MHB Proving Identities Using Axioms of Equality

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The discussion focuses on proving identities using axioms of equality, specifically addressing the proofs of 0=0A and (1/A)(1/B)=1/AB. Participants emphasize the importance of following axioms strictly while also acknowledging the necessity of substitution rules in mathematical proofs. There is a debate on whether the approach to using axioms is too restrictive, with some arguing for a more flexible application that includes established mathematical principles. The conversation highlights the balance between adhering to axioms and the practical use of substitution and other mathematical tools. Ultimately, the thread reveals differing opinions on the methodology of proving identities within the framework of axioms.
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Given the following axioms:

For all A,B,C...we have:

1) A=A

2) A=B <=> B=A

3) A=B & B=C => A=C

4) A=B => A+C= B+C

5) A=B=> AC =BC ( NOTE :Instead of writing A.C or B.C e.t.c we write AB.BC e.t.c)

6) A+B= B+A..........AB=BA

7) A+(B+C) = (A+B)+C............A(BC)=(AB)C

10) A+0=A...............1A=A

11) A+(-A)=0...............A=/=0 => (1/A)A=1

12).......(A+B)C= AC+BC.........

13) 1=/= 0

Then prove: (By using only the axioms above)

1) 0=0A

2) A=/=0 & B=/= 0 => (1/A)(1/B)= 1/AB
 
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solakis said:
Given the following axioms:

For all A,B,C...we have:

1) A=A

2) A=B <=> B=A

3) A=B & B=C => A=C

4) A=B => A+C= B+C

5) A=B=> AC =BC ( NOTE :Instead of writing A.C or B.C e.t.c we write AB.BC e.t.c)

6) A+B= B+A..........AB=BA

7) A+(B+C) = (A+B)+C............A(BC)=(AB)C

10) A+0=A...............1A=A

11) A+(-A)=0...............A=/=0 => (1/A)A=1

12).......(A+B)C= AC+BC.........

13) 1=/= 0

Then prove: (By using only the axioms above)

1) 0=0A

2) A=/=0 & B=/= 0 => (1/A)(1/B)= 1/AB
For (1)
0 = A - A = 1A - 1A = (1 - 1)A = 0A

Hint: For (2) Note that 1/A is the multiplicative inverse of A. So what then is (1/A)(1/B) equal to?

-Dan

Addendum: For (1) I really should be using 1A - 1A = 1A + (-1)A where -1 is the additive inverse of 1.
 
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topsquark said:
For (1)
0 = A - A = 1A - 1A = (1 - 1)A = 0A

Hint: For (2) Note that 1/A is the multiplicative inverse of A. So what then is (1/A)(1/B) equal to?

-Dan

Addendum: For (1) I really should be using 1A - 1A = 1A + (-1)A where -1 is the additive inverse of 1.

In my list of axioms there is no axiom like the one you mention : 0 = A-A

However there is an axiom : A+(-A)=0.

As for (2) that is what we want to prove that the multiplicative inverse of (1/A)(1/B) is 1/AB
 
0 = 0 + 0 (by 10 and 2)
0.A = 0.A + 0.A (by 5 and 12)
0.A + (-0.A) = 0.A + 0.A + (-0.A) (by 4)
0 = 0.A + 0 = 0.A (by 4 and 10)
 
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I use the notation A'=1/A

(*) if C $\neq$ 0 then AC = BC => A = B
proof: Given AC = BC and C $\neq$ 0
(AC)C' = (BC)C' (by 5 and C $\neq$ 0)
A(CC') = B(CC') (by 7)
A1 = B1 (by 11)
1A = 1B (by 6)
A = B (by 10)
end proof

(**) (ab)(cd) = (ab)e = a(be) = a(b(cd)) = a((bc)d) (by 7) and e = cd.

(***) Let AB = 0, we have 0 = 0B, proven above, so AB = 0B (by 3)
Suppose B $\neq$ 0, then A = 0 (by *)
Thus: if A $\neq$ 0 and B $\neq$ 0 then AB $\neq$ 0 (by logics)

(B'A')(AB) = B'((A'A)B) (by 7 and by **, A $\neq$ 0 and B $\neq$ 0)
= B'(1B) (by 11)
= B'B (by 10)
= 1 (by 11)

(B'A')(AB) = 1 = (AB)'(AB) (by 2, 3, and 11 and ***)
(AB)' = B'A' = A'B' (by (*) and 6)
ready
 
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steenis said:
0 = 0 + 0 (by 10)
0.A = 0.A + 0.A (by 5 and 12)
0.A + (-0.A) = 0.A + 0.A + (-0.A) (by 4)
0 = 0.A + 0 = 0.A (by 4 and 10)

[Let us go step by step in your proof:

1) 0=0+0 by using axiom 10

BY using axiom 10 (A+0=A) we get : 0+0=0 instead of 0=0+0 To get 0=0+0 we must use the axiom : A=B<=>B=A

2) 0A=0A+0A BY using axioms 5 and 12

BY using axiom 5 (A=B => AC=BC) we get : 0A=(0+0)A

By using axiom 12 ((A+B)C=AC+BC) we get : (0+0)A= 0A+0A .To get to 0A=0A+0A we have to use the axiom : A=B and B=C => A=C

[0A=(0+0)A and (0+0)A= 0A+0A] => 0A=0A+0A

3) 0A+(-0A) = (0A+0A)+(-0A) By using 4 .This is correct

4) 0 =0A +0 = 0A By 4 and 10 ........How??
 
we have
(a) 0A + (-0A) = 0A + 0A + (-0A) (by 4)

by (11) we have 0A + (-0A) = 0

apply twice to (a): 0 = 0A + 0

0A + 0 = 0A by (10)

0 = 0A (by 3)

you are right, (4) should be (11)
 
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steenis said:
we have
(a) 0A + (-0A) = 0A + 0A + (-0A) (by 4)

by (11) we have 0A + (-0A) = 0

apply twice to (a): 0 = 0A + 0

0A + 0 = 0A by (10)

0 = 0A (by 3)

you are right, (4) should be (11)

Again by 4 we have : 0A+(-0A)= (0A+0A)+(-0A) and not 0A+(-0A)= 0A+0A+(-0A) (put A=0A,B=0A+0A,C=-0A) ,Then by using the axiom 4 we have : 0A+(-0A)= (0A+0A)+(-0A)

And now how are you going to apply the result of the axiom 11 (A+(-A)=0),Which is 0A+(-0A)=0

Becides if you mean when you saying "apply twice to (a)" substitute twice to (a),there is no substitution axiom in my axioms
 
Like you said I only can use the axioms you gave me. Is stead of "use", I use the word apply.
Axiom 11) says A + (-A) = 0
So if you have something like 'X + (-X)' in a "formula" then you can apply axiom 11) and write 0 for X + (-X).
If you have something like 'AQU + (-AQU)', you can also apply axion 11) and write 0 for AQU + (-AQU).
You do the same in arithmetic, you have '3567 + 1132', you can apply/use axiom 6) and you know that 3567 + 1132 = 1132 + 3567
How else would you use axioms?
 
  • #10
steenis said:
Like you said I only can use the axioms you gave me. Is stead of "use", I use the word apply.
Axiom 11) says A + (-A) = 0
So if you have something like 'X + (-X)' in a "formula" then you can apply axiom 11) and write 0 for X + (-X).
If you have something like 'AQU + (-AQU)', you can also apply axion 11) and write 0 for AQU + (-AQU).
You do the same in arithmetic, you have '3567 + 1132', you can apply/use axiom 6) and you know that 3567 + 1132 = 1132 + 3567
How else would you use axioms?
No No you do not get what i mean

We have the formula 0A+(-0A)= (0A+0A)+(-0A).................1

Now from axiom 11 we have : A+(-A)=0 and if we put A=0A we have 0A+(-0A)=0.........2

Now by using axiom 2 we get :0=0A+(-0A)..................3

So now we have : 0=0A+(-0A) and 0A+(-0A)= (0A+0A)+(-0A) and using axiom 3 we have :0=(0A+0A)+(-OA)

How do we continue from here ??
 
  • #11
I see what you mean, and I agree that is can be done in that way. But it is too restrictive. It is like saying:

$a + b = 7$ and $b = 5$ but one cannot deduce $a + 5 =7$ because there is no substitution rule.

I remember these exercises from university, follow the axioms, but we were always allowed to use substitution even it was not implicitly stated that it was allowed.

I leave this for the "logic-experts" to comment on this.
 
  • #12
steenis said:
I see what you mean, and I agree that is can be done in that way. But it is too restrictive. It is like saying:

$a + b = 7$ and $b = 5$ but one cannot deduce $a + 5 =7$ because there is no substitution rule.

I remember these exercises from university, follow the axioms, but we were always allowed to use substitution even it was not implicitly stated that it was allowed.

I leave this for the "logic-experts" to comment on this.

To take your very good example.

Can we prove : if, a+b=7 and b=2,then a+2=7 without using the substitution rule ??

proof:

1) a+b=7........................given

2) b=2........................given

3) b+a =2+a.............by using the axiom (4) A=B=> A+C=B+C

4) 2+a =a+2.............by using the axiom (6) A+B= B+A

5) b+a = a+2............by using the axiom (3) A=B and B=C => A=C

6) a+2 =b+a.............by using the axiom (2) A=B <=> B=A

7) b+a= a+b.............by using the axiom (6) A+B= B+A

8) a+2 =a+b...............by using the axiom (3) A=B and B=C => A=C

9) a+2 =7...............by using again the axiom (3) A=B and B=C => A=CIf we generalize now we can prove the 1st substitution theorem : If ,A+B=C and B=D ,then A+D=C

And the 2nd substitution theorem : If, AB=C and B=D .then AD=C.

And now we can use substitution in proving : 0A=0 and A=/=0,B=/= 0 => (1/A)(1/B) = 1/AB
 
  • #13
solakis said:
To take your very good example.

1) Can we prove : if, a+b=7 and b=2,then a+2=7 without using the substitution rule ??

If we generalize now we can prove the 1st substitution theorem : If ,A+B=C and B=D ,then A+D=C

And the 2nd substitution theorem : If, AB=C and B=D .then AD=C.

2) And now we can use substitution in proving : 0A=0 and A=/=0,B=/= 0 => (1/A)(1/B) = 1/AB
1) it is a good exercise to do once in your lifetime

2) I already did both exercises, I think

It was fun to do and learn this "axiomatic thinking", but I leave it now to other people.
 
  • #14
steenis said:
1) it is a good exercise to do once in your lifetime

2) I already did both exercises, I think

It was fun to do and learn this "axiomatic thinking", but I leave it now to other people.

So you thing this kind of proof strictly based on the axioms ,theorems, and definitions of an axiomatic system it should not be used in ordinary mathematics??

Then how can we be 100% sure for the correctness of each and every proof in any field in mathematics??
 
  • #15
Of course we need axioms and the application of axioms. But we do not need to rebuild everything from the ground when we need something. That is why we invented "substitution", theorems, propositions, lemma's and corollaries and proofs. I repeat: it is meaningful to learn to handle axioms. However, I think your way of using axioms is like doing a game. It is useful to do it once to see how it works. But after that is important to build a building with definitions, theorems and proofs.
 
  • #16
Moderator's note: this thread is off-topic now and is derailing.
 
  • #17
I agree. I have made my point and I have the feeling I am repeating myself.
 

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