Date: Sep 7, 2017 10:38 PM
Author: Dan Christensen
Subject: Re: 0 = 1
On Thursday, September 7, 2017 at 9:40:13 PM UTC-4, conway wrote:
> You are starting to play fair. I AGREE that this idea needs a LOT of work. The fact that you are still here....suggest you know at some level "something" might here. Help me! Also I agree my logic can be "garbled". I am a philosopher. Not a mathematician. Mathematics...(indeed all the sciences) were born in philosophy. This idea makes sense. I think you see that. Until now you have refused to take it serious however because you "believe" that I have not done my homework (I have, just not the kind you like). Give me a serious....honest chance to explain this idea to you.
> N = set
Do you mean N is an arbitrary set?
> A = ANY number in the set
So A is an element of N that is also a number? Might there be non-numbers in N?
> z1 = quantity of value (we can work on a more "formal" definition)
> z2 = quantity of space (we can work on a more "formal" definition)
> A = (z1+z2)
So z1 and z2 are simply any pair of numbers such that A = z1+z2?
If A=2 then could we have z1 = 5 and z2 = -3?
> ANY binary expression of multiplication
> z1 x z2
> z2 x z1
> 3 x 2
> z1 for 3 = 3
> z2 for 3 = 3
Doesn't z1+z2 have to be 3?
> z1 for 2 = 2
> z2 for 2 = 2
> 3(as z1) x 2(as z2)
> 3(as z2) x 2(as z1)...
> z1 for 1 = 1
> z2 for 1 = 1
> z1 for 0 = 0
> z2 for 0 = 1
Makes no sense at all. Examples will not do in this context. We need unambiguous definitions. Maybe something of the form: For all x in R, there exists (unique ?) z1 and z2 in R such that _______________ (Fill in the blank)
But even if you manage to do this, you must show how this relationship somehow determines the real value of x where 0x=1. It's just too bizarre. Every elementary school graduate knows that 0x is always 0 for ANY real number x. There can be no doubt about this. I really can't see where you are going with this. Are you deliberately trying to create confusion and frustration among students?
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