Element 0
Consider integland Z := (Z – 1). Let’s define
0 := 1-1
so that (by axiom 2)
∀x ∈ Z x-x = 0
It also follows (see axiom 3) that:
∀x ∈ Z x-0 = x
and, by substituting x := 1 in axiom 4:
0 ≠ 1
DEFINITION 1 Set A ⊆ Z is called a subminusop ⇐:⇒
∀a b ∈ A a-b ∈ A
Instead of the long and cumbersome term subminusop I’ll be using its abbreviation submop.
THEOREM 0 Element x := 0 is the only x ∈ Z such that the one-element set {x} is a submop.
Also
THEOREM 1 x = y ⇔ x-y = 0
PROOF
- If x=y then x-y = x-x = 0. That’s the first implication.
- If x-y = 0 then
x =7nbsp; x-0 = x – (x – y) = y – (x – x) = y-0 = y
That’s the second implication
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END of PROOF