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You are not logged in. #926 20080425 08:50:31
Re: 0.9999....(recurring) = 1?
George, if you were truly interested, you would look up a term like "ZFC" which I have mentioned countless times. And if were were to spend 5 seconds reading the wikipedia page on ZFC, you would know that ZFC is that "basis" which you asked about.
This is how mathematics was constructed (yes  all mathematics) in the 1920's and it has been so successful that alternate axiom systems have pretty much been left in the dust. There are probably better axiom sets, but no one knows of them, at least not yet. We go with what we have, and a PhD thesis can be made showing a better way. But no one has done that. All you want to do is sit and moan that you don't like it.
That is so incredibly wrong, and I'm fairly certain Einstein just rolled over in his grave. If you really think intuition is the best way to go, the only thing I can tell you is that you haven't always studied mathematics. No matter how much mathematics I study, intuition will always tell me that: is not a real number.
You're quote mining, George. It is yet another thing which shows your not truly interested in learning or debate, but rather just arguing and getting people to agree. The link says:
It is that no physical quantity can be infinite, and this completely agrees with mathematics and the real numbers. There are an infinite amount of real numbers, but every real number is of finite value. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #927 20080501 10:49:13
Re: 0.9999....(recurring) = 1?first we should know that 0.999...... is not a number. [math0.99\1\frac{1}{10^2}math] so Let when we say that 0.999......=1,we just mean that Here is a intitutive proof: we can see that as n goes bigger and bigger,0.999...... goes more and more close to 1,but it's not accurate in mathematics.I will give a precise proof. obviously 0.999......<1,if 0.999...=A where A between 0 and 1,then we can easily find some N such that a contridition derived. #928 20080501 12:02:14
Re: 0.9999....(recurring) = 1?
No, it's not obvious. In fact, it is not even true. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #929 20080502 01:29:55
Re: 0.9999....(recurring) = 1?moxiuming, please fully understand the concept of infinity...... Last edited by Dragonshade (20080502 01:33:47) #930 20080502 03:13:59
Re: 0.9999....(recurring) = 1?
It's true that if you could find some number x such that 0.999... < x < 1, then 0.999... would undeniably have to be less than 1. Why did the vector cross the road? It wanted to be normal. #931 20080813 07:07:23
Re: 0.9999....(recurring) = 1?hmm this is a vey interesting subject, indead! π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930... #932 20080813 11:52:15
Re: 0.9999....(recurring) = 1?
um.. no? Multiplying both dies of 1/3=0.333 gives you 1 = 0.999..." which says they are equal. It does not say they are unequal. So how would that suggest they are unequal? A logarithm is just a misspelled algorithm. #933 20080814 05:52:46
Re: 0.9999....(recurring) = 1?Hmm yeah... I guess I was thinking about the problems where you have to determine if an equation is true such as 2x1=2, x=1 which would be false... I am just wondering how 0.999... can be equal to 1 if they are not the same, because in theory 0.999... would get infinitely closer to 1 but would never actually reach 1. Such as in Physics when someone touches someone else, in theory we are really never touching them because the atoms electron clouds would prevent the nucleui from thouching each other, but we do get an electrical signal to the brain signifying touch... Another example is when an oblect moves halfway to another object then halfway again, then halfway again to infintity but the object can in theory always move another half closer to the other object... 0.25(0.5)=0.125; 0.125(0.5)=0.0625 then it would move on to infinity... I guess in theory 0.999... is not equal to 1, but in reality we set it equal to 1 to make computing easier, just like in theory we never actually touch each other but in reality we feel as if we do because of what our brain is telling us... Last edited by MooseofDoom (20080814 05:56:40) π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930... #934 20080814 06:56:14
Re: 0.9999....(recurring) = 1?You're thinking of 0.999... as if it started as 0.9, then it changed to 0.99, then 0.999, and so on, continually growing more nines and getting closer and closer to 1. Why did the vector cross the road? It wanted to be normal. #935 20080814 12:39:27
Re: 0.9999....(recurring) = 1?
This is incorrect. You are trying to use it as an example of things that get "infinitely" close, but it is not. The two nuclei always have a positive, finite, distance apart.
As mathsyperson said, this is not what happens with 0.999... Nothing is "moving", and nothing is "changing". It just... is.
No. 0.999... = 1 in theory. Two real numbers are not equal when they are a finite distance apart. This is actually, in part, how they are defined. 0.999... is not a finite distance from 1, and thus, it equals 1. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #936 20080815 13:34:07
Re: 0.9999....(recurring) = 1?1 = 0.9 + 0.1 Last edited by John E. Franklin (20080815 14:00:57) igloo myrtilles fourmis #937 20080815 23:23:26
Re: 0.9999....(recurring) = 1?Pretty much. The 1 appears at the end of the zeroes, but in the infinite case the zeroes don't have an end and so the 1 can't be there. Why did the vector cross the road? It wanted to be normal. #938 20080816 07:20:54
Re: 0.9999....(recurring) = 1?infinity is a paradox, try to explain it and u get more paradoxes! #939 20080816 14:22:15
Re: 0.9999....(recurring) = 1?Wow this is a crazy concept to grasp... π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930... #940 20080816 16:46:02
Re: 0.9999....(recurring) = 1?Just don't push the button if it ain't quite one, cause then your dumbweed brain has contributed enough to mankind. I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy... #941 20080816 18:40:10
Re: 0.9999....(recurring) = 1?
To add to that, if you were to ever put a 1 at any point in the chain of zeroes, you would thereby be ending the chain of zeroes, and so the number of zeroes would be finite; so any argument along the lines of 'an infinite number of zeroes, then a one' makes no sence. The Beginning Of All Things To End. The End Of All Things To Come. #942 20080816 18:57:14
Re: 0.9999....(recurring) = 1?Add to that That Is Correct! I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy... #943 20080817 09:36:34
Re: 0.9999....(recurring) = 1?Don't forget that you can have igloo myrtilles fourmis #944 20080817 19:14:20
Re: 0.9999....(recurring) = 1?that is a terrible argument that doesn't describe the same thing, to say 7water7 with 7 being the banks, you would have to instead say that the river is infinitely wide not infinitely long. which ofcourse is not possible. Last edited by lucadeltodesco (20080817 19:14:38) The Beginning Of All Things To End. The End Of All Things To Come. #945 20081027 15:33:11
Re: 0.9999....(recurring) = 1?Hi guys. I just submitted one article to philosophy of science. Basically I argue arbitary fraction cannot exist. X'(yXβ)=0 #946 20081027 20:02:17
Re: 0.9999....(recurring) = 1?What is the title of the journal? And do let us know if it gets published. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #947 20081213 12:19:44
Re: 0.9999....(recurring) = 1?Hi. New member here. I have been discussing this problem with some students of mine and have reached a bit of a wall. I have a way of thinking about this and I hope someone can point me in right direction. #948 20081213 12:57:45
Re: 0.9999....(recurring) = 1?Sorry I don't have time to answer you in full. But I can mention these points: "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #949 20081215 19:08:09
Re: 0.9999....(recurring) = 1?
If you keep taking math classes, you will eventually deal with series and convergence in a rigorous setting. If you have taken calculus classes, you may have already seen some theorems about series, though probably not proved them. In any case, there is a theorem:
Now "rearrangement" is rigorously defined, but I am attempting to avoid getting lost in the definitions, and it pretty much means exactly what you did in your post. In other words, both ways you summed the terms converge to the same number. In essence, you proved that 8.9999 = 9.
The existence of this function with special properties (if you're not familiar, either ask or wikipedia) can only occur when two sets of finite size have the same number of elements. And having this map will mean that the sets have the same number of elements. We say that these two statements are "equivalent", and what we mean is that they represent the same idea.
Now to show you that "infinity" = "infinity + 1", let's take the natural numbers and find a 11 map that is also onto and maps into the natural numbers as well as the number 0. It shouldn't be too hard to prove that this map is both 11 and onto. Just ask if you want to see it. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #950 20081227 16:17:47
Re: 0.9999....(recurring) = 1?mrphysics, i used the similar method to locate the last 9 in 0.999... X'(yXβ)=0 