Thursday, October 9, 2008

Why is 0 times everything 0?

Why is 0 times everything 0?

To answer this question, I first have to give you the field axioms and equation logic:

aa^(-1)=1=a^(-1)a if a!=0

Equation logic:

Reflexivity: s = s
Symmetry: if s=t, t=s.
Transitivity: if s=t and t=v, s=v.
Substitution: if s=t and v is a member of the field, s*v = t*v.
Substitution: if s=t and v is a member of the field, s+v = t+v.

There's another rule of logic that's irrelevant to what we're doing here.

These axioms and logic rules define everything you can do with numbers in the various fields (complex numbers, real numbers and rational numbers, for example. The set of integers is not a field because you cannot mathematically invert any given integer and get an integer back.

Okay, let's use the above axioms and equation logic to prove that, for any number A that is part of a field, A*0=0.

1: 0=0 (reflexivity)
2: 0+0=0 (additive identity)
3: A*(0+0) = A*0 (logical substitution; multiplication is well-defined).
4: A*0 + A*0 = A*0 (multiplicative distribution)
5: A*0 + A*0 + -A*0 = A*0 + -A*0 (logical substitution; addition is well-defined)
6: A*0 + (A + -A)*0 = 0*(A + -A) (multiplicative distribution)
7: A*0 + 0*0 = 0*0 (additive inverse)
8: let 0*0 = b. (simplification, and therefore:)
9: A*0 + b = b
10: A*0 + b + -b = b + -b (logical substitution; addition is well-defined)
11: A*0 + 0 = 0 (additive inverse)
12: A*0 = 0 (additive identity)
13: QED (pompousness)

There you have it. For any number A in a field that obeys the above axioms and rules of logic, A*0 = 0.


Kerry said...

Because 0 groups of anything is 0. Good grief.

Grumpator said...

My eyes glazed over for most of this post. I do like #13.