Proving Identities Using Axioms of Equality

Time to move on.In summary, the given axioms establish the properties of equality and addition, multiplication, and distributivity of real numbers. From these axioms, it can be proven that 0 multiplied by any number is equal to 0, and that the multiplicative inverse of a nonzero number multiplied by the multiplicative inverse of another nonzero number is equal to the multiplicative inverse of their product. These proofs rely on the use of substitution and other logical techniques.
  • #1
solakis1
422
0
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
 
Mathematics news on Phys.org
  • #2
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.
 
Last edited by a moderator:
  • #3
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
 
  • #4
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)
 
Last edited:
  • #5
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
 
Last edited:
  • #6
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??
 
  • #7
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)
 
Last edited:
  • #8
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
 
  • #9
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.
 

FAQ: Proving Identities Using Axioms of Equality

What is the purpose of proving identities using axioms of equality?

The purpose of proving identities using axioms of equality is to demonstrate the equality between two mathematical expressions. This allows us to simplify complex expressions and solve equations more easily.

What are the axioms of equality?

The axioms of equality are the reflexive property, symmetric property, transitive property, substitution property, and the addition and multiplication properties.

How do you prove an identity using axioms of equality?

To prove an identity using axioms of equality, you must manipulate one side of the equation using the axioms until it is equal to the other side. This shows that the two expressions are equivalent.

Why is it important to use axioms of equality when proving identities?

Using axioms of equality provides a logical and systematic approach to proving identities. It ensures that each step of the proof is mathematically valid and can be easily understood by others.

What are some common mistakes to avoid when proving identities using axioms of equality?

Some common mistakes to avoid when proving identities using axioms of equality include not applying the correct property, making an incorrect assumption, and not simplifying both sides of the equation equally.

Similar threads

MHB Proving Equivalence of 0A=0, (-A)B=-AB, (-A)(-B)=AB
Replies
1
Views
813
MHB Proving the Equality of A and B: A+B=0 or A=B
Replies
2
Views
1K
MHB Does Commutativity Hold for Matrices A and B with a Specific Matrix C?
Replies
3
Views
2K
I Are there any two pairs of integers with the same result in a specific function?
Replies
7
Views
1K
MHB Prove $(a+2b+3c+4d)a^ab^bc^cd^d < 1$
Replies
1
Views
833
MHB Triangle Equality Proof: $\frac{a}{b}=1+\sqrt{2\left(\frac{c}{b}\right)^2-1}$
Replies
1
Views
1K
MHB Finding all Possible Values of $x$ in Triangle ABC
Replies
1
Views
981
MHB Proofing Theorems with Axioms: Help Answer One!
Replies
11
Views
3K
Use mathematical logic to prove this proposition
Replies
1
Views
901
I Cancellation law multiplication natural numbers
Replies
17
Views
4K

Hot Threads

Recent Insights

Back
Top