0.9999....(recurring) = 1?

SMBoy · 2012-11-23 02:51:25

A = ...999

B = ...001

A + B = 1

If A was to Equal 1 on it's own then... A + B = Would Equal 1.1

Maelwys · 2012-11-23 02:56:34

SMBoy wrote:

A = ...999
B = ...001
A + B = 1
If A was to Equal 1 on it's own then... A + B = Would Equal 1.1

First off "A + B = Would Equal 1.1" is blatantly false. A + B = 1. ...001, by your notation.

Secondly, that assumes that there is a "last digit". In your notation, the 9's clearly stop. There is a "last nine" in A, and in the same spot in B there is a corresponding 1. The problem is that this contradicts with the very definition of "infinity". If there are an infinite number of 9s, there can be no "last nine". That means that there is no corresponding location in B to put the 1, so B just becomes an infinite string of 0s. And 0.000.... (with an infinite number of 0s) = 0. So if B is 0, and A + B = 1, then A = 1.

Calligar · 2012-11-23 13:21:43

Okay, I don't even know where to begin arguing with this...

anonimnystefy wrote:

No, 3 is allowed, 0.(9) isn't. There cannot be recurring 9's after the decimal point.

So I know more specifically about what your talking about, I'd like to see your response to...

bob bundy wrote:

Why not?

Stefy, supposing what you say is true, then what is the point of this argument in the first place? Why are there hyper real numbers even? You completely invalidate ... everything this controversy has to do with.

Here's a case where this could come in. 1/3 + 2/3 = 1. In decimal form, this can be argued to be 0.¯9 (not saying this is 1/3 and 2/3s tho). The issue directly has to do with decimal, because it is in base 10. This is impossible to accurately represent this in decimals in base 10 because the closest you can get is 0.¯3 and 0.¯6. Just simply looking at 0.¯3, and 0.¯6, what does 0.¯3 + 0.¯6 equal (only look at it like that is now, do not assume this is 1/3 and 2/3s, as I'm trying to represent how people get to this)? Even though it is repeated infinitely, it is irrelevant, because in the "end," it still never equals 1. Now to clear up any confusion, I am not saying this is equal to 1/3 + 2/3s, otherwise it would be 1. I'm only trying to represent (if you are able to look at it from a different mindset and not assume its 1/3 and 2/3s), a way to get to 0.¯9. In other words, 0.¯3 and 0.¯6 are not a fraction of 1, they are just numbers on their own. Now, I still don't know exactly what you are talking about, so I still would like to see what you say to to, "why not?"

Maelwys wrote:

SMBoy wrote:
A = ...999
B = ...001
A + B = 1
If A was to Equal 1 on it's own then... A + B = Would Equal 1.1
First off "A + B = Would Equal 1.1" is blatantly false. A + B = 1. ...001, by your notation.
Secondly, that assumes that there is a "last digit". In your notation, the 9's clearly stop. There is a "last nine" in A, and in the same spot in B there is a corresponding 1. The problem is that this contradicts with the very definition of "infinity". If there are an infinite number of 9s, there can be no "last nine". That means that there is no corresponding location in B to put the 1, so B just becomes an infinite string of 0s. And 0.000.... (with an infinite number of 0s) = 0. So if B is 0, and A + B = 1, then A = 1.

I think what SMBoy is saying Maelwys is that if you add that to 0.9, or 0.99, or 0.999, regardless how many 9s there are, it will equal 1. But if you add the same thing to 1, you will get 1.1, or 1.01, or 1.001, etc. More showing how the 2 answers are different and not equal in that sense.

Last edited by Calligar (2012-11-23 20:29:23)

SMBoy · 2012-11-23 22:51:16

"I think what SMBoy is saying Maelwys is that if you add that to 0.9, or 0.99, or 0.999, regardless how many 9s there are, it will equal 1. But if you add the same thing to 1, you will get 1.1, or 1.01, or 1.001, etc. More showing how the 2 answers are different and not equal in that sense."

Another way of looking at & Solving the Problem is 0.999... = (The Start of the Universe) and 1 = (The End of the Universe ? )

Calligar · 2012-11-23 23:10:33

Yes, well I have a lot of experience with this too. It is probably my most argued thing to do with math, maybe the most argued thing in math in general. I actually haven't argued it that heavily on this forum, but I have argued this with people so much and amazingly am not tired of it yet. I want to argue this further because I feel that 0.¯9 ≠ 1, however, proving this is the issue. You can argue anything, but there is always a counter argument. Any proof of it can be argued from a different angle. The same can arguably be said true for those who argue against the idea that 0.¯9 = 1. Though I'm sorry, I can never ... personally accept any number equaling a different number, as it defies its very definition. Though as I said earlier, I will not continue to argue it either, as for that reason, it is futile to go forward. However, I do not mind following it either, as maybe sometime, they might eventually change this in mathematics finally (one can always hope...).

I should note though, I don't quite understand the start of the universe, end of the universe thing, mind explaining it?

SMBoy · 2012-11-24 00:06:13

Quote " I should note though, I don't quite understand the start of the universe, end of the universe thing, mind explaining it?'

If ...999 was to equal 1 then the .9's would End? But as we Know Reach the End of the Universe ? & Find a Wall ? ...

Then There must be Something behind the Wall !! Example = .........999999999 [1]Wall] .........999999999

Calligar · 2012-11-24 00:21:15

Oh, I think I might see, because 0.¯9 never ends, but 1 ends, is that correct?

Math Is Fun Forum

#1076 2012-11-23 02:51:25

Re: 0.9999....(recurring) = 1?

#1077 2012-11-23 02:56:34

Re: 0.9999....(recurring) = 1?

#1078 2012-11-23 13:21:43

Re: 0.9999....(recurring) = 1?

#1079 2012-11-23 22:51:16

Re: 0.9999....(recurring) = 1?

#1080 2012-11-23 23:10:33

Re: 0.9999....(recurring) = 1?

#1081 2012-11-24 00:06:13

Re: 0.9999....(recurring) = 1?

#1082 2012-11-24 00:21:15

Re: 0.9999....(recurring) = 1?

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