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October 11th, 2017, 03:10 PM   #11
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Quote:
 Originally Posted by zylo Axioms for rational numbers: (BM for example) Q(Z) of Z consists of all ordered pairs (a,b) with a,b $\displaystyle \epsilon$ Z and b $\displaystyle \neq 0$, and std definitions of equality, multiplication, and division of rational numbers. What is the point of a sloppy, confusing, paraphrase?
Zylo

Your post is clearly "flaming" and against the rules. If you really wish to discuss this with me, you may begin by leaving polite replies.

October 11th, 2017, 03:11 PM   #12
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Quote:
 Originally Posted by Denis C'mon Zylo...dontcha know that 4*0 = 0+0+0+0 ?
Denis

Your post is clearly "flaming" and against the rules. If you really wish to discuss this with me, you may begin by leaving polite replies

October 11th, 2017, 05:58 PM   #13
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Quote:
 Originally Posted by Conway51 Denis Your post is clearly "flaming" and against the rules. If you really wish to discuss this with me, you may begin by leaving polite replies
I've scoured the site and can find no reference to these rules you mention.

October 11th, 2017, 06:29 PM   #14
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Quote:
 Originally Posted by romsek I've scoured the site and can find no reference to these rules you mention. Please provide a link to said rules so we all can read them.
Romsek

Lol...is that so...perhaps I should have clarified...it is against my personal rules. Should you wish to discuss the idea as opposed to me.... you may begin by posting a polite reply regarding the idea ...

October 11th, 2017, 06:31 PM   #15
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Quote:
 Originally Posted by Conway51 Romsek Lol...is that so...perhaps I should have clarified...it is against my personal rules. Should you wish to discuss the idea as opposed to me.... you may begin by posting a polite reply regarding the idea ...
If 0 times 1 is sometimes 1 and sometimes 0, how can you base a rational system of computation on that?

October 11th, 2017, 06:36 PM   #16
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Quote:
 Originally Posted by Maschke If 0 times 1 is sometimes 1 and sometimes 0, how can you base a rational system of computation on that?
It is derived from the table....if 1 is paired with a zero that 1 is 0
If a 1 is paired with a 1 that 1 is 1

An alternative idea is to consider
Z1=multiplicative property of 0
Z2=multiplicative identity property of 1

In this way no table is necessary. Allowing that only one property at a time may be used for zero ...either z1 or z2

Thank you

I think I may have misunderstood your last question. If I can show that there is a unique solution for each "product" of the expression ( A x 0 )...then rationality still exists

Last edited by Conway51; October 11th, 2017 at 07:09 PM.

 October 12th, 2017, 07:48 AM #17 Senior Member   Joined: Sep 2016 From: USA Posts: 416 Thanks: 230 Math Focus: Dynamical systems, analytic function theory, numerics Once again Conway the problem is that if you are able to define any method for inverting 0 then you are not working in a field. This is a fact which is unavoidable. I still can't follow what you are describing for multiplication by 0 but I would bet the rent that your multiplication and addition operations are one of the following: 1. not well defined 2. not commutative 3. not distributive These 3 properties are far more important than being able to divide by 0. If you still don't understand, please specify the following so I can show you an example which violates 1,2, or 3. Using your system please fill in the following values: $1 \times 2 = ?$ $2 \times 1 = ?$ $1 \times 0 = ?$ $0 \times 1 = ?$ $0 \times 2 = ?$ $2 \times 0 = ?$ Thanks from Conway51
October 12th, 2017, 10:17 AM   #18
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Quote:
 Originally Posted by SDK Once again Conway the problem is that if you are able to define any method for inverting 0 then you are not working in a field. This is a fact which is unavoidable. I still can't follow what you are describing for multiplication by 0 but I would bet the rent that your multiplication and addition operations are one of the following: 1. not well defined 2. not commutative 3. not distributive These 3 properties are far more important than being able to divide by 0. If you still don't understand, please specify the following so I can show you an example which violates 1,2, or 3. Using your system please fill in the following values: $1 \times 2 = ?$ $2 \times 1 = ?$ $1 \times 0 = ?$ $0 \times 1 = ?$ $0 \times 2 = ?$ $2 \times 0 = ?$

SDK

Perhaps because you are still having trouble understanding me, is the reason why you do not understand yet why this is valid in a field. I can guarantee you that commutative and distributive properties exist with out change. In fact....

Field Axioms -- from Wolfram MathWorld

I have used this link many times........

I shall fill in the values as you have requested...perhaps then we will begin to communicate better...

1 x 2 = 2
2 x 1 = 2
1 x 0 = 0 (if and only if 0 is z1)
0 x 1 = 0 (if and only if 0 is z1)
0 x 2 = 0 (if and only if 0 is z1)
2 x 0 = 0 (if and only if 0 is z1

now...

1 x 0 = 1 (if and only if 0 is z2)
0 x 1 = 1 (if and only if 0 is z2)
2 x 0 = 2 (if and only if 0 is z2)
0 x 2 = 2 (if and only if 0 is z1)

I truly hope we are making progress in our communication....

I still insist that your issues is with multiplicative inverses. You seemed very concerned about division and multiplication being "inverse" properties...."except by zero".........I can show that they are by "zero"...should you wish we can start doing the math with fractions.....

Now I ask you to do some homework...as you have asked me...

use the original table...follow the steps....perform the following expressions...

A x B =
where A and B =/=0
A x B =
where A and B = 0
A x B =
where only A or B = 0

Thank you

October 12th, 2017, 11:41 AM   #19
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Quote:
 Originally Posted by Conway51 SDK Perhaps because you are still having trouble understanding me, is the reason why you do not understand yet why this is valid in a field. I can guarantee you that commutative and distributive properties exist with out change. In fact....
Yet its a FACT that every element in a field except for 0 is invertible. The proof of this fact is trivial. Suppose you have a ring where 0 is invertible and let $x = 0^{-1}$, then $x\cdot 0 =1$ by definition. Also, $x \cdot 1 = x$ since 1 is the multiplicative identity. Now we have
$1 = x\cdot 0 = x\cdot (1-1) = x - x = 0$
which I hope you agree is a problem.

Quote:
 0 x 1 = 0 (if and only if 0 is z1)
now...
Quote:
 0 x 1 = 1 (if and only if 0 is z2)
This says your multiplication is not well defined.

Quote:
 I still insist that your issues is with multiplicative inverses. You seemed very concerned about division and multiplication being "inverse" properties...."except by zero"
Because in a field every element except 0 is invertible. This is essentially the definition of a field. As shown above 0 can't be invertible and this is true for any commutative ring with identity, not just fields.

October 12th, 2017, 02:03 PM   #20
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Quote:
 Originally Posted by SDK Yet its a FACT that every element in a field except for 0 is invertible. The proof of this fact is trivial. Suppose you have a ring where 0 is invertible and let $x = 0^{-1}$, then $x\cdot 0 =1$ by definition. Also, $x \cdot 1 = x$ since 1 is the multiplicative identity. Now we have $1 = x\cdot 0 = x\cdot (1-1) = x - x = 0$ which I hope you agree is a problem. now... This says your multiplication is not well defined. Because in a field every element except 0 is invertible. This is essentially the definition of a field. As shown above 0 can't be invertible and this is true for any commutative ring with identity, not just fields.

SDK

0 raised to any negative number is undefined. Therefore (x) in your equation is equal to "undefined". Therefore your equations really are non-sense.

https://math.stackexchange.com/quest...equal-infinity
http://mathforum.org/library/drmath/view/55594.html

However I do see what you were trying to explain to me (I think). What I am trying to explain to you is this...

(x * 0 = 0 ) if and only if 0 is "chosen" as z1

You chose to use zero as z2 in your equation. Therefore your "previous" equation would yield a false sum.....hence your argument....but it would only do so if I chose 0 as z2 otherwise (A * 0 =/= A) BY DEFENITION....

however...you may "chose" to use zero as z2

(x * 0 = x )

so...

if I use 0 as z1 then...

x * 0 = 0

therefore

0 = x * 0 = x(1-1)= x-x = 0

You seem to be taking issue that I can "chose" the product of the expression ( A * 0 ) to fit my arguments....I can...

You claim that this is what makes it "ill defined" and inconsistent. If... however I can show a unique solution for both "products" then there is NO inconsistency. Further the unique solutions for both "products" is exactly what makes these definitions "well" defined.

Thank you very much for your sincere replies.

Last edited by Conway51; October 12th, 2017 at 02:15 PM.

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