Integland 0

Integland 1

Integland 2

Integland 3

Parity

Axioms 4 5 (see Integland 0) are concerned with inequalities:

• 1-x  ≠  x
• if  1-x = x-1  then x=1

It is due to the fact that viagra online generic is only effective to stimulate penile erection, it is useless when taken and administered in a woman’s body. Kamagra oral jelly is the latest invention of the levitra prices downtownsault.org medical science to beat impotency. Apart from this one point this magical weed canadian viagra no prescription has no reason to be not popular among the ED patients. What do you know about GMO’s (genetically-modified organisms)? GMO’s have been linked to many different health problems prescription for viagra across America.

for all  x ∈ Z.

The latter axiom above admits the following equivalent version:

if  Neg(x) = x   then  x=0.

On the other hand the former axiom implies another two equivalent versions of itself:

• x+x  ≠  1
• x+(x-1)  ≠  0

Thus, as we already have seen it in Integland 1:

0  ≠  1

Our goal in this note is to get a complete picture with respect to some such inequalities. It is described by the following special case of the classical Euclid theorem about the integer division and remainder:

THEOREM 10 (Parity)   For every integer  z ∈ Z  there exists exactly one integer  x ∈ Z,  and exactly one value  e = 0 or 1,  such that:

z  =  x + x + e

PROOF   We will need several steps (observations).

Observation 0   Let  e f ∈ {0 1}.  Then

e-f  ∈  {Neg(1) 0 1}

Observation 1   Let  t ∈ Z.  Integer  t + t + Neg(1)  is of the form  x+x+e  (as in the theorem).

Indeed:   t + t + Neg(1)  =  (t-1) + (t-1) + 1

Observation 2   Integer  1  is of the form  x+x+e.

Indeed:   1 = 0+0+1

Observation 3   The difference of arbitrary two integers of the form  x+x+e  is of the same form.

Indeed:   Let  x+x+e  and  y+y+f  be two integers of the said form. Then:

(x+x+e) – (y+y+f)  =  (x-y) + (x-y) + (e-f)

If  e-f  is  0 or 1  then the claim holds.  If  e-f = Neg(1)  then apply observations 0 1.

Observation 4 (the first half of the theorem)   The set of all integers of the form  x+x+f  coincides with  Z.

Indeed:   Apply axiom 6 to observations 2 3.

Observation 5   Let  x y ∈ Z  be such that

x+x = y+y

Then  x = y

Indeed:   if  x+x = y+y  then:

x-y = y-x

hence

x-y = Neg(x-y)

Thus  x-y = 0,  i.e. x=y.

Observation 6   Let  x y ∈ Z  be such that

x+x+1 = y+y+1

Then  x = y

Observation 7   There does not exist any   x y ∈ Z  such that

x+x+1 = y+y     (not possible!)

Indeed:   if  x+x+1 = y+y   then

1 – (y-x)  =  y-x

in a contradiction to axiom 4 (substitute  x  of axiom 4 by  y-x).

Observation 8 (uniqueness–the second and last part of the theorem)   Let

x y e f  ∈  Z

be such that  e f  ∈  {0 1},  and

x + x + e  =  y + y + f

Then   x = y   e = f.

Indeed   This is what observations 5 6 7 tell us.