Foundations of Arithmetic
Marcus Tornea
January 24, 2021
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Introduction
The idea of theorems is central to all known mathematics—alongside definitions, they are what develops any branch of mathematics into an intricate
system worthy of their category. To illustrate, some theorems we know at
present are only there to constitute proofs of even more complicated and
meaningful theorems. However, anyone willing to trace this development
backwards may encounter a fundamental problem of somewhat philosophical
significance. Theorems are by definition, statements for which there exists
a proof, but proofs themselves are based on other theorems previously obtained. In other words, even before we prove the first theorem, there are
certain mathematical truths that we already know prior to doing so, which
is problematic for the idea of there being a first theorem at all. Even more
so is the unavoidable predicament that searching for a proof of every mathematical truth we encounter is bound to be an endless cycle of finding proofs
within proofs.
The only obvious solution to this problem is the acceptance of certain mathematical statements without proof, which by our conscious decision, shall
serve as the basis of all mathematical reasoning. These statements are called
axioms, or as you may have already heard, postulates. Essentially all of
mathematics rests on a particular group of axioms, though there are other
axiom groups convenient only for a single branch of mathematics, should one
be interested in studying only such. In particular, we shall expose a certain
axiom system that suffices to lay foundations only for arithmetic over the
natural numbers. We call this Peano Arithmetic.
Basic truths about arithmetic such as the associative, commutative and distributive laws of addition and multiplication are what we deem essentially
rudimentary that we almost never notice their usage underlying our dealings
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with high school algebra. A few empirical examples already suffice to convince us that these laws hold. However, we may have not so far seen their
actual proofs as we deem them too basic for our attention. Now is the time.
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Axioms
Below we expose the current formulation of Peano Arithmetic that shall begin
with a few prerequisites. We must know beforehand the range of entities that
the axioms refer to in the first place, among which is the constant 0. We do
not know what 0 is (at least for now, we will pretend so). It is merely an
arbitrary symbol that names some entity in the universe of discourse for the
axiom system at hand.
Mario
λ
α
β
0
Luigi
ψ
D
C
A universe of discourse is, in simple terms, the implicit range of entities
that we intend to talk about, or study. It constructs the setting in which
our logical arguments take place. For someone interested in studying the
algebraic properties of real numbers, the universe of discourse could be R,
for instance.
Generally, a universe of discourse could be any collection of entities, as long
as it does not violate any axioms established so far. For instance, the set of
natural numbers N cannot be the universe of discourse for any talk about
the real numbers, since N does not satisfy the law of additive inverses, which
otherwise holds for R. But for our current situation, we have not yet introduced any axioms for Peano Arithmetic, so the associated universe of
discourse could be any random collection at the moment, just like the figure
above. Only when we introduce new axioms should the possible universes
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become more exact. Similarly, we will soon discover that the universe of
Peano Arithmetic is N (at least by convention).
Other than 0, there are three more symbols in the language of Peano Arithmetic that resemble operations: operation symbols + and ·, that will soon
resemble addition and multiplication of two entities, and another symbol S
called the successor operation, that will resemble an “add one” function; i.e.
S(0) = 1, S(1) = 2, and so on. We do not know this yet, however, so you may
freely dismiss the remark just now; the symbols 0, +, ·, and S are undefined
and will remain so, but only by introducing new axioms do we establish their
intended properties. Do note that since +, ·, and S are operations, which
dictate relationships between entities in the universe of discourse, they are
not themselves entities in the universe. All that said, we are now ready to
introduce the first axiom.
Axiom 2.1. For all n, S(n) 6= 0.
By “for all n”, we are referring to all entities in the universe, over which
the variable n ranges. Also note that the outputs of the operations are also
themselves entities therein, therefore S(0), an output of S, is some entity
in this universe, which we know by the axiom, is not equal to 0. As such,
S(S(0)) is also an entity not equal 0, just as S(S(S(0))), and so on.
However, this axiom does not prevent us from stipulating that
S(S(0)) = S(0).
Note that just as we can configure the universe of discourse to contain whatever we want as long as no axioms were violated, we can also configure the
operations to behave however we want under the same condition. If we stipulate the above, and apply S to both sides of the equation, we obtain
S(S(S(0))) = S(S(0)),
but since S(S(0)) = S(0) as stipulated, we have S(S(S(0))) = S(0). One
can demonstrate that S(S(S(S(0)))) is also equal to S(0), and the same
occurs no matter how many times S was applied. This means that as far
as our axioms are concerned, our universe of discourse can at the very least
contain only two distinct objects, namely 0 and S(0), and be a valid universe
nonetheless.
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S(0)
0
Such a universe is trivial and insufficient to model our goal of arithmetic
over natural numbers; we want this universe to somehow contain an infinite
amount of distinct entities to reflect the infinite set of natural numbers, so
we call for the need to introduce more axioms.
Axiom 2.2. For all m and n, if m 6= n, then S(m) 6= S(n).
Now, this axiom tells us that distinct inputs to S yield distinct outputs, so
if S(0) and 0 are distinct by the first axiom, then so are S(S(0)) and S(0);
thus we can no longer make the problematic stipulation that we made before,
a promising sign. Moreover, since S(S(0)) and 0 are also distinct by Axiom
2.1, the universe must now contain at the very least, three mutually distinct
objects: 0, S(0), and S(S(0)).
A similar argument shows that S(S(S(0))) is also distinct from the rest; we
have S(S(S(0))) 6= 0 as given by axiom 2.1; moreover, since S(S(0)) 6= 0,
we have S(S(S(0))) 6= S(0), and since S(S(0)) 6= S(0) as well, we have
S(S(S(0))) 6= S(S(0)). Now, our universe contains at the very least, four
mutually distinct objects. In fact, we can keep going, showing that the next
application of S yields an entity distinct from the ones so far obtained. But
there is no limit to how many times one could successively apply S, so one
can keep asserting the existence of a new distinct entity, after another. Suddenly, the addition of this axiom just spawned an infinite bounty of mutually
distinct entities in our universe, just as promised.
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S(S(S(0)))
S(0)
0
S(S(0))
. . . so on
S(S(S(S(0))))
We now casually borrow some abbreviational shorthands from the Arabic
symbols 1, 2, 3, 4, 5, 6, 7, 8, and 9 to make our notation concise:
1 := S(0)
2 := S(S(0)) = S(1)
3 := S(S(S(0))) = S(2)
4 := S(3)
5 := S(4)
6 := S(5)
7 := S(6)
8 := S(7)
9 := S(8).
Our notational convention may also adopt some kind of “overflow” system
so that 10 := S(9), 100 := S(99), and so on. Soon, we will be able to name
our universe of discourse the set of natural numbers.
It appears however that something is hindering us from doing so. Recall that
the universe of discourse can contain whatever we want and S could perform
however we wish as long as no axioms so far established were violated. Consider the following universe:
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3
1
0
2
. . . so on
A
B
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namely the previous one just with two new distinct objects A and B that
are different from the rest, and such that S(0) = 1, S(1) = 2, S(2) = 3 and
so on like the usual, but additionally, S(A) = B and S(B) = A. In this
setup, one can check for sure that both axioms still hold. Moreover, other
than the “chain” formed by the objects 0, 1, 2, 3, . . . with the S operation
“linking” them together, the objects A and B appear to form a closed loop
that S cycles around, particularly one that is completely separate from the
chain linking 0, 1, 2, 3, . . . . Now we have the opposite of the problem we had
before—nothing stops our universe from containing excess stuff other than
what we intended. Clearly, more axioms are needed to restrict our universe
to contain only the chain.
Before that, we need some technical prerequisites. You will encounter some
notation that reads “P (m, n1 , . . . , nk )”. For our purposes, we will interpret
this as an abbreviation for an equation where the variables m, n1 , . . . , nk
appear, that can be substituted for entities as we wish. For example, we
can define P (m, n) as a convenient shorthand for “m + n = n + m” and
thus P (0, k) and P (S(m), n) become shorthands for 0 + k = k + 0 and
S(m) + n = n + S(m), respectively.
This convention shall only be for our own purposes. In truth, P (m, n1 , . . . , nk )
can stand for a larger class of statements (called “well-formed formulas”) of
which equations are merely a subclass. However, this document wishes not to
overwhelm the reader, as we will not encounter these more general formulas
herein, even in later sections. Still, we shall introduce the following principle
in the most general form.
Axiom Schema 2.3. Principle of Induction. If P (m, n1 , . . . , nk ) is a
well-formed formula, then the following is an axiom:
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for all n1 , . . . , and nk , if
1. P (0, n1 , . . . , nk ), and
2. for all m, if P (m, n1 , . . . , nk ) then P (S(m), n1 , . . . , nk ),
then for all m, P (m, n1 , . . . , nk ).
In other words, if 0 satisfies a property and for all m that satisfy the same,
S(m) does so as well, then all entities m in the universe satisfy the property.
Notice that it is an axiom schema—a scheme to generate multiple axioms
following some format. For each equation P (m, n1 , . . . , nk ), there is one
induction axiom. Reflecting the unlimited number of ways to construct an
equation, there is essentially an infinite number of induction axioms, of which
below are two examples. Pay attention to how the following axioms are
derived from the general axiom schema.
If
1. 0 + 0 = 0, and
2. for all m, if 0 + m = m, then 0 + S(m) = S(m),
then for all m, 0 + m = m.
In this example, we constructed an induction axiom relative to the equation
P (m) : 0 + m = m. Below we construct another one associated with the
equation P (k, m, n) : m + (n + k) = (m + n) + k instead:
For all m and n, if
1. m + (n + 0) = (m + n) = 0, and
2. for all k, if m + (n + k) = (m + n) + k, then m + (n + S(k)) =
(m + n) + S(k),
then for all k, m + (n + k) = (m + n) + k.
These induction axioms will appear in later sections. Those already familiar
with the process of induction may be able to identify the induction basis,
hypothesis, and the induction step within these examples.
Now, the motivation behind this principle is that since 0, 1, 2, 3, . . . are linked
by S in a single chain, i.e. S(0) = 1, S(1) = 2, S(2) = 3, and so on, then if 0
satisfies the property, then so does S(0) = 1, and since 1 satisfies the same,
so does S(1) = 2, and so on, essentially establishing the property for each
entity in the entire chain, in some sort of “domino effect”. However, if the
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universe of discourse is like the previous one with an excess pair A and B
around which S cycles in isolation from everything else, then this pair will be
untouched by the domino effect, and thus A and B cannot certainly be proven
to satisfy the property. But this contradicts the supposed conclusion of the
induction, that the property is certainly true for all entities m. Therefore,
such a universe would violate the principle of induction.
Essentially, the principle of induction eliminates outliers like A and B from
the universe of discourse, thereby leaving only the entities 0, 1, 2, 3, . . . , sequenced by S. We now call this universe the set of natural numbers and the
“entities” shall be referred to accordingly, as natural numbers.
0, 1, 2, 3, 4, 5, 6, 7, . . .
Recall that S is named the successor operation. Its role of succession is fully
realized at this point. As such, we are now prepared to axiomatize arithmetic
by utilization thereof.
Axiom 2.4. For all n, n + 0 = n.
Axiom 2.5. For all m and n, m + S(n) = S(m + n).
The addition operator is described fully with the above two axioms. That
is, all the information we need to prove all the properties of addition over
the natural numbers is encapsulated in these two concise axioms. We now
finally arrive at our first theorem:
Theorem 2.6.
1 + 1 = 2.
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Proof.
1 + 1 = 1 + S(0)
= S(1 + 0)
= S(1)
= 2.
by Axiom 2.5
by Axiom 2.4
Another similar theorem below demonstrates the recursive nature of the axioms of addition. Do you notice that they dictate some kind of recurring
algorithm for computing sums?
Theorem 2.7.
3 + 4 = 7.
Proof.
3 + 4 = 3 + S(3)
= S(3 + 3)
= S(3 + S(2))
= S(S(3 + 2))
= S(S(3 + S(1)))
= S(S(S(3 + 1)))
= S(S(S(3 + S(0))))
= S(S(S(S(3 + 0))))
= S(S(S(S(3))))
= S(S(S(4)))
= S(S(5))
= S(6)
= 7.
You are invited to identify which axioms were used for which lines of the
proof.
At this point, we can now demonstrate our awaited proof of associativity and
commutativity rules for addition.
Theorem 2.8. For all k, m, and n, k + (m + n) = (k + m) + n.
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Proof. Suppose that P (n, k, m) is a sentence abbreviation that stands for
“k + (m + n) = (k + m) + n”. According to the axiom schema of induction,
the following is an axiom:
for all k and m, if
1. k + (m + 0) = (k + m) + 0, and
2. for all n, k + (m + n) = (k + m) + n implies that k + (m + S(n)) =
(k + m) + S(n),
then for all n, k + (m + n) = (k + m) + n.
Let k and m be natural numbers. By the first axiom of addition,
k + (m + 0) = k + m
= (k + m) + 0;
thereby proving the first condition of the induction axiom (those familiar
with induction would call this the induction basis). Now, let n be a natural
number such that
k + (m + n) = (k + m) + n.
This is the induction hypothesis. We have:
k + (m + S(n)) = k + S(m + n)
= S(k + (m + n))
= S((k + m) + n)
= (k + m) + S(n)
by the induction hypothesis
Axiom 2.5 backwards.
Thereby proving the final condition of the induction axiom (the induction
step). As a result, we thus have for all n,
k + (m + n) = (k + m) + n.
Theorem 2.9. For all m and n, m + n = n + m.
Proof. We will use induction twice in this argument. First, we want to prove
the following claim:
Claim. For all m, 0 + m = m.
If P (m) stands for “0 + m = m”, we obtain an induction axiom as follows:
if
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1. 0 + 0 = 0, and
2. for all m, 0 + m = m implies that 0 + S(m) = S(m),
then for all m, 0 + m = m.
Notice that there is no “for all” statement at the beginning. This is because
P (m) has no other variables available for substitution other than m, unlike
all examples we’ve shown so far. In such cases, the prefix is omitted. The
induction basis is obvious, and is an instance of Axiom 2.4. Now let m such
that 0 + m = m (induction hypothesis). We have:
0 + S(m) = S(0 + m)
= S(m)
by the induction hypothesis
thereby proving the induction step, and so proving the claim.
Now let P (n, m) stand for “m+n = n+m”; the following becomes an axiom:
Axiom 2.10. For all n, n · 0 = 0.
Axiom 2.11. For all m and n, m · S(n) = m · n + m.
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