for n+1
[I27.7] 0 + a = a = a + 0 From [L27.6] and [A21]
[L27.8] 0 + a = a + 0 From [L5.2]
Now we show that 1 commutes with any number by induction.
1 + (n + 1)
= 1 + s(n) From [L23.1] on (n+1) with n for a
= s(1 + n) From [A22] with 1 for a and n for b
= s(n + 1) From inductive assumption that 1 commutes with n, known true for n=0
= n + s(1) From [A21] with n for a and 1 for b
= n + (1 + 1) From [L23.1] on s(1) with 1 for a
= (n + 1) + 1 From [L25.1]
[L28] 1 + a = a + 1 Summary of the above with a for n+1
Finally, we use induction again to show that any two numbers commute.
a + (n + 1)
= (a + n) + 1 From [L25.1]
= (n + a) + 1 From inductive assumption that a commutes with n, known true for n=1 [L28]
= n + (a + 1) From [L25.1]
= n + (1 + a) From [L28]
= (n + 1) + a From [L25.1]
[L29] a + b = b + a Summary of the above with b for n+1
As a final note for addition, since we have demonstrated that
(a+b)+c=a+(b+c), we can omit the parentheses when adding multiple
terms without creating any ambiguity.
[A30] a + b + c = (a + b) + c = a + (b + c)
Repeated application of this rule can be used for addition with
four or more terms without parentheses.
By combining this rule with [L29] commutative law,
we can see that we can take an expression with multiple terms
added together, such as
a + b + c + d + e
and rearrange and group the terms any way we want.
The associative rule also makes it easy to calculate our addition facts.
We already know that 1=0+1, 2=1+1, 3=2+1 etc from our definitions [A7]
with [L23.1].
That lets us fill in the first row of our addition fact table.
We can then calculate all of the n+2 values based on the n+1 values,
and repeat ad infinitum for the rest of the numbers.
n + 2 = n + (1 + 1) = (n + 1) + 1
n + 3 = n + (2 + 1) = (n + 2) + 1
n + 4 = n + (3 + 1) = (n + 3) + 1
Wikipedia has
proofs of associativity and commutativity
of addition, which are similar to mine but
actually a little more concise,
and
here
is a proof of commutativity that does not rely on associativity -
but I wanted to think through these derivations
myself and present them here in-line with the rest of my exposition.
Identity
At this point we know that a+0=a [A21] and 0+a=a [L27.6],
or in other words adding zero to any number (on either side,
since we showed addition is commutative) yields that number.
This is an interesting enough fact that we will give this
number a special name: the
Identity for addition.
It's easy to show that there is only one identity for addition.
Assume two identity values e and f.
Consider the expression e+f.
Because e is an identity, e+f=f.
Because f is an identity, e+f=e.
Therefore e=f.
[L31] Since this is true for any two identities,
all are in fact the same one identity.
Algebra
We have built up our concepts in layers, like building a house:
we set a foundation with zero and the successor function,
put in some rim joists with the natural numbers,
and laid on some flooring with the addition operator and
its identity element.
We have created a little structure from our concepts.
Whereas a house is a physical structure, this is an
algebraic structure.
It turns out that this algebraic structure is useful enough
that mathematicians have given this kind of structure a name:
a
monoid.
A monoid has these characteristics (with our case in parentheses):
- It has a set of elements (the natural numbers).
- It has a binary operation on those elements (the + operator).
- The operation is associative (+ is associative).
- The operation is closed (adding two natural numbers always
produces another natural number).
- It has an identity element (zero).
There are a few rules from the above section that we will use often enough
that we want to reference them by name rather than lemma number.
We use the first letter of the name of the characteristic, followed
by the operator character.
[a+] a + (b + c) = (a + b) + c [L26] Associativity of addition
[c+] a + b = b + a [L29] Commutativity of addition
[i+] a + 0 = 0 + a = a [A21], [L27.6] Identity for addition
Subtraction
At this point we have the ability to perform addition, which allows us
to calculate a value for x in such equations as
x = a + b.
But we don't yet have the ability to solve for x in the equation
a + x = b.
We want to add an operation that is the opposite of addition.
In other words, if we start with a and add b to it, we want to be
able to take the result and perform another operation using b
in order to get back to a.
An operator that has this characteristic is called an inverse.
We are going to define an operation that is the inverse of addition.
We will call that operation subtraction,
and we will use the dash character (
-) as the operator.
Before we defined addition, we already had the successor function [A2]
and we defined the numbers [A7] in terms of the successor function.
We defined addition with two axioms [A21] and [A22], then showed that
adding 1 to any number is the same [L23] as applying the successor function.
Including the successor function and the definitions of the numbers in
terms of the successor function, we really had four pieces going into
the definition of addition.
We could follow the same path and define a predecessor function that is
the inverse of the successor function, but instead we will skip that step
and work in terms of adding and subtracting 1 instead of successor and
predecessor functions.
We define our subtraction operator (
-) recursively,
similarly to how we defined the addition operator, using an additional
axiom [A41.1] in place of defining a predecessor function p(x):
[A41] a - 0 = a
[A41.1] (a + 1) - 1 = a
[A42] a - (b + 1) = (a - b) - 1
So let's see how this works:
3 - 0 = 3 From [A41]
3 - 1 = (2 + 1) - 1 = 2 From [A41.1], and since 3 is the successor to 2 (i.e. 3=2+1)
3 - 2 = 3 - (1 + 1) = (3 - 1) - 1 = 2 - 1 = (1 + 1) - 1 = 1
Associative
We want to prove the associative laws for subtraction so we know how
we can transform various combinations of parentheses and operators.
We already know about
a + (b + c),
so there are three other possible combinations of + and - with the
parentheses in the same position:
a - (b + c)
a + (b - c)
a - (b - c)
We start with
a - (b + c).
[L43.1] a - (b + n) = (a - b) - n Inductive assumption, true for n=1 from [A42]
a - (b + (n + 1))
= a - ((b + n) + 1) From [a+]
= (a - (b + n)) - 1 From [A42]
= ((a - b) - n) - 1 From [L43.1] on (a-(b+n))
= (a - b) - (n + 1) From [A42] with (a-b) for a and n for b
[L43.2] a - (b + c) = (a - b) - c Above lines summarized, with c for n+1
Next we do
a + (b - c), which we do by induction
after first doing
a + (b - 1).
(a + (n + 1)) - 1
= ((a + n) + 1) - 1 From [a+]
= a + n From [A41.1] with a+n for a
= a + ((n + 1) - 1) From [A41.1] with n for a
[L44] (a + b) - 1 = a + (b - 1) Above lines summarized, with b for n+1
[L45.1] a + b = a + (b - 0) From [A41] with b for a
[L45.2] a + b = (a + b) - 0 From [A41] with (a+b) for a
[L45.3] a + (b - 0) = (a + b) - 0 From [L45.1] and [L45.2] by [A4]
[L45.4] a + (b - n) = (a + b) - n Inductive assumption, true for n=0 by [L45.3]
a + (b - (n + 1))
= a + (b - (1 + n)) From [c+] with n for a and 1 for b
= a + ((b - 1) - n) From [L43.2] on b-(1+n)
= (a + (b - 1)) - n From [L45.4] with b-1 for b
= ((a + b) - 1) - n From [L44]
= (a + b) - (1 + n) From [L43.2] with a+b for a, 1 for b, n for c
= (a + b) - (n + 1) From [c+] with n for a and 1 for b
[L45.5] a + (b - c) = (a + b) - c Above lines summarized, with c for n+1
Finally we tackle
a - (b - c),
which we build up to through quite a few lemmas.
[L46.1] 0 - 0 = 0 [A41] with 0 for a
[L46.2] (0 + 1) - 1 = 0 [A41.1] with 0 for a
[L46.3] 1 - 1 = 0 From [L27.2] on 0+1
[L46.4] n - n = 0 Inductive assumption, true for n=1 from [L46.3]
(n + 1) - (n + 1)
= (n + 1) - (1 + n) From [c+]
= ((n + 1) - 1) - n) From [L43.2] with n+1 for a, 1 for b, n for c
= n - n From [A41.1] on (n+1)-1 with n+1 for a
= 0 From [L46.4]
[L46.5] a - a = 0 Above lines summarized, with a for n+1
a - b
= a - (b + 0) From a+0=0 with b for a
= a - (b + (n - n)) From a-a=0 with n for a
= a - ((b + n) - n) From [L45.5] with b for a, n for b and c
= a - ((n + b) - n) From commutative+
= a - (n + (b - n)) From [L45.5]
= (a - n) - (b - n) From [L43.2]
[L47] a - b = (a - n) - (b - n)
Substituting a = (c + n), b = (d + n) in [L47] yields
[L48.1] (c + n) - (d + n) = ((c + n) - n) - ((d + n) - n) = c - d
[L48.2] c - d = (c + n) - (d + n) [L48.1] last and first parts
(a - n) + n
= n + (a - n) From [c+]
= (n + a) - n From [L45.5]
= (a + n) - n From [c+]
= a + (n - n) From [L45.5]
= a + 0 From [L46.5]
= a From [i+]
[L49] (a - n) + n = a Above lines summarized
a - (b - c)
= (a + c) - ((b - c) + c) From [L48.2] with a for c, b-c for d, c for n
= (a + c) - b From [L49] with c for n
= (c + a) - b From [c+] on a+c
= c + (a - b) From [L45.5]
= (a - b) + c From [c+]
[L50] a - (b - c) = (a - b) + c Above lines summarized
We now have all of our rules of association for addition and subtraction.
The following four equations, repeated from above, show all eight
possible combinations of + and - operators and grouping of three
variables.
[L26] a + (b + c) = (a + b) + c
[L43.2] a - (b + c) = (a - b) - c
[L45.5] a + (b - c) = (a + b) - c
[L50] a - (b - c) = (a - b) + c
Earlier we saw that, because of [L26], we can write
a + b + c
and know that it is unambiguous.
But that is not true if we write
a - b - c, because
the statement
(a - b) - c = a - (b - c)
is not in general true.
In order to be able to write fewer parentheses, we arbitrarily choose
to have
a - b - c mean the same thing as
(a - b) - c.
[A51] a - b - c = (a - b) - c
We have specified that the middle variable (b in our equation),
following the
- operator, should be
grouped with the variable on its left,
so we call the
- operator left-associative;
but we generally say it is not associative,
meaning it does not associate both ways as does addition.
Unlike addition, subtraction is not commutative,
and it has no identity.
More precisely, we could say that zero is a
right identity for subtraction,
but since it is not also a left identity,
it is not a simple identity and we usually don't mention it.
Negative Numbers
You may already have noticed that adding the subtraction operator
to our structure has created a bit of a problem:
we are now able to write expressions which we can not evaluate
within our structure.
For example, the expression
2 - 4 can not be reduced to
a single natural number.
When we reduce this equation according to our rules, we eventually
get to the point where we need to solve for
0 - 1,
and we have no rule to reduce that any further.
In other words, our system is no longer a closed system:
to state the problem more precisely,
the natural numbers are not closed under subtraction.
A pet peeve of mine: elementary school math teachers who tell their
students "You cannot subtract 5 from 3."
This statement is misleading in its imprecision, since it can be solved with
the use of negative numbers.
Math is a precise field.
The correct statement should include that qualification:
"You cannot subtract 5 from 3 using the counting numbers we are studying."
Likewise for other incorrect statements such as
"You can not divide 3 by 2" and
"You can not take the square root of -4."
We would like to be able to solve any equation we can write with our
subtraction operator, so we will define new numbers that we can use
for that purpose.
We call these numbers negative numbers.
We choose to write them using the same digits as we write our natural
numbers, with a leading
- character, such as -1 and -2.
In our house-building analogy, so far we have built a little
house from the foundation upwards,
and now we realize we need some more support in order to finish subtraction.
Adding negative numbers is like adding another room to that house:
in order to have a solid structure, we need to extend our foundation.
To save on design work,
we are going to reuse the same basic plan as we used
when we built up the natural numbers.
This is like using the same blueprint for the second room
of our house as for the first,
except in mirror image because we find symmetry pleasing.
Here is a little diagram:
+-----+ +-----+
/ 3 \ / 3 \
+----+ +----+----+ +----+----+
| 2 | | 2 | | 5 | 2 |
+------+ +----+-+ +----+-+ +-+----+----+-+
| 1 | | 1 | | 1 | | 4 | 1 |
+------+ +------+ +------+ +------+------+
1. Natural 2. Addition 3. Subtraction 4. Negative Numbers
Numbers on Naturals Oops! 5. Addition on Negatives
3. Completion of Subtraction
Thus we go back to the beginning of our derivation of natural numbers.
To distinguish our original numbers from our newly defined negative
numbers, we will call all of the numbers generated by our successor
function (that would be all numbers 1 and above) the positive numbers.
We will call the collection of all of these numbers
(positive, negative and zero) the integers.
We will call the characteristic of being
"positive" and "negative" the sign of the number.
Since we want our rules to apply to all integers, we start by stating
that in any of our previous assumptions and derivations, a variable
name can refer to any integer unless the specific proof or assumption
states otherwise (such as for induction proofs).
We started by defining a successor operator s(x) [A2],
and we now define a corresponding predecessor operator p(x)
that generates our negative numbers in a way which is symmetric to s(x):
[A61] given x, p(x) generates another number, where p(x) is not the same as x
In all of our original assumptions and following proofs, we now state
that variable names in those assumption refer to any integer.
We define the predecessor function as the inverse of the successor function
and vice-versa.
In other words:
[A62.1] p(s(a)) = a
[A62.2] s(p(a)) = a
We define our negative numbers in the same way as we defined
our natural (positive) numbers [A7]:
[A63.1] -1 = p(0)
[A63.2] -2 = p(-1)
[A63.3] -3 = p(-2)
etc. to negative infinity.
We take our no-duplicates assumption [A8] on the successor function
and state it for the predecessor function:
[A64] For any x, repeated application of the predecessor function
any number of times will never generate x.
For the relational operators, we can derive their meaning relative to
the predecessor operator:
s(a) > a [A9]
p(s(a)) > p(a) Apply p(x) to both sides [L6.6]
a > p(a) From [A62.1]
[L65] p(a) < a From [A9]
Addition
We add to our definition of Addition ([A21] and [A22]) to handle
negative numbers,
and we extend our induction assumption [A24] to negative numbers:
[A71] a + p(b) = p(a + b)
[A72] If an equation is true for a known value of n,
and it can be demonstrated to be true for n+(-1) for any n when true for n,
then it is true for all natural numbers x where x < n.
For each of our original assumptions through addition, we have now
added similar assumptions to handle our negative numbers.
All of our assumptions are completely symmetrical:
take any of the original assumptions, replace successor by predecessor,
replace 1 by -1, and exchange < with >,
and you will get the equivalent assumption for our negative numbers.
Because all of our other proofs in those sections are based on those
assumptions, the symmetric proofs for negative numbers follow from
the symmetric assumptions in exactly the same way as for the natural
numbers.
Thus all of the results and conclusions in those sections
are valid for addition of negative numbers:
commutative, associative, identity, algebra.
We list the results of one lemma here,
leaving the details of the derivation as an exercise to the reader:
[L73] a + -1 = p(a)
We derive a couple of other useful results:
p(s(a)) = a [A62.1]
p(a + 1) = a [L23.1]
(a + 1) + -1 = a [L73]
a + (1 + -1) = a [a+]
(1 + -1) = 0
[L74] -1 + 1 = 0 [c+]
(1 + -1) = 0 [L74]
n + -n = 0 Inductive assumption, true for n=1 [L74]
(n + -n) + (1 + -1) = 0 From [i+] because (1 + -1) = 0
(n + 1) + (-n + -1) = 0
(n + 1) + (-(n+1)) = 0 From p(x) defn
[L75] a + -a = 0 Above lines summarized, with a for n+1
The above statement says that, for any element a in our set of natural
numbers, there is an element -a (a negative number, negative a)
which can be added
to that natural number to produce zero (our identity element).
We call negative a the
inverse element
of a, and likewise a is the inverse element of -a.
-a + a = 0 [L75]
(-a + a) - a = 0 - a Subtract a from each side
-a + (a - a) = 0 - a [L45.5]
[L76] -a = 0 - a [L46.5] and [i+]
a + -a = 0 [L75]
(a + -a) - -a = 0 - -a Subtract -a from each side
a + (-a - -a) = 0 - -a [L45.5]
a = 0 - -a [L46.5]
[L76.1] a = -(-a) [L76]
a + -b
= a + (0 - b) [L76]
= (a + 0) - b [L45.5]
= a - b [i+]
[L77] a + -b = a - b
Subtraction
As with addition, we note that we can
create a set of symmetric assumptions using negative numbers in place
of positive numbers,
so that all of our results and conclusions of subtraction on positive
numbers also work on negative numbers.
For improved symmetry with the definition of addition,
we restate our assumptions defining subtraction to use the
successor and predecessor functions,
and we add a symmetric assumption that covers negative numbers.
We no longer need
(a+1)-1=0 [A41.1]
as an assumption for subtraction,
because it is equivalent to
p(s(a))=a) [A62.1].
Since these assumptions are just a rewriting of our original
assumptions for subtraction, all of our derivations remain the same.
[A41] a - 0 = a Repeat of original [A41]
[A81] a - s(b) = p(a - b) [A42] restated in terms of s and p
[A82] a - p(b) = s(a - b) Symmetric assumption to [A81]
Algebra
With the addition of negative numbers to our structure,
our set is closed with respect to subtraction.
We now have a set (the integers)
with an associative binary operator (+) with an identity (0)
and inverse elements (the negative numbers).
This algebraic structure is called a
group.
Because our operator (addition) is commutative,
our algebraic structure is an
abelian group.
The group, however, ignores the subtraction operator.
Multiplication
Once we start using addition for real tasks, we find that we are often
adding the same number many times, such as 3+3+3+3.
Because this is so common, we would like to define a shortcut -
a new operator - that means the same thing.
We call this operation multiplication.
There are various conventions for how the multiplication operator
is written: x, * and dot are common, and in some cases a convention
is adopted that two variables written next to each other with no
operator between them are to be multiplied.
Most computer programming languages use the asterisk character (*),
and I will use that here.
In order to have as much symmetry as we can, and to minimize our design work,
we will define multiplication using a similar approach as we did when
we defined addition:
[A101] a * 0 = 0
[A102] a * (b + 1) = (a * b) + a
[A103] a * (b - 1) = (a * b) - a
We could equivalently have used a slightly different formulation
for [A103] in which we add -1 rather than subtracting 1,
as supported by [L77]:
a * (-1)
= a * (0 - 1) [L76]
= (a * 0) - a [A103]
= 0 - a [A101]
= -a [L76]
[L104.1] a * -1 = -a Above lines summarized
a * (b + -1)
= a * (b - 1) [L77]
= (a * b) - a [A103]
= (a * b) + -a [L77]
= (a * b) + (a * -1) [L104.1]
[L104.2] a * (b + -1) = (a * b) + (a * -1) Above lines summarized
If the second operand is negative, we can factor that out and we see
that it changes the sign of the result.
a * -n = -(a * n) Inductive assumption, true for n=1
a * -(n + 1)
= a * (-n - 1)
= (a * -n) - a
= -(a * n) - a
= 0 - (a * n) - a
= 0 - ((a * n) + a)
= 0 - (a * (n + 1))
= -(a * (n + 1))
[L104.3] a * -b = -(a * b) Above summarized, with b for n+1
[L104.4] -a * b = -(a * b) Swap a with b and use [c*]
-a * -b = -(-a * b) [L104.3]
= -(-(a * b)) [L104.3] again
= a * b [L76.1]
[L104.5] -a * -b = a * b Above lines summarized
Identity and Zero
By setting b=0 in [A102], we see that 1 is a right-identity for
multiplication:
a * (0 + 1) = (a * 0) + a From [A102] with 0 for b
a * 1 = 0 + a From [i+] on LHS, [A101] on RHS
[L105] a * 1 = a
We show by induction that zero multiplied on either side gives zero:
[L106.1] 0 * 0 = 0 From [A101] with 0 for a
[L106.2] 0 * n = 0 Inductive assumption, true for n=0
[L106.3] 0 * (n + 1) = (0 * n) + 0 From [A102] with 0 for a, n for b
[L106.4] 0 * (n + 1) = 0 + 0 From [L106.2]
[L106.5] 0 * (n + 1) = 0
[L106.6] 0 * a = 0 Above summarized with a for n+1
By doing the same proof using [A103] we can conclude that [L106.6]
holds for all integers.
We show that 1 is a left identity:
1 * 1 = 1 From [L105] with a=1
1 * n = n Inductive assumption, true for n=1
1 * (n + 1)
= (1 * n) + 1 From [A102] with a=1 and b=n
= n + 1 From Inductive assumption
[L106.8] 1 * a = a Above summarized, with a for n+1
Since 1 is both a left identity and a right identity,
we can drop the handedness and just refer to it as an identity.
With addition we had one special number, 0, which when added to any
number yielded that number.
With multiplication we see that we have two special numbers:
the number 1 is an identity for multiplication,
but 0 is also special, since anything multiplied by 0 yields 0.
We choose to use the word "zero", when associated with a specific
operation such as multiplication, to mean a value that, when given
as an operand to that operator, always yields zero.
Our multiplication operator has only one zero, but other systems
and operators may have more than one zero.
By the same argument [L31] as for the additive identity, we can
see that there is only one multiplicative identity and only one
multiplicative zero.
Distributive
We show that multiplication is distributive over addition by induction:
[L107.1] a * (b + 0) = a * b = (a * b) + 0 = (a * b) + (a * 0)
a * (b + 1) = (a * b) + a [A102]
a * (b + 1) = (a * b) + (a * 1) From [L105] on rightmost a
[L107.2] a * (b + n) = (a * b) + (a * n) Inductive assumption, true for n=1
a * (b + (n + 1))
= a * ((b + n) + 1) From [a+]
= (a * (b + n)) + a From [A102]
= ((a * b) + (a * n)) + a From [L107.2]
= (a * b) + ((a * n) + a) From [a+]
= (a * b) + (a * (n + 1)) From [A102]
[L107.3] a * (b + c) = (a * b) + (a * c) Above summarized, with c for n+1
The above proof can be repeated using -1 instead of 1 (by [L104.2]),
so [L107.3] covers all integers.
Using the same proof steps using [A103] rather than [A102]
demonstrates that multiplication distributes over subtraction as well.
Since by [L77] subtraction is the equivalent of adding the negative
of a number, this is consistent.
[L107.4] a * (b - c) = (a * b) - (a * c)
2 * 1 = 2 = 1 + 1
2 * n = n + n Inductive assumption, true for n=1
2 * (n + 1)
= 2 * n + 2
= (n + n) + (1 + 1)
= (n + 1) + (n + 1)
2 * a = a + a
1 * b = b
(0 * 1) * b = (0 * b) + b
(n + 1) * b = (n * b) + b Inductive assumption, true for n=0
(n + 2) * b = (n * b) + b + b Inductive assumption, true for n=0
((n + 1) + 1) * b
= (n + 2) * b
= (n * b) + b + b
= ((n + 1) * b) + b
(a + 1) * b = (a * b) + b
Associative
We show multiplication is associative by induction:
[L108.1] (a * b) * 0 = 0 = a * 0 = a * (b * 0)
[L108.2] (a * b) * 1 = a * b = a * (b * 1) From [L105] on each side
[L108.3] (a * b) * n = a * b = a * (b * n) Inductive assumption, true for n=1
(a * b) * (n + 1)
= ((a * b) * n) + (a * b)
= (a * (b * n)) + (a * b) From [L108.3]
= a * ((b * n) + b) From [L107.3] with b*n for b, b for c
= a * (b * (n + 1)) From [A102] with b for a, n for b
[L108.4] (a * b) * c = a * (b * c) Above lines summarized, with c for n+1
As with the distributive law, we can replace 1 by -1 to show that
our conclusion covers negative numbers as wel.
Commutative
m * n = n * m Inductive assumption, true for m=0 or 1 and n=0 or 1
(m + 1) * (n + 1)
= (m + 1) * n + (n + 1) From [A102]
= (m * n) + m + (n + 1) From [(a+1)*b = a*b+b]
= (n * m) + n + (m + 1) From Inductive assumption and [a+]
= (n + 1) * m + (m + 1) From [same as two lines up]
= (n + 1) * (m + 1) From [A102]
[L109] a * b= b * a
As with addition, the fact that multiplication is associative [L108.4]
means that, if we have an expression that is a string of values multiplied
together, we can drop the parentheses from the expression without
creating any ambiguity; and the fact that it is commutative means that
we can rearrange all of those multiplied values to any order we want.
Algebra
We have added a second operator to our repertoire
that, like addition, is an associative binary operator with an identity.
With two such operators, where one distributes over the other, we have a
ring
(for a more precise definition, follow the link).
In the same way that group ignores subtraction,
the ring ignores the division operator.
As with addition,
there are a few rules from the above section that we will use often enough
that we want to reference them by name rather than lemma number.
[a*] a * (b * c) = (a * b) * c [L108.4] Associativity of multiplication
[c*] a * b = b * a [L109] Commutativity of multiplication
[z*] a * 0 = 0 * a = 0 [L106.6] Zero for multiplication
[i*] a * 1 = 1 * a = a [L106.8] Identity for multiplication
[d*] a * (b + c) = (a * b) + (a * c) [L107.3] Distributivity of multiplication over addition
Division
As when we defined subtraction to be the inverse operation of addition,
we want an inverse operation to multiplication so that we can solve for
x
in equations such as
a * x = b.
We call our inverse operation division.
As with multiplication, there are a number of common ways this
operation is expressed.
For use in this presentation, we choose to use the slash character (/)
to represent the division operation.
We want division and multiplication each to be the inverse of the other,
as is the case with addition and subtraction,
so we have two candidate definitions:
[A120.1] (a * b) / b = a for all a and b except b=0
[A120.2] (a / b) * b = a for all a and b except b=0
Our definitions exclude zero because we already have a rule that says
anything times zero is zero, so we know
a priori that we can't make
these new rules work for all a when b is zero.
The fact that we can't divide by zero is the first time we have
encountered a special case in our structure, where we have to add a
qualification to one of our rules stating that you can't do something
rather than extending our structure to make it possible to do that.
When, in building our structure of numbers, we realized that we could
not answer the question "what is 3 - 5?", we expanded the structure to
allow us to answer tha question ("negative 2"). In this case, we can't
answer the question "what is 5 / 0?", but, for the first time,
instead of trying to expand
our structure to be able to answer that question, we make the statement
"you can't do that".
As we will see later, the further we go in defining our structure,
the more such exceptions and caveats we need to make.
We check that the two assumptions above are compatible by starting with one
and converting it into the other.
(a * b) / b = a [A120.1]
((a * b) / b) * b = a * b Right-multiply both sides by b
(c / b) * b = c Previous line with c for a*b; this is [A120.2]
We can quickly get some useful lemmas by plugging in a few different
values for a and b:
[L121] a / 1 = a From [A120.1 or 2] with b=1, after a*1=a
[L122] b / b = 1 From [A120.1] with a=1, after b*1=b
[L123] (1/b)*b = 1 From [A120.2] with a=1
[L124] 0 / b = 0 From [A120.1] with a=0, after 0*b=0
[L124.2] a / a = 1 From [A120.1] with a=1 and b=a
If we are looking at the equation
[I125] a = c / b
what does that mean?
If we assume
[A126] c = a * b
then [I125] becomes
[I127] a = (a * b) / b
which is [A120.1]. This is true by definition, so our assumption [A126]
is a valid assumption to use in solving [I125].
What we are saying here is that the solution (a) to [I125] is the value
that, when multiplied by b, gives c.
[L128] If a = c / b, then c = a * b, and vice-versa (from [I125] and [A126])
Associative
As we did with subtraction,
we want to prove the associative laws for division so we know how
we can transform various combinations of parentheses and the
multiplication and division operations.
We already know about
a * (b * c),
so there are three other possible combinations of * and / with the
parentheses in the same position:
a / (b * c)
a * (b / c)
a / (b / c)
[I129.1] a / (b * c) = d Given
a = d * (b * c) From [L128]
a = (d * c) * b From [a*] and [c*]
a / b = d * c From [L128]
[I129.2] (a / b) / c = d From [L128]
[L129.3] a / (b * c) = (a / b) / c From [I129.1] and [I129.2]
[I130.1] a * (b / c) = d Given
a * (b / c) * c = d * c Multiply both sides by c
a * b = d * c Reduce b /c * c = b by [A120.1]
[I130.2] (a * b) / c = d From [L128]
[L130.3] a * (b / c) = (a * b) / c From [I130.1] and [I130.3]
[I131.1] a / (b / c) = d Given
a = d * (b / c) From [L128]
= (d * b) / c From [L130.3]
a * c = d * b From [L128]
c * a = d * b From [c*]
(c * a) / b = d From [L128]
c * (a / b) = d From [L130.3]
[I131.2] (a / b) * c = d From [c*]
[L131.3] a / (b / c) = (a / b) * c From [I131.1] and [I131.2]
We now have all of our rules of association for multiplication and division.
The following four equations, repeated from above, show all eight
possible combinations of * and / operators and grouping of three
variables.
Note that this table is identical to the table of rules of association
for addition and subtraction, with * instead of + and / instead of -.
[a*] a * (b * c) = (a * b) * c
[L129.3] a / (b * c) = (a / b) / c
[L130.3] a * (b / c) = (a * b) / c
[L131.3] a / (b / c) = (a / b) * c
We derive a few more useful lemmas.
a / b
= (a * 1) / b From [i*]
= a * (1 / b) From [LL130.3]
[L132] a / b = a * (1 / b) Summary of the above lines
1 / (a / b)
= (1 / a) * b From [L131.3]
= b * (1 / a) From [c*]
= b / a From [L132]
[L133] 1 / (a / b) = b / a Summary of the above lines
(a / b) * (c / d)
= ((a / b) * c) / d) From [L130.3]
= (c * (a / b)) / d) From [c*]
= ((c * a) / b) / d) From [L130.3]
= (c * a) / (b * d) From [L129.3]
= (a * c) / (b * d) From [c*]
[L134] (a / b) * (c / d) = (a * c) / (b * d) Summary of the above lines
(a / b) / (c / d)
= ((a / b) * 1) / (c / d) From [i*]
= (a / b) * (1 / (c / d)) From [L130.3]
= (a / b) * (d / c) From [L133]
= (a * d) / (b * c) From [L134]
[L135] (a / b) / (c / d) = (a * d) / (b * c) Summary of the above lines
Rational Numbers
You may have noticed in the above section about the division operation
that we discussed things like
1 / a without commenting on the
fact that our number system, which up to now includes only integers,
does not in general include the numbers that can represent that.
The proper sequence would have been to introduce rational numbers first,
but I wanted to finish the discussion about the properties of the division
operation before discussing rational numbers.
With that out of the way, let's turn to rational numbers.
We can easily build a table for specific values of a, b and c for equation
[I125] by taking all pairs of integer values for a and b, generating c as their
product, and defining the value of c/b to be a for all of those triplets.
For example, 2*3=6, therefore 6/3=2.
Our division table does not include all possible
combinations of
c/b,
so there are some division equations for which the
answer can not be found in our tables.
For example, 3/2 does not appear in our table because,
in our system of numbers up to this point, which is all integers,
there is no number that, when multiplied by 2, yields 3.
In order for our numbers to be closed under division, we have to add some
new numbers, which are the numbers needed to solve the equation
c/b when
there is no integer number
a such that
a*b=c.
We call these numbers rational numbers,
because they are the ratio of two integers,
and we choose to represent them
as a fraction using the division operator.
In other words, when we ask what is the answer to the equation
c/b,
we are simply defining the answer to be
c/b and stating that
that value is a number.
We will then examine how to manipulate these numbers.
We have defined rational numbers as numbers of the form
c/b.
We also know from our table-based enumeration of division equations that,
for any number c which can be written as
a*b, the value of
the division equation
c/b is a.
We define the value of our rational number that we write as
c/b
to be consistent with the known solutions of our division equations written
the same way.
Thus the value of the rational number 6/3 is defined to be 2, etc.
Algebra
With division as the inverse of multiplication, the multiplicative identity 1,
and rational numbers, our ring is now a
field.
This is as far as we will go with algebra. When we continue with
exponentiation to derive real numbers and then complex numbers,
those structures are still fields.
Operator Precedence
Up to now, we have been using parentheses to ensure that the order of
application of operators in an expression is unambiguous. We noted
earlier that we don't need those parentheses in an expression that
consists solely of a number of values added together, and likewise that we
don't need parentheses in an expression that consists solely of a number
of values multiplied together. This is nice because it reduces the amount
of writing we need to do.
We can further reduce the need for parentheses by defining a rule that
tells us which operations to evaluate first when there are no parentheses
to guide us. When we start with an operation and then define a second
operation as the repeated application of the first operation,
we can think of that second operation as being more powerful than the
first operation. We then give priority to the more powerful operator,
defining our rule of precedence to be that, in an expression in which
the order of evaluation would otherwise be ambiguous, we will evaluate
the more powerful operators first.
We define addition (+) and subtraction (-) to be at the first level,
and multiplication (*) and division (/) to be at the second level and
higher power than the first level.
Thus, for example, the expression
a + b * c will be
equal to
a + (b * c), and the expression
a / b - c will be equal to
(a / b) - c.
In cases where there are multiple operators of the same power,
we define the order of evaluation to be left to right.
Thus, for example, the expression
a / b * c will be
equal to
(a / b) * c, and the expression
a - b + c will be equal to
(a - b) + c.
Exponentiation
Up to this point the structure we have built is pretty clean. With
rational numbers and our four operators (+, -, *, /), we have a system
that is closed and mostly complete and consistent, with the only exception
being that we can't divide by zero. Other than that one exception,
operations are well-defined, we have a nice set of rules including our
commutative, associative, and distributive rules, and we have a host of
identities and lemmas we can apply to our rational numbers.
Once we add exponentiation, things get a lot messier: we will have
expressions that have multiple values, bigger swaths of undefined
operations, and many places where our lemmas and rules of manipulation
no longer apply. It might seem like it's hardly worth trading our nice
clean rational numbers for this mess. But despite all of the rough
edges, there are enough useful things you can do with real and complex
numbers that it is worth carefully defining where those rough edges are
and avoiding them. So, let's forge ahead.
As with addition, once we start using multiplication for real problems,
we often find we want to multiply the same number together many times,
such as
3*3*3*3.
As we did when defining multiplication, we define a new operator that
means the same as repeated multiplication.
We call this new operation exponentiation.
In programming languages this is sometimes written using the up-arrow (^)
as an operator, but since this is HTML we have the luxury of using the
standard notation, which is to write the exponent as a superscript.
For example the expression
34
means 3 multiplied by itself
4 times, or
3 * 3 * 3 * 3.
We call the number on the left the base, and the superscript number
the exponent.
The operation of exponentiation is also referred to as taking a base to a power,
where the power is the exponent.
In line with our precedence rules by which we evaluate higher-power
operations first, we will evaluate exponentiation before multiplication,
division, addition, and subtraction, when there are no parentheses to
otherwise indicate the order of evaluation.
From [a*] we know we can group repeated multiplication any way we want,
so for example
3 * 3 * 3 * 3 = (3 * 3 * 3) * 3 = (3 * 3) * (3 * 3).
Using our new superscript notation, we can write this as
34 = (33) * (31) = (32) * (32).
More generally, we can see these things
from our definition of exponentiation and [a*]:
[L201.1] a(b + c) = ab * ac
[L201.2] a1 = a
[L201.3] (ab)c = a(b * c)
[L201.4] (ab)c = ab*c = ac*b = (ac)b From [L201.3] and [c*]
We can figure out how to deal with
(a * b)n
by starting with n=2:
(a * b)2
= (a * b) * (a * b)
= a * b * a * b From [a*]
= a * a * b * b From [c*]
= a2 * b2
[L201.5] (a * b)2 = a2 * b2 Summary of above lines
Then we use induction for the general case:
Assume (a * b)n = an * bn for some n
(a * b)(n + 1)
= (a * b)n * (a * b)1 From [L201.1]
= (an * bn) * (a * b) From [L201.5]
= an * a * bn * b From [a*] and [c*]
= a(n + 1) * b(n + 1)
True when n=2 from [L201.5], so by induction true for all positive n
[L201.5] (a * b)n = an * bn
Unlike addition and multiplication, we can quickly see from
counterexamples that exponentiation is neither commutative:
23 = 2 * 2 * 2 = 8
32 = 3 * 3 = 9
8 != 9, so 23 != 32
nor associative:
2(32) = 2(3 * 3) = 29 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 512
(23)2 = (2 * 2 * 2)2 = 82 = 8 * 8 = 64
512 != 64, so 2(32) != (23)2
These initial lemmas are based on our intuitive definition of exponentiation
as repeated multiplication, which provides obvious answers only in the case
where the exponent is a counting number (strictly positive integer).
Let's extend our definition to cover other numbers in our algebra.
[A202.1] d = b + c Starting assumption
[I202.2] b = d - c
ad = ab * ac From [A202.1] and [L201.1]
ad / ac = ab * ac / ac Assuming ac!=0
[I202.3] ab = ad / ac
[L202.4] a(d - c) = ad / ac Substitute b from [I202.2]
We can't divide by zero, so the above is not valid when
ac
is zero. When is that expression zero? From the definition of
exponentiation, this expression represents repeated multiplication of
a. What number when multiplied by itself is zero?
There is only one such number: zero. So [L202.4] is not valid when
a = 0, but it is valid for any other base.
Let's look at two special cases of [L202.4].
a0 = a(1 - 1) From [L46.5], a!=0
= a1 / a1 From [L202.4]
= a / a From [L201.2]
= 1 From [L124.2]
[L203] a0 = 1 Above lines summarized, a!=0
a-b = a(0 - b) From [L76], a!=0
= a0 / ab From [L202.4], b!=0
= 1 / ab From [L203]
[L204] a-b = 1 / ab Above lines summarized, a!=0, b!=0
[L204.1] a-1 = 1 / a From [L204] with b = 1, and [L201.2]
The above extends our exponentiation operator to all integer exponents
and all bases other than zero. What about rational exponents?
Remember that our goal is to define a set of consistent and useful
operations. To that end, we want to ask ourselves how we can define
exponentiation using a rational exponent such that it is consistent with
the rest of our algebra.
Rational numbers are equivalent to division using integers, which is
the inverse of multiplication. Our exponentiation rule [L201.3]
includes multiplication, from which we can derive a rule for division.
a = a1 [L201.2]
= a(b / b) From [L124.2], b!=0
= a(b * 1/b) From [L132]
= a(1/b * b) From [c*]
= (a1/b)b From [L201.3]
[L205] (a1/b)b = a Summary of the above lines
What the above says is that the value of
a1/b is the
number that, when raised to the power b, is equal to a.
For example, the number
a1/2 is the number that,
when raised to the power 2, is equal to a.
We call
a1/b the b-th root of a.
The case where b is 2 or 3 is common enough that we define special names:
we call
a2 a squared and
a1/2 the square root of a;
we call
a3 a cubed and
a1/3 the cube root of a.
Previously when we added a new operation to represent repeated application
of an earlier operation (addition as repeated counting and
multiplication as repeated addition), we did not encounter closure
problems until we added an inverse operation to the newly added
operation (subtraction, division).
As we will see below, this is not the case for exponentiation:
here we will run into closure problems even without an inverse
operation. But to keep the flow the same as with the other operators,
I will discuss the inverse operation before getting back to closure.
Logarithms
As when we defined division to be the inverse operation of multiplication,
we want an inverse operation to exponentiation so that we can solve for
x
in equations such as
ax = b.
We call our inverse operation logarithm.
There is a curious hole in math terminology about logarithms.
Our other operations all have names: we talk about performing
addition, multiplication, or exponentiation.
We do addition by adding two addends to get a sum.
But we don't "do logarithm": we "take a logarithm".
The word logarithm refers to one of the elements in
that operation, similar to how the word exponent
refers to one of the elements in the operation of exponentiation.
There seems to be no single word for logarithms that corresponds
to the operation names such as addition, multiplication, and exponentiation.
Talking about logarithms is like talking about sums rather than addition.
[A221.1] loga(ab) = b for all a and b except a=0 or b=0
[A221.2] alogab = b for all a and b except a=0 or b=0
We can derive a few lemmas for log.
[L222.1] loga(a) = loga(a1) = 1 [L201.2] and [A221.1] with b=1
[L222.2] loga(1) = loga(a0) = 0 [L203] and [A221.1] with b=0
[L222.3] loga(1/a) = loga(a-1) = -1 [L204.1] and [A221.1] with b=0
[I223.1] loga(ac) = c [A221.1] using c instead of b
[I223.2] loga(ad) = d [A221.1] using d instead of b
[I223.3] loga(ac) + loga(ad) = c + d Add left sides and right sides of [I223.1] and [I223.2]
[I223.4] loga(ac+d) = c+d [A221.1] using c+d instead of b
[L223.5] loga(ac+d) = loga(ac) + loga(ad) Transitive equals on [I223.3] and [I223.4]
[I224.1] loga(ac) + loga(ad) = c + d Subtract left sides and right sides of [I223.1] and [I223.2]
[I224.2] loga(ac-d) = c-d [A221.1] using c-d instead of b
[L224.3] loga(ac-d) = loga(ac) - loga(ad) Transitive equals on [I224.1] and [I224.2]
[I225.1] loga(ac+d) = loga(ac*ad) [L201.1]
[I225.2] loga(ac+d) = loga(ac) + loga(ad) [L223.5]
[I225.3] loga(ac*ad) = loga(ac) + loga(ad) Transitive equals on [I225.1] and [I225.2]
[L225.4] loga(x*y) = loga(x) + loga(y) Substitute x for ac and y for ad
[I226.1] loga(ac-d) = loga(ac/ad) [L202.4], ad!=0
[I226.2] loga(ac-d) = loga(ac) - loga(ad) [L224.3]
[I226.3] loga(ac/ad) = loga(ac) - loga(ad) Transitive equals on [I226.1] and [I226.2]
[L226.4] loga(x/y) = loga(x) - loga(y) Substitute x for ac and y for ad, y!=0
Principal Values
Previously, we noted that, when we added division to our
algebraic structure, we had to add a small complication in that
we can't divide by zero.
When we add square root (or, more generally, exponentiation with any non-integer exponent),
we run into another kind of special case
where we have to take additional care: multivalued functions.
We note that every number has two square roots:
for example, the square root of 4 is 2 or -2, because either of
those numbers, when multiplied by itself, is equal to 4.
With multivalued functions like square root,
we can run into trouble
if we are not careful about choosing which value to use.
Here's an example of this problem:
(4
1/2)
2 = 4
4
1/2 * 4
1/2 = 4
2 * 4
1/2 = 4 Substitute 2 as the first square root
2 * -2 = 4 Substitute -2 as the second square root
-4 = 4 Wrong!
The bad substitution in the above sequence may be easy to spot and understand,
but as we go further into building our algebra, problems of this
nature become subtler and harder to recognize.
We can reduce the probability of running into this kind of problem by
carefully selecting which of these multiple values to use. When we
have one preferred value for a multivalued function, we call that the
principal value
of the function.
For example, the principal value of sqrt(4) is 2.
Irrational Numbers
The ancient Greeks knew that
21/2
(the square root of two) is not a rational number.
There are
a
lot of proofs of this.
I happen to like this one that demonstrates that all roots
(square root, cube root, and others) that are not integers
are not rational.
Assume ab = c (b=2 for square root, b=3 for cube root, etc)
and a = d/e, e!=1
where d/e is reduced to the lowest form, so they have no prime factors in common.
Then ab = (d/e)b = db/eb = c = c/1
But db has no prime factors that are not in d,
and eb has no prime factors that are not in e,
so db and eb have no prime factors in common,
and the fraction can not be reduced at all,
and in particular can not be reduced to c/1,
therefore it can not be equal to c.
Since there is no rational number satisfying the original assumption,
any solution must not be a rational number,
except in the case that e=1, which means the root is an integer.
In order for our numbering system to be closed under exponentiation,
we need to extend our numbers to include these values that are not
rational numbers. We call them irrational numbers.
When we added negative numbers and rational numbers, that was after
we had added not only an operation defined by repetition, but also
its inverse. In this case, we had to extend our numbers to provide
closure even without having yet added that inverse operation.
A brief aside about infinity:
before adding irrational numbers, our set of numbers was always
countably infinite, which means there was always a way to map the
entire set of numbers onto the counting numbers.
For example, we can count off all the integers, both positive and negative,
by ordering them like this: 0, 1, -1, 2, -2, 3, -3, and so on.
We can count off all the rational numbers by ordering them according to
the sum of the numerator and denominator and alternating positive and
negative, like this: 0, 1/1, -1/1, 1/2, -1/2, 2/1, -2/1, 1/3, -1/3,
2/2, -2/2, 3/1, -3/1, 1/4, and so on, then removing duplicates
(any fraction that is not reduced).
But once we add all the irrational numbers
we can no longer come up with a counting order like this, which is why
we say the set of all irrational numbers is
uncountable.
For a proof of this assertion, look up Canter's
diagonalization argument.
Decimal Notation
When we introduced rational numbers, such as 1/2, we defined their values
in terms of the division operation, but did not provide any other
representation. This was perhaps acceptable, as we can easily manipulation
rational numbers in order to answer questions about them.
With irrational numbers, it is not quite so easy. How can we tell, for example,
which of
21/2,
31/3,
or 723/510 is the largest?
We would like a representation that allows us to do real-world
calculations with these values.
When counting up with integers, we use a place-notation system in
which each digit, as we move to the left, represents a value that is
ten times as much as the digit just to its right. For example,
1234 means 1 * 1000 + 2 * 100 + 3 * 10 + 4.
We extend this sequence by defining each place to the
right of the ones digit as having a place value of one tenth of the digit
to its left. In order to unambiguously know which place is the ones place,
we put a decimal point (.) just to the right of the ones digit
(we in America, that is; in some other parts of the world people
use a comma (,) instead).
For example, 0.5678 means 5 * 1/10 + 6 * 1/100 + 7 * 1/1000 + 8 * 1/10000.
We can convert fractions to decimal form such as
a.bcde by remembering that that means
a + b/10 + c/100 + d/1000 + e/10000
723/510 = (510 + 213) / 510
= 510/510 + 213/510
= 1 + 213/510
= 1 + 10 * 213/510 / 10
= 1 + 2130/510 / 10
= 1 + (2040 + 90)/510 / 10
= 1 + 2040/510 / 10 + 90/510 / 10
= 1 + 4/10 + 10 * 90/510 / 100
= 1 + 4/10 + 900/510 / 100
= 1 + 4/10 + (510 + 390)/510 / 100
= 1 + 4/10 + (510/510 + 390/510) / 100
= 1 + 4/10 + 1/100 + 390/510 / 100
= 1 + 4/10 + 1/100 + 10 * 390/510 / 1000
= 1 + 4/10 + 1/100 + 3900/510 / 1000
= 1 + 4/10 + 1/100 + (3570 + 330)/510 / 1000
= 1 + 4/10 + 1/100 + (3570/510 + 330/510) / 1000
= 1 + 4/10 + 1/100 + 7/1000 + 330/510 / 1000
= 1.417 + more digits from 330/510 / 1000
Figuring out the decimal representation for a number such as
21/2 is not quite as straightforward,
but we can start by the brute-force approach of trial and error
to get an estimate.
12 = 1, 1<2
22 = 4, 4>2, so our number must start with 1
1.12 = 1.21
1.22 = 1.44
1.32 = 1.69
1.42 = 1.96
1.52 = 2.25 so our number must start with 1.4
1.412 = 1.9881
1.422 = 2.0164 so our number must start with 1.41
1.4112 = 1.990921
1.4122 = 1.993744
1.4132 = 1.996569
1.4142 = 1.999396
1.4152 = 2.002225 so our number must start with 1.414
From this much we can determine that
21/2
is less than
723/510. We don't have an exact answer,
but for real world questions we often don't need to go to
very many decimal digits to get the answer.
Our decimal notation is a sum of fractions, so any finite decimal number
can be converted to a rational number. Conversely, irrational numbers can
not be exactly represented as a decimal number, we can only approximate
them when using decimal notation.
If we want to maintain an exact representation of an irrational number
such as
2, we have to keep it in that notation
or something similar.
Imaginary Numbers
Adding irrational numbers extends our numbers to include the value of
21/2 and other fractional roots of positive
numbers, but it doesn't cover everything. In particular, our numbers
don't yet include a value for the expression
-11/2.
This is the square root of negative 1, which is equal to the number that,
when multiplied by itself, equals negative 1.
But any positive number multiplied by itself is a positive number, and
from [L104.5] any negative number multiplied by itself is also
a positive number, so we don't have any numbers that are candidates to
be the square root of negative 1. In order to have exponentiation be closed
for negative bases, we need to extend our numbers.
We need to add a set of numbers that, when multiplied by themselves,
produce negative numbers.
When we added negative numbers, we used our existing counting numbers with
an added character (-) in front to indicate a negative number.
We will do something similar here, using our existing counting numbers with
an added character, in this case the letter
i, following the number to
indicate the new kind of numbers we are adding.
We define
1i (or just
i) to
be the number such that
i2 = -1,
and given a number
a, we define
ai = a * i
(which is consistent with a common convention of defining
ab = a * b).
We need to pick a name to distinguish these new numbers from what we had
before, and "the square root of negative one" is too unwieldy, so we pick
a shorter name and call them imaginary numbers.
When we defined negative numbers, we might have instead called them
imaginary numbers, because you can't have negative lengths or a negative
number of apples in the real world, so those numbers are not real, right?
In the sense that they are highly useful for certain mathematical
calculations, imaginary numbers are no more "imaginary" than negative
numbers. It is unfortunate that we are stuck with a name that causes
some people to get distracted from thinking about these new numbers
as simply the next step in expanding our numbering system to be closed
under exponentiation.
To distinguish them from our newly added imaginary numbers, we go back
and lump together our previously defined rational and irrational numbers
and call those real numbers.
Having made the distinction between real and imaginary numbers, we note
that we can have imaginary rational numbers, such as
(1/2)i,
or imaginary irrational numbers, such as
21/2i,
as well as negative imaginary numbers such as
-4i
or negative irrational imaginary numbers such as
-21/2i.
If we work through the mechanics of addition and subtraction with
imaginary numbers, we find that they work the same as real numbers but
with that extra
i everywhere. To put it another way, imaginary
numbers are closed under addition and subtraction.
This is not the case with multiplication: imaginary numbers are not
closed under multiplication, since
i * i = -1, which
is not an imaginary number.
Similarly, imaginary numbers are not closed under division, since
i / i = 1, which is not imaginary.
Complex Numbers
Since we defined imaginary numbers as being a different set of numbers from
real numbers, we can't convert from one to the other, so if we try to add
a real number
a and an imaginary number
bi
together, we can't reduce that, so we just write it as
a + bi.
We call this kind of number a complex number,
and since a or b could be zero, we note that all real numbers and all
imaginary numbers are complex numbers.
We are, in a sense, cheating when we use the + symbol to enumerate
the real and imaginary parts of a complex number, because, as just stated,
we can't actually do anything with that operator to reduce the number.
In that sense, we could have used any special character in that location.
But we choose to use the + sign because it turns out the rules we have
that deal with the + operator on real numbers also work with complex numbers:
commutative, associative, and distributive rules all work consistently
when applied to complex numbers when we use a + sign between the real and
imaginary parts.
As with square root, complex numbers come with multivalued functions,
some with an infinite number of solutions.
It's easy to get bad results if you're not careful, so it's important
to define a principal value for these functions and consistently use it.
Cartesian Coordinates
Since real and imaginary numbers can't be reduced to each other and are
thus orthogonal, we can represent them on the plane. We choose real to
be the X axis and imaginary to be the Y axis.
With this cartesian environment, we can represent complex numbers in
polar coordinates using the standard conversion:
(r, θ) = (sqrt(x2 + y2), arctan(y/x),
where x is the real part and y is the imaginary part
(and with the appropriate sign adjustments for quadrants
other than I).
Converting the other way, we have
(x, y) = (r * cos(θ), r * sin(θ)).
Sometimes we refer to a complex number as
z,
where we can decompose it either by real and imaginary parts,
written as
x = Re(z), y = Im(z),
or by polar coordinates,
written as
r = |z|, θ = Arg(z),
where
|z| is the
magnitude of
z
and
Arg(z) is the
argument of
z.
More precisely,
arg(z) is the argument of
z,
and
Arg(z) is the principal argument of
z.
arg(z) is a multi-valued function equal to
Arg(z) + n*2*π for all integer values of
n.
We can treat our complex numbers as vectors in the two dimensional
complex plane, so that adding two complex numbers can be displayed
in our plane as vector addition.
More interesting is multiplication, where we can see that when we
use polar coordinates we get this nice result:
(r1,θ1) * (r2,θ2) = (r1*r2, θ1+θ2).
(r1,θ1) * (r2,θ2) = (r1*cos(θ1) + r1*sin(θ1)i) * (r2*cos(θ2) + r2*sin(θ2)i)
= r1*(cos(θ1) + sin(θ1)i) * r2*(cos(θ2) + r2*sin(θ2)i)
= r1*r2 * (cos(θ1) + sin(θ1)i) * (cos(θ2) + r2*sin(θ2)i)
= r1*r2 * (cos(θ1)*cos(θ2) + cos(θ1)*sin(θ2)i + sin(θ1)*cos(θ2)i + sin(θ1)*sin(θ2)*i2
= r1*r2 * ((cos(θ1)*cos(θ2) - sin(θ1)*sin(θ2)) + (cos(θ1)*sin(θ2) + sin(θ1)*cos(θ2))i)
= r1*r2 * (cos(θ1+θ2) + sin(θ1+θ2)i)
= r1*r2*cos(θ1+θ2) + r1*r2*sin(θ1+θ2)i
= (r1*r2, θ1+θ2)
[L301] (r1,θ1) * (r2,θ2) = (r1*r2, θ1+θ2) The above summarized
Euler's Formula
Here is Euler's Formula:
eiθ = cos(θ) + i*sin(θ)
Feynman calls this
"one of the most remarkable, almost astounding, formulas in all of mathematics"
and refers to it as an "amazing jewel".
As described in an article at
Brilliant,
Euler's Formula can be derived using the series expansions of sin(x), cos(x), and e
x:
cos(x) = 1 - x2/2! + x4/4! - ...
sin(x) = x - x3/3! + x5/5! - ...
ex = 1 + x + x2/2! + x3/3! + ...
so:
ei*x = 1 + i*x + (i*x)2/2! + (i*x)3/3! + (i*x)4/4! + (i*x)5/5! + ...
= 1 + i*x - x2/2! - i*x3/3! + x4/4! + i*x5/5! - ...
= (1 - x2/2! + x4/4! - ...) + i*(x - x3/3! + x5/5! - ...)
= cos(x) + i*sin(x)
In the section on Cartesian Coordinates above, we noted that any complex
number can be represented in polar coordinates using r and theta, but
we didn't have a good place to put the
i.
With Euler's Formula, we can now unambiguously represent any complex number
z = x + i*y as
|z| * ei*arg(z)
where
|z| is the magnitude of
z
and
arg(z) is the argument of
z.
Complex Exponentiation
Given
w = u + i*v and
z = x + i*y,
how do we calculate
wz?
We would like
wz to satisfy the
rules of exponentiation that we derived for real numbers, such as
ka+b = ka * kb.
We will assume that we can apply this rule to complex exponentiation
and see how that works out.
From the discussion of Euler's Formula above we know that we can represent
any nonzero complex number
w as
|w|*ei*arg(w),
and we can represent the real number
|w| as
eln(|w|).
Let's see where that takes us.
wz = (|w|*e(i*arg(w)))z Expand w
= (eln(|w|)*ei*arg(w))z Use exp form for magnitude of w
= (eln(|w|)+i*arg(w))z ea * eb = ea+b
= e(ln(|w|)+i*arg(w))*z (ea)b = ea*b
= e(ln(|w|)+i*arg(w))*(x+i*y) Expand z to real and imaginary parts
= eln(|w|)*x + ln(|w|)*i*y + i*arg(w)*x + i*arg(w)*i*y (a+b)*(c+d)=ac+ad+bc+bd
= e((ln(|w|)*x - arg(w)*y) + i*(ln(|w|)*y + arg(w)*x) i2=-1 and rearrange terms
[L310] wz = e((ln(|w|)*x - arg(w)*y) + i*(ln(|w|)*y + arg(w)*x) The above summarized
This gives us a number of the form
r * ei*θ
where
r = e((ln(|w|)*x - arg(w)*y)
and
θ = ln(|w|)*y + arg(w)*x,
both of which we can evaluate.
Note that the above result includes
arg(w) in two places,
once multiplied by
x and once multiplied by
y.
arg is a multi-valued function, and thus complex exponentiation is
also multi-valued for all exponents except zero.
If we are raising to a real power, then
y is zero, so [L310] reduces to
wx = e((ln(|w|)*x) + i*(arg(w)*x) [L310] with y=0
= |w|x * ei*arg(w)*x For real x and all w
This equation says the magnitude of the result is the magnitude of
w
raised to the
x power
and the
arg of the result is the arg of
w multiplied by
x. If, for example,
we are squaring and thus
x is 2, we square the magnitude of the number and
double the angle.
This result is consistent with our earlier observation that, when multiplying two
complex numbers, we can multiply the magnitudes and add the angles.
If
y is zero and
x is an integer, then
ei*arg(w)*x
gives the same result for all of the multiple values of
arg(w), so the overall function
is single-valued. If
x is not an integer, this is not the case.
For example, if
x is 1/2, then we get two different answers by plugging in
Arg(w) and
Arg(w) + 2*π. These are the two square roots of
a number: they always have the same magnitude and differ in angle by π.
If we consider the path that would be traced out for powers of some fixed
w as we
change the real exponent, we can see that it generates a circle or a spiral.
Here is a nice visualization of
zx from
Suitcase of Dreams
for when
|z|>1:
If we are raising to an imaginary power, then
x is zero, so [L310] reduces to
[L311] wi*y = e(-arg(w)*y + i*ln(|w|)*y) [L310] with x=0
Let's evaluate
ii.
We use [L311] with
w=i and
y=1:
ii = e(-arg(w) + i*ln(|w|)) [L311] with w=i and y=0
= e-π/2 * ei * 0 |w|=1, ln(1) is 0
= e-π/2 Imaginary part drops out completely!
= 0.207879...
Surprisingly,
ii is a real number, a little larger than one fifth.
At least, that's one answer. We can use any of the answers
e-π/2 + k*2π
for any integer
k.
We see that we can represent any nonzero complex number in the form
ei*z, given
z = x + i*y.
ei*z = ei*(x+i*y)
= ei*x + i*i*y
= e-y + i*x
= e-y * ei*x
One interesting thing we can do now is to extend Euler's Formula
from real theta to complex theta, which allows us to define
sin and
cos for the entire complex plane:
ei*z = cos(z) + i*sin(z)
e-i*z = cos(z) - i*sin(z) cos is an even function, sin is an odd function
ei*z + e-i*z = 2*cos(z)
cos(z) = 1/2 (ei*z + e-i*z)
ei*z - e-i*z = 2*i*sin(z)
sin(z) = 1/(2*i) (ei*z - e-i*z)
Euler's Identity
We evaluate Euler's Formula with theta set to pi:
ei*π = cos(π) + i*sin(π)
= -1 + 0
= -1
We add one to both sides to get the typical presentation,
ei*π + 1 = 0.
Not only does this identity tie together five of the key values of algebra
(e, π,
i, 1, and 0), it does it with one each of the key operations
we derived above (equality, addition, multiplication, exponentiation).
That's a pretty sweet equation.
Final Closure
Throughout this presentation, we have expanded our system of numbers as we
defined new operators and discovered our system of numbers was not closed
under the new operators. But with complex numbers, we have reached a point
where we don't need to define any new number types. Complex numbers are
sufficient to solve all algebraic equations.
This is one of the interpretations of the
Fundamental Theorem of Algebra,
but the proofs are pretty difficult, so I'm not going to try to prove it here.
