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For all a,b,c we have:

1) a+b = b+c

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.

For any numbers a,b

10) either a<b or a>b or a=b

11) if a<b and b<c then a<c

12) if a<b then a+c<b+c for any c

13) if a<b and c>0 then ac<bc for any c.

Then by using only the axioms stated above prove:

A) a0 = 0 ,B) 0<1

In trying to prove A i followed the proof shown below:

1) 0+0=0 .......................by using axiom 6

2) (0+0)x =0x.................by multiplying both sides by x

3)0x+ox =0x.....................by using axiom 5

4) (0x+0x) +(-0x) =0x+(-0x) ..................by adding (-0x) to both sides

5) 0x +[0x+(-0x)]= 0x+(-0x)....................by using axiom 2

6) 0x +0 = 0 .....................by using axiom 7

7) 0x = 0 ..........................by using axiom 6

For B ,I could not show a proof based only on the axioms stated above

Is my proof for A, correct 100%?