0.99999~ = 1 ?

Hi Briterican,I'm with Catholic Scientist on the simple approach, I just look at it a slightly different way:(1/3)x3 == 1

1/3 = 0.3 + 0.1/3
= 0.3 + 0.03 + 0.001/3
= 0.3 + 0.03 + 0.003 + ... + 3x10^-n + 1x10^-n/3
= 0.33333333~ and there is always a remainder of 1 to divide by 3 wherever you stop.

Now multiply that by 3(1/3)x3 = (0.33333333~)x3 = 0.99999999~ == 1This is really a product of using base 10, not anything magic with the math.If we used base 9 then 1/3 would be 0.3: However 1/2 would be 0.44444~ My personal favorite repeating decimal fraction is 7ths: Now 1/7 + 6/7 = 0.999999~ == 1
and 2/7 + 5/7 = 0.999999~ == 1
and 3/7 + 4/7 = 0.999999~ == 1... now if we used base 7 ...Enjoy.

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Hi Son Goku,Yes, π is a fun number, no matter which base you use.Another one that shows up in a lot of natural systems is the golden ratio Which is the ratio of a rectangle where you make a square from the short side and the remaining rectangle is the same proportion. The greeks used this as a foundation in their architecture for "ideal" shapes.Page not found | Geophysical Institute

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Cross-section of nautilus shell showing the growth pattern of chambers governed by the golden ratio.

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It is also the ratio of the "diagonal" of a pentagram to the side ... (now you can graphically draw a perfect pentagram without ruler or calculator).We have a member on this forum that uses it as his ID.Enjoy.

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Hi Straggler First off, 0.999~ is a product of using base 10. I don't think you would have any difficulty with saying: Or with saying: Put them together and you have: 0.999... - Wikipedia

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One reason that infinite decimals are a necessary extension of finite decimals is to represent fractions. Using long division, a simple division of integers like 1⁄3 becomes a recurring decimal, 0.333, in which the digits repeat without end. This decimal yields a quick proof for 0.999 = 1. Multiplication of 3 times 3 produces 9 in each digit, so 3 0.333 equals 0.999. And 3 1⁄3 equals 1, so 0.999 = 1.[1]

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Of course, this also means that 6.999R ≡ 7.0 ... Not necessarily, as there are many instances where you can do Limit analysis where the result is and each limit on it's own is infinite, yet the result is a finite number.Enjoy.

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Hi Jon,I had some trouble with Son Goku's proof as well. As this seems to be begging the question. Sure (1+1)/2 = 2/2 = 1 but do we have that?Is the 0.999~ in (3) the same as the 0.999~ in (4)? Every time you do it out to the same number of decimals you get different numbers with different remainders, and the average number (4) is always between (3) and 1, where you would expect it to be.I would think that a stronger proof would be to subtract 0.999~ from 1, or 1 from 0.999~, and what you get is a string of 0's, no matter where you stop. Or the frame shift proof: Note that this later method works for any repeating decimal: Enjoy.Edited by RAZD, : added

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Hi Jon, Well, you are starting with a mathematical concept, rather than objective evidence, and math does not need to conform to reality, just be internally consistent.For instance try to imagine a precisely equivalent 2nd of anything. What we get is 1 + ~1 = ~2 at best, and try to ignore the ~ parts.Second, the 0.999~ repeating decimal is purely an artifact of using the decimal system, an intellectual concept in itself.Enjoy.

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My dear Rrhain, Yes, I most certainly did, because it is not proven to be the same in the original, and we are just asked to take it on faith. That is not how I do mathematical proofs. And just asserting it doesn't make it so.I'm well aware of the reality here, all I've pointed out is that the proof offered was incomplete.Enjoy.

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Hi Dr ACuriously you have just proven that you have not read my post in context.Enjoy.

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I haveEnjoy.

Hi lyx2no2, The proof was intended to show that 1 ≡ 0.999~ by assuming that it wasn't, and then showing that this results in a contradiction. In the process it uses another version of 0.999~ and the problem is that if one is not 1 then the other isn't either and it remains half way between. One can't use the conclusion as part of the proof eh?A much simpler process is take 0.999~, multiply it by two (=1.999~), where the 9's are exactly aligned from the decimal, and subtract the original (≡ 1), QEDEnjoy

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Hi cavediver, it may just be a mathematical semantic thing. This also holds for all the frame shift proofs as well.I'm aware that functionally they are similar, I just find them conceptually different, as one seems (unnecessarily) more manipulated than the other, and it seems that the main argument is about how you phrase the problem, rather than the actual solutions.My dad makes a comment about mathematicians understanding the problems very well, but not understanding how to communicate their understanding to those who do not understand maths, so I try for the simplest paths possible.It's quite interesting to see the variety of ways people have gone about this, and I'm rather astounded that this thread has persisted so long.It might be interesting to see how people think which explanations are the ones that best convince them -- Huntard and Straggler and any other lurkers who had trouble with this issue.enjoy.

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