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 Skip Navigation LinksMath Help > Basic Math > Arithmetic Rules

In order to get started in Mathematics, you must accept a set of axioms as "given" -- that is, not needing proof.  You should expect there to be enough axioms to be sufficient to derive theorems that are "useful" to the mathematician.  But you should not expect too many axioms.  That is, you should expect the axioms to be consistent with one another -- that is, you shouldn't be able to derive the contradiction of one of the axioms as a logical consequence of the other axioms.  Also, you should discard axioms that can be derived from other axioms.  This last fact sets up a choice: where one proposed axiom can be derived from another, then either one can be tossed out of the list of axioms, and reformulated as a "theorem".  So the same set of facts can be derived from various different (but consistent) sets of axioms.  I show the axioms in RED.  All other statements, in blue, are derived from those axioms using principles of logic.

Contents of this section:

Skip Navigation Links.

Did you just want to know how to do basic Arithmetic?  Then click here:

Arithmetic for sixth through ninth graders -- a less rigorous (that is, more practical) lesson on the rules of arithmetic.

For more advanced (or more curious) students

Proving the basic facts about arithmetic -- the Peano Postulates

Properties of Equality

First, we define three basic properties as follows:

a = a
If a = b then b = a
If a = b and b = c, then a = c.

These are called the reflexive, symmetric, and transitive properties of equality.  A relation that has these three properties is called an "equivalence relation".

One more axiom, which I call the substitution axiom, states,

If a = b, and a function f(a) exists, then f(a) = f(b).

In practice, the way we use these properties of equality is to invoke the "substitution principle", which is,

If a = b, and f(a) = g(a), then f(a) = g(b)

In other words, if a = b, then b can be substituted in place of a in any expression without changing the value of that expression.

Here's the proof using the substitution axiom followed by the transitive property of equality:

Let a = b, and f(a) = g(a).
g(a) = g(b)
f(a) = g(b)

Principles of addition, subtraction and negation

Addition is defined this way:

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

This definition consists of three axioms, called the identity, commutative, and associative properties of addition.

One very useful result that can be derived from these axioms is known as the "add to both sides" rule, which states that if a = b, then a + c = b + c.  Here's the proof, using the substitution principle, showing how we can add "c" to both sides of an equation:

a = b
a + c = a + c
a + c = b + c

There are two ways to approach the definition of subtraction.  One is to define the additive inverse first.  If b is a number, then -b is its additive inverse.  Then a - b is defined as a + (-b).  The other approach to subtraction is to define subtraction in terms of addition, and then use that definition to define the additive inverse.  I favor the first, because it seems more natural to define subtraction as an inverse operation to addition than to take the more circuitous route of defining an additive inverse first.

So I define subtraction as:

If and only if a + b = c, then c - b = a.

Then, I define define negation as:

-a = 0 - a

To show that (-a) + a = 0, which is called the additive inverse property, the definitions of negation and then subtraction are used as follows:

-a = 0 - a
(-a) + a  =  0

To show that 0 = -0, let a be zero in the definition of negation, then apply the definition of subtraction, and finally the additive identity principle:

-0 = 0 - 0
-0 + 0 = 0
-0 = 0

Now it is possible to derive an alternative definition of subtraction, which is that subtraction is adding the negative, or in symbols, a - b = a + (-b).

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

This result was obtained by using the "add to both sides" rule, then the  identity property of addition.

Now we can derive important associative rule involving subtraction, which is that a + (b - c) = (a + b) - c.

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

Now, I will show that (-a) + (-b) = -(a + b).  This is known as the "distributive property of negation over addition".  Since it's one of the longer examples, I number the lines.  Here it is:

(1)  (-a) + a  =  0
(2)  (-b) + b  =  0
(3)  0 + (-b) + b  =  0
(4)  (-a) + a + (-b) + b  =  0
(5)  ((-a) + (-b)) + (a + b) = 0
(6)  (-a) + (-b) = 0 - (a + b)
(7)  (-a) + (-b) = -(a + b)

Lines 1 and 2 use the additive inverse property.  Line 3 is the additive identity.  In line 4 I substitute (-a) + a in place of 0.  In line 5 I use the associative and commutative properties of addition to rearrange and regroup the terms.  Line 6 uses the definition of subtraction again.  Line 7 uses the definition of negation.

At this point, I'll show that a = -(-a) by applying additive inverse, the definition of subtraction, then the definition of negation:

(-a) + a  = 0
a = 0 - (-a)
a = -(-a)

Now here is another associative rule involving subtraction, which is a - (b + c) = (a - b) - c.  This is different from the associative rule for addition in that one of the addition operations is changed to a subtraction operation.

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

Finally, a commutative rule for subtraction: a - b = -(b - a).  Like the associative rule of subtraction, this commutative rule for subtraction is different from the commutative rule of addition because of the insertion of the negation operator.  You need to remember that b-a is the negative of a-b.  Here's the proof, using the following principles in this order: subtraction is adding the negative, commutative, double negation, distribution of negation, and subtraction is adding the negative.

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

These rules of substitution, subtraction, negation, association, etc. are the basic tools in the standard toolbox of every mathematician and student.  Within the constraints of these rules, you may freely substitute and interchange terms, and distribute and factor negations.  For example,

b - d + a - c

can be treated as if an invisible plus-sign is attached to the first term, and the plus- and minus- signs of the other terms are "glued" to the symbols that follow them.  Using that thinking, this expression can be rearranged as:

a + b - c - d

Notice that terms can be freely rearranged as long as the sign (plus- or minus-) of each term is preserved across the rearrangement of terms.

Principles of Multiplication

Multiplication is defined this way:

a × 1 = a
a × b = b × a
(a × b) × c = a × (b × c)
a × (b + c) = (a × b) + (a × c)

This definition consists of four axioms, called the identity, commutative, and associative properties of multiplication, and the distributive property of multiplication over addition.

There are three more properties that are very important for multiplication.  They are:

0 × a = 0,
a × (-1) = -a
, and
a × (b - c) = (a × b) - (a × c)

Each one can be proved using the properties that have already been presented.  I'll prove them in order, starting with the simplest one: 0 × a = 0

0 × a = (0+0) × a
0 × a = (0 × a) + (0 × a)
(0 × a) - (0 × a) = (0 × a)
0 = 0 × a

Now I will show that a × (-1) = -a, using 0 × a = 0, the distributive property and the definition of subtraction.

(a × (-1)) + (a × 1)  =  a × (1 + (-1))
(a × (-1)) + (a × 1)  =  a × 0
(a × (-1)) + (a × 1)  =  0
a × (-1)  =  -(a × 1)
a × (-1)  =  -a

The distributive property applies to subtraction, too.  In other words,

a × (b - c) = (a × b) - (a × c)

Here's the proof:

a × (b + (-c)) = (a × b) + (a × (-c))
a × (b - c) = (a × b) + (a × (c × (-1)))
a × (b - c) = (a × b) + ( (a × c) × (-1))
a × (b - c) = (a × b) + (- (a × c))
a × (b - c) = (a × b) - (a × c)

The product of two negative numbers is the same as the product of the same two positive numbers:

a × b = (-a) × (-b)

Here's the proof:

(1)  0 = 0
(2)  x(0) = 0(-y)
(3)  x(y + (-y)) = (-x + x)(-y)
(4)  xy + x(-y) = (-x)(-y)+ x(-y)
(5)  xy = (-x)(-y)

Equation 2 follows because 0 × a = 0.  Equation 3 expresses 0 on each side as a + (-a).  Then the distributive property is used, and finally x(-y) is subtracted from both sides.


Axioms of Real Arithmetic

In the list, below, the * asterisk means multiplication, and ^= means not equal.

The real numbers satisfy, and are defined by, the following 15 axioms

a + (b + c) = (a + b) + c (associativity of +)

a + b = b + a (commutativity of +)

a + 0 = a (identity of +)

a + (-a) = 0 (inverse of +)

a * (b * c) = (a * b) * c (associativity of *)

a * b = b * a (commutativity of *)

a * 1 = a (identity of *)

a ^= 0 ==> a * (1/a) = 1 (inverse of *)

a * (b + c) = a * b + a * c (* distributes over +)

a ≤ b or b ≤ a (≤ is total)

(a ≤ b and b ≤ a) ==> a = b (≤ is antisymmetric)

(a ≤ b and b ≤ c) ==> a ≤ c (≤ is transitive)

a ≤ b ==> a + c ≤ b + c (substitution principle of addition)

(a ≤ b and 0 ≤ c) ==> a * c ≤ b * c (substitution principle of multiplication)

And finally the completeness axiom:

Any non-empty set of real numbers that is bounded above has a least upper bound.

Some authors also point out "closure" as a separate axiom -- addition, multiplication, and the least upper bound all result in real numbers.  One author (Keisler) also points out the "root axiom", that for every real number a>0 there is a real number b>0 such that bn=a.

Internet references

Elementary Calculus: An Approach Using Infinitesimals, by H. Jerome Keisler.

Test Preparation Review: CBEST Test, HOBET Test

Related pages in this website

Back to the Basic Principles section

Arithmetic for sixth through ninth graders -- a less rigorous (that is, more practical) lesson on the rules of arithmetic.

Proving the basic facts about arithmetic for the natural numbers -- the Peano Postulates

Upper Bound -- definition of "upper bound" and "least upper bound" of either sets or functions

Definition of Interval -- a subset satisfying certain properties of a totally connected set such as the set of real numbers.  "Totally connected" means there exists a "≤" operator that is reflexive (a property in all sets), and antisymmetric, reflexive, and total (three of the thirteen properties listed above).


The webmaster and author of this Math Help site is Graeme McRae.