Those who dislike mathematics should skip to the next post.

Algebra class. 0 = x^2 -11x +30 = (x-5) * (x-6). We are now told that, if the product of two numbers is zero, one of them must be zero. so x-5 or x-6 is zero, and x = 5 or x = 6.

I remember this, and remember thinking, ‘can you prove that?’, but decided it was true, and let it go. At the time.

A few months ago, insomniac, this problem re-presented itself to me, and I decided to attempt a proof. I studied groups/rings/fields in UofT, so I used the axioms of a field to prove the assertion, a*b=0 implies a=0 or b=0, for any field. The proof does not require commutativity of multiplication, so it applies to Hamiltonian Quaternions, as well as to real numbers, rational numbers, algebraic numbers, and integers mod a prime number, and other Fields in the mathematical sense.

I will not go into the tedious details of what an ‘operation’ is. It is a mapping from F*F into F. Instead I’ll simply say, a+b is always in F if a and b are, and is always defined and always unique. Similarly, the operation * follows the same rule: a*b is always in F if a and b are, is always defined and is always unique.

The relationship “=” is supposed to be reflexive, symmetric, and transitive. That means, for any a, b, c in F we have: a=a; and a=b and b=c implies a=c; and a=b implies b=a.

Now for the axioms of a field F:

F is a set, and has members. There are operations on its members, which we will write as + and *. These operations work much as you’ve seen numbers add and multiply, but we’ll have abstract axioms that could apply to totally different operations (rotations in space, permutations, et cetera) on different sets holding different members entirely.

Associative law of addition: a + (b + c) = (a + b) + c.

Commutative law of addition: a + b = b + a. (we won’t actually need this if we assume left- and right- identity elements).

Identity element for +: We will write this element as 0 and it has the property that

0 + a = a for all a in F. We can either assume commutativity, or another element *0*

with the property a + *0* = a for all a in F. If you consider the value of 0 + *0*, depending on which identity element you remove, you prove they are equal.

Inverse element for addition. For all a in F, there is another unique element *a* say, such that a + *a* = 0. You can either assume commutativity, or assume a left- and right- additive inverse, and the expression a(left inverse) + a + a(right inverse), with the associative law, can be evaluated to be either inverse. In any case we will from now on write -a for the additive inverse of a. You should remember that b – a is really b + -a. We don’t exactly have subtraction, we have adding the additive inverse.

Multiplication. This runs quite similar to addition with two small twists.

a * b is always defined. a * b = b * a and we won’t actually need this axiom (commutative law of multiplication).

a * (b * c) = (a * b) * c for all a, b, c, in F (associative law of multiplication).

There is an identity element for multiplication, which I will write as 1. It has the property that

1 * a = a for all a in F. Again, we can either assume commutativity or assume both a left-identity and a right-identity and prove they are equal simply by multiplying them together.

So I can write 1 for the multiplication identity element and 0 for the addition identity element without ambiguity.

As in addition, there is a multiplicative inverse for all elements in F **except 0**.

For any a not zero, there is a unique element z say, such that a * z = 1.

We can either assume commutativity of multiplication, or assume both a left-inverse and a right-inverse, and by evaluating (left-inverse-of-a) * a * (right-inverse-of-a) using the two interpretations given by the associative law for multiplication, and prove the left inverse is the same as the right inverse.

I will write, 1/a for the inverse of a. If I accidentally write b/a, what I really mean is b * 1/a. Again, we have multiplication by the inverse, not division as a new operator.

The second thing unusual about multiplication is, it interacts with addition in a very specific way.

a * (b + c) = (a * b) + (a * c). or, a*(b+c) = a*b + a*c where we ‘read’ the * as happening before the +. This is the distributive law of multiplication over addition.

We are now ready to prove a small lemma we’ll need in a few more lines.

Claim: for any x in F, x * 0 = 0.

Proof: x = x * 1 (multiplicative identity). 1 = 1 + 0 (additive identity 0, special case).

so x = x * 1 = x * (1 + 0) = x * 1 + x * 0 (distributive law). Restating, x = x + x * 0.

Now we add -x to the left of both sides: -x + x = -x + x + x * 0; which gives 0 = x * 0. QED.

Now assume a*b=0. we are to prove a=0 or b=0.

if a=0 we are done. If it does not, we have 1/a such that 1/a * a = 1.

multiply the original equation, a * b = 0, by 1/a on each side. This gives us

1/a * (a * b) = 1/a * 0. By the associative law, the left side equals (1/a * a) * b which is 1 * b or b. The right hand side, by our lemma, is a case of x * 0 and is zero.

So we have, if a * b = 0 and a is not 0, then b = 0. QED.