On Arithmetic
It seems obvious that 2 + 2 ≠ 5…
A rigorous proof that 2 + 2 ≠ 5 is not particularly trivial. In order to be concise, we need to begin with some definitions:
Which is seems more complicated than it is.
- “∀” means “for all”.
- “∈” means “element of” “belongs to” or “in”.
- “ℕ” is a symbol for the natural numbers (0, 1, 2, 3 … etc.).
- “¬” means “not”.
- “⇔” means “if and only if”.
The only other part here that hasn’t been described is “S()”, the successor function. The successor of a number is the number after itself, for example, S(0) = 1, and S(3) = 4.
The axioms above in English for reference are as follows:
- 0 is a natural number
- For every natural number x, the number after x is also a natural number.
- For any two natural numbers x and y, if x = y, then S(x) = S(y) and if S(x) = S(y) then x = y
- 0 does not come after any natural number.
- Any natural number plus zero is itself
- For any two natural numbers x and y, x + S(y) = S(x + y)
This is sufficient to prove 2 + 2 ≠ 5. For brevity we have left out some axioms of equality.
We define 2 as S(S(0)) and 5 as S(S(S(S(S(0))))).
To prove that 2 + 2 ≠ 5, we need to prove ¬(2 + 2 = 5). ¬(2 + 2 = 5) is equivalent to 2 + 2 = 5 ⇒ ⊥ where “⇒” means “implies” and “⊥” means “falsehood” or “contradiction”. So if we can prove that 2 + 2 = 5 implies a contradiction or falsehood, then we have our proof.
We start with our definition of 2 and 5:
Then we apply our axioms of addition:
We use axiom 3 to reduce our problem by removing one layer of successor functions each time.
For which we already know by axiom 4 that ¬(S(0) = 0), or as we have seen before, S(0) = 0 ⇒ ⊥.
Therefore:
or:
Qed.
Originally published at andymac-2.github.io.