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## Error message

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## Latest Messages

Dec 7
\textbf{Proposition:} All connected space is metrically convex.

Dec 7
\textbf{Proposition:} All connected space is metrically convex.

Dec 7
\textbf{Proposition:} All connected space is metrically convex.

Dec 7
\textbf{Proposition:} All connected space is metrically convex.

Dec 7
\textbf{Proposition:} All connected space is metrically convex.

Dec 7
\textbf{Proposition:} All connected space is metrically convex.

Dec 7
Jussi: it was the old classic, "missing a minus sign". Fixed now! Thanks kindly! Joe

Dec 7
If 0.999... \ne 1 then what is the value of $(1 - 0.999...)$?

Dec 7
Due to Cantor’s diagonal argument(!) you cannot actually construct that first table of "decimal expansions" of \textbf{all} numbers that contain $d_{nn}$ for arbitarily large nn. It would be better to take a countable list of decimals between 0 and 1, taking the terminating version where possible. For example 0.5 and 0.4999... are the same number - we would choose 0.5. Taking the first number in our list, if the first decimal is a '3' we will write out a two (if the first decimal is not a '3' we shall write out a three). So our number will differ from the first number in our list. We repeat the same action with the second decimal in our second number in the list. This will differ from the first number (in the first decimal) and the second number (in the second decimal). Repeat for each number in the list. If the list is finite, all numbers will have been exhausted but our diagonal number will differ from all of them. Should you wish to keep adding numbers to the list, the diagonalization process will continue to generate a number which is not on the list. Conclusion: the real numbers cannot be countable.

Dec 3
In the following publication, I have systematically debunked all the so-called "proofs" that 0.999... is the same as 1: https://www.filesanywhere.com/fs/v.aspx?v=8b6b6a8a596275a7a7a9 As for thiago's question, there is a problem with the 10x "proof". For starters, it was never a valid proof. Example: x = 0.999... 10x = 10(0.999...) 9x = 9(0.999...) x = 0.999... If you input anything into the algorithm, you should get the same output. Now, only ignorant academics (and there are many on planet Math!), would do something as stupid as input a non-number (0.999...) into the algorithm. Actually they are capable of doing a lot more stupid things! A little closer examination will reveal that in their false proof, they have already assumed that x=1. How so? Well, add 9x to the LHS of first line and 9 to the RHS: 10x = 9.999... Observe that 9x=9 implies that x=1, long before you arrive at the so-called conclusion. However, modern academics who are not much smarter than baboons, would easily miss this detail. Of there is no valid construction of the real numbers and 0.999... or any other non-terminating representation is the result of a dysfunctional mind. I have no doubt the pathetic cowards who run this site will delete my comments before anyone can learn something worthwhile.

Nov 30
I'll look into that shortly, thanks for the alert! Joe

Nov 29
Ok.

Nov 28
The reference for my paper is: The parbelos, a parabolic analog of the arbelos, Amer. Math. Monthly 120 (2013) 929-935.

Nov 26
Hi admins, making a new entry gives 100 points, deleting it also gives 100 points!! I have lately deleted one of my entries (getting 100 p) and made two test entries and deleted them (thus getting 400 p). Please correct this system bug and subtract 500 points from me! Jussi