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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.

And the definition: a/b= a(1/b)

Prove,based only on the axioms and the definition stated above the following:

A) a/b + c/d = (ad + bc)/bd if b and d are different from zero

B) (a/b)/(c/d) = (a/b)(d/c) if b and d are different than zero.

In trying to prove A the following result is shown.

1) \(\displaystyle a\neq 0\)......................assumption

2)\(\displaystyle b\neq 0

\).........................assumption

3) bd =0 ..........................hypothesis for contradiction

4) (bd)1/d = 0(1/d)...............by multiplying both sides by 1/d

5) 1/d(bd) = (1/d)0..................by axiom 3 (commut)

6) [(1/d)d]b = 0..................by axioms 3,4 and the theorem A0= 0

7) 1b = 0...........................by axiom 9

8) b=0...............................by axioms 3, 8

9) \(\displaystyle b\neq 0 \) and b=0 ...........contradiction

10) \(\displaystyle bd\neq 0\)

11) (bd)(1/bd) =1......................axiom 9

(a/b +c/d) =

= (a/b+c/d)1=.............................by axiom 8

= (a/b+c/d)((bd)(1/bd))=.................by axiom 9

=[(a/b+c/d)(bd)]1/bd=.....................by axiom 4

=[((a(1/b)) +(c(1/d)))(bd)]1/bd =..............by the definition

=[(bd)(a(1/b) + (bd)(c(1/d))]1/bd=...............by axioms 3 and 5

=[(a(1/b))(bd) + (c(1/d))(bd)]1/bd=..............by axiom 3

=[((a(1/b))b)d + ((c(1/d))d)b]1/bd=............. by axioms 3 and 4

=[(b((1/b)a))d + (d((1/d)c))b]1/bd=.................by axiom 3

=[( (b(1/b))a)d +((d(1/d))c)b]1/bd= by axiom 4

=[ (1a)d + (1c)b]1/bd=...........................by axiom 9

=[ad + cb]1/bd=...................................by axioms 3 and 8

= (ad + cb)/bd ................................ny definition

Can perhaps somebody show a shorter proof?

For B I have not try yet

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.

And the definition: a/b= a(1/b)

Prove,based only on the axioms and the definition stated above the following:

A) a/b + c/d = (ad + bc)/bd if b and d are different from zero

B) (a/b)/(c/d) = (a/b)(d/c) if b and d are different than zero.

In trying to prove A the following result is shown.

1) \(\displaystyle a\neq 0\)......................assumption

2)\(\displaystyle b\neq 0

\).........................assumption

3) bd =0 ..........................hypothesis for contradiction

4) (bd)1/d = 0(1/d)...............by multiplying both sides by 1/d

5) 1/d(bd) = (1/d)0..................by axiom 3 (commut)

6) [(1/d)d]b = 0..................by axioms 3,4 and the theorem A0= 0

7) 1b = 0...........................by axiom 9

8) b=0...............................by axioms 3, 8

9) \(\displaystyle b\neq 0 \) and b=0 ...........contradiction

10) \(\displaystyle bd\neq 0\)

11) (bd)(1/bd) =1......................axiom 9

(a/b +c/d) =

= (a/b+c/d)1=.............................by axiom 8

= (a/b+c/d)((bd)(1/bd))=.................by axiom 9

=[(a/b+c/d)(bd)]1/bd=.....................by axiom 4

=[((a(1/b)) +(c(1/d)))(bd)]1/bd =..............by the definition

=[(bd)(a(1/b) + (bd)(c(1/d))]1/bd=...............by axioms 3 and 5

=[(a(1/b))(bd) + (c(1/d))(bd)]1/bd=..............by axiom 3

=[((a(1/b))b)d + ((c(1/d))d)b]1/bd=............. by axioms 3 and 4

=[(b((1/b)a))d + (d((1/d)c))b]1/bd=.................by axiom 3

=[( (b(1/b))a)d +((d(1/d))c)b]1/bd= by axiom 4

=[ (1a)d + (1c)b]1/bd=...........................by axiom 9

=[ad + cb]1/bd=...................................by axioms 3 and 8

= (ad + cb)/bd ................................ny definition

Can perhaps somebody show a shorter proof?

For B I have not try yet

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