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#55
Apr704, 09:18 AM

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What are you trying to prove there anyway, now you've edited it? cos looking at it you can't really tell.



#56
Apr704, 09:21 AM

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More then thet, any element, that can be represented by infinitely many elements, its exact place in the real line is unknown, for example: 3/9 place is wellknown 3/10 place is unknown. 


#57
Apr704, 09:26 AM

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Why is 1/3's place known? How do you know where 1 is? Or zero? The real numbers aren't actually physically a line, Organic. You are confusing the representation of something with the something... Oh, no, you're going to talk about x and model(x) again aren't you?
Actually the statement above is trivially true because it is of the form A=>B whre A is false.... 


#58
Apr704, 09:27 AM

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Therefore ab = d < e, but both d and e > 0. 


#59
Apr704, 09:28 AM

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Do you know what a number line is? It is not something physical... 


#60
Apr704, 09:31 AM

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but that isn't deducible from your hypothesis: just because ab<e does not state that a and b are both less than e. (take e=1 a=b=100,000,000). so it's a further pointless assumption. Try writing out the statement of the lemma again, and its proof making sure all the hypotheses are written correctly and that it is not vacuous (which it was first time) and seeing as the statement was for all e, then you've just shown a=b=0 


#61
Apr704, 09:33 AM

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#62
Apr704, 09:34 AM

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good, then the model of the real numbers that is in mathematics is cauchy sequences, and 0.9999...=1 in that model, and as the discussion started about that model that's the end of the story.



#63
Apr704, 09:46 AM

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When we writing ab < e we mean that d < e. 


#64
Apr704, 09:49 AM

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#65
Apr704, 09:52 AM

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let e=d>2, then you have d/2>d and d>0. now think for a second. d/2>d => d>2d => 0>d ,yet d>0. Want to rethink that at all. 


#66
Apr704, 09:57 AM

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#67
Apr704, 09:59 AM

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breaking, infinitely many, finite, that's you wishing something to be true that isn't, you are thinking unmathematically (perhaps intuiitive and physically in your opinion, but that isnt' mathematics). 


#68
Apr704, 10:04 AM

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Here is your initial post: Suppose a and b are different real numbers and for any e>0 we know ab<e then a not= b. No restriction on e>d. at all and it says for any e>0 doesn't it? Once more you move the goal posts half way through your argument when someone points out where it's gone wrong. So want to start from the beginning and clearly write out what it is you are trying to prove again? Because so far you're not doing very well. I mean what was the point of it anyway? 


#69
Apr704, 10:10 AM

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Math language is only a rigorous agreement between people, no more no less. I have found that the current agreement includes lot of weak point in it, where one of them is the infinity concept. 


#70
Apr704, 10:17 AM

P: 1,210

Suppose a and b are distinct real numbers and for any e>0 we know ab<e then a not= b.
If a and b are distinct real numbers then for any e > ab = d > 0. proof: If a is not b then ab>0. Let d be the difference. Let e = d/2 then ab=d or ab = d/2 > 0, hence d > 0 and also ab/2 > 0. therefore nonzero/2 > 0. 


#71
Apr704, 10:17 AM

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No, the idea of infinity is well understood, but apparently not by you. Neither, it seems, is the idea of axioms and definition.
All of the 'problems' you've come across have been because of your own refusal to accept the definitions that are there. (Cantor, Natural numbers, axiom of infinity, convergence, real numbers). There are some deep and troubling issues in mathematics that we don't understand and have to live with. They cause no practical problems. Your findings aren't these, though. If you want to say things like 'there is no objective thing like maths' at least take the time to learn some of it, you might, well, learn something. 


#72
Apr704, 10:18 AM

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