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Given the following axioms:

For all a,b,c we have:

1) a+b = b+a

2) a+(b+c)=(a+b)+c

3) ab = ba

4) a(bc) = (ab)c

5) a(b+c) =ab+ac

NOTE,here the multiplication sign (.) between the variables have been ommited

6) There ia a number called 0 such that for all a,

a+0 =a

7)For each a, there is a number -a such that for any a,

a+(-a) = 0

8)There is a number called 1(diofferent from 0) such that for any a,

a1 = a

9)For each a which is different than 0there exists a number called 1/a such that;

a.(1/a)= 1.

10) exactly one of a>b,b>a or a=b holds

11) if a>b ,b>c then a>c

12) if c>0 ,a>b then ac>bc

13) if a>b then a+c>b+c for any c

The definitions:

14) a/b = a(1/b)

15) $a\geq 0\Longrightarrow |a|=a$ and $ a<0\Longrightarrow |a|=-a$.

Then by using only the axioms and the definitions above prove:

A) If a>0 and b>0 then a>b iff aa>bb

B) If $x\neq 0$ then (|x||x|)/x=x for all x

For all a,b,c we have:

1) a+b = b+a

2) a+(b+c)=(a+b)+c

3) ab = ba

4) a(bc) = (ab)c

5) a(b+c) =ab+ac

NOTE,here the multiplication sign (.) between the variables have been ommited

6) There ia a number called 0 such that for all a,

a+0 =a

7)For each a, there is a number -a such that for any a,

a+(-a) = 0

8)There is a number called 1(diofferent from 0) such that for any a,

a1 = a

9)For each a which is different than 0there exists a number called 1/a such that;

a.(1/a)= 1.

10) exactly one of a>b,b>a or a=b holds

11) if a>b ,b>c then a>c

12) if c>0 ,a>b then ac>bc

13) if a>b then a+c>b+c for any c

The definitions:

14) a/b = a(1/b)

15) $a\geq 0\Longrightarrow |a|=a$ and $ a<0\Longrightarrow |a|=-a$.

Then by using only the axioms and the definitions above prove:

A) If a>0 and b>0 then a>b iff aa>bb

B) If $x\neq 0$ then (|x||x|)/x=x for all x

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