Integland practice, cnt. 1

Element 0

Consider integland  Z := (Z  –  1).  Let’s define

:=  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

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