# Axiomatic Approach 2

#### solakis

##### Active member
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

Last edited:

#### Prove It

##### Well-known member
MHB Math Helper
Re: axiomatic approach 2

Axiom 1 is wrong, so I didn't bother reading anything else.

#### hmmm16

##### Member
Re: axiomatic approach 2

Axiom 1 is wrong, so I didn't bother reading anything else.
What do you mean that axiom 1 is "wrong"? Unless I have made a mistake I would have:

$\frac{(ad+bc)}{bd}=(ad+bc).\frac{1}{bd}$ by the definition.

$(ad+bc).\frac{1}{bd}=\frac{1}{bd}(ad+bc)$ by axiom 3.

$\frac{1}{bd}(ad+bc)=\frac{1}{bd}ad+\frac{1}{bd}bc$ by axiom 5.

$\frac{1}{bd}ad+\frac{1}{bd}bc=\frac{ad}{bd}+\frac{bc}{bd}$ by the definition.

$\frac{ad}{bd}+\frac{bc}{bd}=\frac{a}{b}+\frac{c}{d}$ as required.

Last edited:

#### Prove It

##### Well-known member
MHB Math Helper
Re: axiomatic approach 2

Last time I checked, a + b = b + c is only true when a = c.

#### hmmm16

##### Member
Re: axiomatic approach 2

Last time I checked, a + b = b + c is only true when a = c.
Oh right I read it as (and I just assume that it is supposed to be) $a+b=b+a$.

I suppose that could still be an axiom, just says everything is equal though.

#### solakis

##### Active member
Re: axiomatic approach 2

What do you mean that axiom 1 is "wrong"? Unless I have made a mistake I would have:

$\frac{(ad+bc)}{bd}=(ad+bc).\frac{1}{bd}$ by the definition.

$(ad+bc).\frac{1}{bd}=\frac{1}{bd}(ad+bc)$ by axiom 3.

$\frac{1}{bd}(ad+bc)=\frac{1}{bd}ad+\frac{1}{bd}bc$ by axiom 5.

$\frac{1}{bd}ad+\frac{1}{bd}bc=\frac{ad}{bd}+\frac{bc}{bd}$ by the definition.

$\frac{ad}{bd}+\frac{bc}{bd}=\frac{a}{b}+\frac{c}{d}$ as required.

By what axiom you conclude that :

$\frac{ad}{bd}=\frac{a}{b}$ AND

$\frac{bc}{bd}=\frac{c}{d}$ ??

#### hmmm16

##### Member
Re: axiomatic approach 2

By what axiom you conclude that :

$\frac{ad}{bd}=\frac{a}{b}$ AND

$\frac{bc}{bd}=\frac{c}{d}$ ??
$\frac{ad}{bd}=ad.\frac{1}{bd}$ by the definition the using the inverse axiom

$=ad.\frac{1}{d}\frac{1}{b}=\frac{a}{b}$ hence result

#### solakis

##### Active member
Re: axiomatic approach 2

$\frac{ad}{bd}=ad.\frac{1}{bd}$ by the definition the using the inverse axiom

$=ad.\frac{1}{d}\frac{1}{b}=\frac{a}{b}$ hence result
Is the inverse axiom ,axiom No 9 ??

By what axiom you conclude that:

$\frac{1}{bd} =\frac{1}{b}\frac{1}{d}$ ?

Last edited:

#### solakis

##### Active member
In trying to prove problem B i could go as far as :

(a/b)/(c/d)=

=$\frac{a}{b}\frac{1}{\frac{c}{d}}$= ........................by using the definition

=$\frac{a}{b}\frac{1}{c\frac{1}{d}}$..........................by using again the definition

That is as far as i can go

Can anybody help??

#### topsquark

##### Well-known member
MHB Math Helper
Re: axiomatic approach 2

Is the inverse axiom ,axiom No 9 ??

By what axiom you conclude that:

$\frac{1}{bd} =\frac{1}{b}\frac{1}{d}$ ?
Hint: (Actually a rewritten form of your question)
Show that $$(bd)^{-1} = d^{-1}b^{-1}$$

To start: What is $$(bd)^{-1}d$$?

-Dan

#### solakis

##### Active member
Re: axiomatic approach 2

Hint: (Actually a rewritten form of your question)
Show that $$(bd)^{-1} = d^{-1}b^{-1}$$

To start: What is $$(bd)^{-1}d$$?

-Dan
You mean $$(bd)^{-1}bd$$,because i cannot see what is :

$$(bd)^{-1}d$$?

#### topsquark

##### Well-known member
MHB Math Helper
Re: axiomatic approach 2

You mean $$(bd)^{-1}bd$$,because i cannot see what is :

$$(bd)^{-1}d$$?
I apologize. I was working on the idea that this is a linear vector space and it doesn't appear to be (unless you want the vectors to act as the scalars as well, which is a construction I don't have a lot of the details for. I know I know..an example of this is the Real number system. Hey, I'm a Physicist. )

What I was going for is this:
$$\displaystyle (bd)^{-1} = d^{-1}b^{-1}$$

I can prove that there is no contradiction with the Axioms you posted, but I can't seem to actually prove it. I know I've seen a proof of this somewhere though.

-Dan

Addendum: It's easy enough to prove that -(a + b) = -b + -a by using the distribution axiom. Maybe there is a generalization you can get to work based on it.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
(a/b) = (a/b) 1 1 = (a/b) c(1/c) d(1/d) = (a/b) c(1/d) d(1/c) = (a/b) (c/d) (d/c)

(a/b) (1/(d/c)) = (a/b) (c/d) (d/c) (1/(d/c))

(a/b)/(d/c) = (a/b) (c/d)

Sorry, didn't look up all the axiom numbers.

#### solakis

##### Active member
(a/b) = (a/b) 1 1 = (a/b) c(1/c) d(1/d) = (a/b) c(1/d) d(1/c) = (a/b) (c/d) (d/c)

(a/b) (1/(d/c)) = (a/b) (c/d) (d/c) (1/(d/c))

(a/b)/(d/c) = (a/b) (c/d)

Sorry, didn't look up all the axiom numbers.
Your solution is a great example of what generaly happens in nearly all branches in mathematics

Proofs with statements which are not justified.
It has been nearly half an hour ,and yet i cannot make out whether your proof is correct and which are the axioms that it is based.

Please be so kind to help me by citing the appropriate axioms involved in your ptroof.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
 let c and d be different from 0 (a/b) = (a/b) 1 1 unity for multiplication (a/b) = (a/b) c(1/c) d(1/d) existence of multiplicative inverse (a/b) = (a/b) c(1/d) d(1/c) associativity + commutativity of multiplication (a/b) = (a/b) (c/d) (d/c) definition of fraction (a/b) (1/(d/c)) = (a/b) (c/d) (d/c) (1/(d/c)) multiply both sides by (1/(d/c)) (a/b) (1/(d/c)) = (a/b) (c/d) 1 multiplicative inverse (a/b) (1/(d/c)) = (a/b) (c/d) unity for multiplication (a/b)/(d/c) = (a/b) (c/d) definition of fraction

Actually, I haven't proven the existence of 1/(d/c), since (d/c) might be 0.
Perhaps you can prove that?

Last edited:

#### solakis

##### Active member
 let c and d be different from 0 (a/b) = (a/b) 1 1 unity for multiplication (a/b) = (a/b) c(1/c) d(1/d) existence of multiplicative inverse (a/b) = (a/b) c(1/d) d(1/c) associativity + commutativity of multiplication (a/b) = (a/b) (c/d) (d/c) definition of fraction (a/b) (1/(d/c)) = (a/b) (c/d) (d/c) (1/(d/c)) multiply both sides by (1/(d/c)) (a/b) (1/(d/c)) = (a/b) (c/d) 1 multiplicative inverse (a/b) (1/(d/c)) = (a/b) (c/d) unity for multiplication (a/b)/(d/c) = (a/b) (c/d) definition of fraction

Actually, I haven't proven the existence of 1/(d/c), since (d/c) might be 0.
Perhaps you can prove that?
Thanks
Now i can follow

But there one or two steps that should be explained in more details

Do you agree that the following step in your proof:

 (a/b) = (a/b) c(1/c) d(1/d) existence of multiplicative inverse

should be :

(a/b) =(a/b)[(c(1/c))(d(1/d)]

If you do,then how do you apply associativity and commutativity of multiplication to come up with:

(a/b) = (a/b) c(1/d) d(1/c)

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Thanks
Now i can follow

But there one or two steps that should be explained in more details

Do you agree that the following step in your proof:

 (a/b) = (a/b) c(1/c) d(1/d) existence of multiplicative inverse

should be :

(a/b) =(a/b)[(c(1/c))(d(1/d)]

If you do,then how do you apply associativity and commutativity of multiplication to come up with:

(a/b) = (a/b) c(1/d) d(1/c)
Strictly speaking you are right (if you add another closing parenthesis), but it's usually not considered necessary for an operation that is associative.
Since multiplication is associative, parentheses can be generally left out.

#### solakis

##### Active member
.
Since multiplication is associative, parentheses can be generally left out.
if parenthese are left out you have no associativity,because in the axiom.a(dc)=(ab)c it is the parentheses that show associativity.

But if you mean associativity and commutativity, then it is a different story.

But then this must be proved.

For example for three variables we must show that:

abc =acb = cab= bca=bac

In our case the variables are five and the story is longer.

For any No of variables the proof must be impossib;le.

Once this is proved can be used as theorem in any subsequent proof.

Hence in your proof you cannot drop out parentheses because we have not proved yet the above theorem

Otherwise the proof is not based only on the axioms stated ,as the OP demanded

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I only meant associativity.
Commutativity is an entirely different matter.

Associativity means that for any a,b, and c: (ab)c=a(bc).
If this is the case we can also simply write abc instead.

#### solakis

##### Active member
I only meant associativity.
Commutativity is an entirely different matter.

Associativity means that for any a,b, and c: (ab)c=a(bc).
If this is the case we can also simply write abc instead.

In a non commutative multiplicative Group with identity element 1 and inverse of $a^{-1}$ ,how would you prove that:

$b^{-1}a^{-1}=(ab)^{-1}$ without using parentheses

#### Klaas van Aarsen

##### MHB Seeker
Staff member
In a non commutative multiplicative Group with identity element 1 and inverse of $a^{-1}$ ,how would you prove that:

$b^{-1}a^{-1}=(ab)^{-1}$ without using parentheses
In this case you need parentheses.
You can leave parentheses out for a single type of operation that is associative.
If another type of operation is involved, you may need parentheses.

#### solakis

##### Active member
In this case you need parentheses.
You can leave parentheses out for a single type of operation that is associative.
If another type of operation is involved, you may need parentheses.
Why we cannot have the following proof without using parentheses??

$b^{-1}a^{-1}$= $b^{-1}a^{-1}ab(ab)^{-1}$=

=$b^{=1}1b(ab)^{-1}$= $1(ab)^{-1}$=$(ab)^{-1}$

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Why we cannot have the following proof without using parentheses??

$b^{-1}a^{-1}$= $b^{-1}a^{-1}ab(ab)^{-1}$=

=$b^{=1}1b(ab)^{-1}$= $1(ab)^{-1}$=$(ab)^{-1}$
Sure you can.
This is correct.
Note that you did leave out a couple steps with 1's.
But that's okay, since there is only one unity of multiplication and it is also commutative for multiplication.

#### solakis

##### Active member
Sure you can.
This is correct.
Note that you did leave out a couple steps with 1's.
But that's okay, since there is only one unity of multiplication and it is also commutative for multiplication.
Why ,then, all the books i have gone thru ,in the above proof or similar proofs,use parentheses??

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Why ,then, all the books i have gone thru ,in the above proof or similar proofs,use parentheses??
Heh.
Do you have references for that?

To be honest, I'd use one extra set of parentheses myself.
Not because they are mathematically necessary, but to make it more clear what I'm doing.