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0.999...=1 and why people believe it is false.


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Recently, I've had a couple arguments with people disputing the fact that 0.999...=1, a couple on the internet and a couple in real life.

 

 

 

It seems to me that this is one of the most commonly disputed mathematical facts, along with the Monty Hall problem, and it is also by far one of the simplest.

 

Even after repeated proofs I've sent to people, they still appear to intuitively distrust this.

 

 

 

I've come to the conclusion that most of these people are ignorant that 0.999.... indicates that the 9's go on forever, infinitely, and thus never reach an end point, and thus, in fact, do equal one.

 

 

 

 

 

To dissuade people from changing the topic and trying to argue that 0.999 !=1, I've included a couple of proofs.

 

 

 

[hide=]1.

 

 

 

1=9/9(fractional laws)

 

=9*(1/9)(because 9*1=9, thus 9/9)

 

=9*0.1111...(because 1/9=0.111....)

 

1=0.999.....(because 9*0.111...=0.999....)

 

 

 

2.

 

x=0.999...

 

10x=9.999...

 

10x-x=9.999...-0.9999...

 

so

 

9x=9

 

x=1

 

therefore 0.999....=1[/hide]

 

 

 

Can anyone think of an alternate explanation for this phenomen? Why would otherwise rational people, who can accept easily that 0.333....=(1/3) and that 0.333...*3=1 (by fractional definitions), be unable to accept this particular mathematical fact?

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I believe graphing y = 1/x+4 will give you an example of such. x =/= -4, so it is able to get to any number as close as possible, without ever actually touching the -4 line. Putting in -4 will divide by zero - impossible. But doing -3.99999... will not destroy the equation.

 

 

 

That was probably explained terribly. But I hope you got my message overall.

 

 

 

 

 

ALSO: .9999... / 3 = .3333....

 

1 / 3 = .3333... + .0000....1

 

 

 

The latter is impossible, so I assume for the sake of easyness we just round to .3333....

 

 

 

 

 

Or just use fractions.

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Most people I know dont think .999.... is equal to one because they dont get it. They just go "No .999 is .999 not 1"

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0.999... != 1, however, the LIMIT of 0.999... = 1. Anything involving infinity is a limit, not an actual number.

 

 

 

Yes, but the limit of a sequence is a fixed value, not an infinite process, so saying this is misleading.

 

 

 

I believe graphing y = 1/x+4 will give you an example of such. x =/= -4, so it is able to get to any number as close as possible, without ever actually touching the -4 line. Putting in -4 will divide by zero - impossible. But doing -3.99999... will not destroy the equation.

 

 

 

 

 

 

Yes, at any given stop, at any given stage of the expansion, for any given finite number of 9s, there will be a difference between 0.999...9 and 1. That is, if you do the subtraction, 1 0.999...9 will not equal zero. But the point of the "..." is that there is no end; 0.9999... is infinite.

 

 

 

1 / 3 = .3333... + .0000....1

 

 

 

This is just untrue; 1/3 is exactly equal to 0.333.....; if you include the 0.0000...1 you won't equal 1/3. This can be shown using infinite series.

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0.999... != 1, however, the LIMIT of 0.999... = 1. Anything involving infinity is a limit, not an actual number.
Yes, but the limit of a sequence is a fixed value, not an infinite process, so saying this is misleading.
The limit of a sequence represents a fixed value, and in this case that fixed value is equal to 1 - but the limit itself is not equal to 1.

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I like the second shown proof, and robert's method

 

 

 

this was the second thing I learned in pre calc, right after there are more numbers between 0 and 1 then there are numbers in the series 1, 2, 3, 4, 5, 6, 7, ....

 

 

 

edit--

 

 

 

another good way of looking at it is the fact that with an infinite amount of 9s the gap becomes infinitely small and is therefore 0

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The first proof is terrible. I'd reword it to connect the bits about 1/9 and 1/3 because right now they're meaningless and the last statement "so 0.9999...=1" seemingly comes from nowhere. In otherwords, it isn't proven with that proof. You just proved "3*0.333....=1" and "9*0.111...=9*(1/9)=(9*1)/9" not "0.9999...=1".

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wait, so does that mean 1.999... is equal to two?

 

I don't quite get this.

 

 

 

yes

 

 

 

to try to give the simplest explanation possible(look at proof 2 in the first post its neat)

 

 

 

.9 is .1 away from equaling 1

 

.99 is .01 away from eqauling 1

 

.999 is .001 away from equaling 1

 

 

 

so an infinite amount of 9s would have an infinite amount of zeroes before a 1; subtracting .999... from 1 will give a difference of an infinite amount of zeroes. 0.000...=0 so 1=.999...

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Well, 0.9999... Exists only in theory, correct? Therefore, it does equal one (using your logic, and evidence you have given) But in one sense it will always be less then 1. So yes and no.

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Well, 0.9999... Exists only in theory, correct? Therefore, it does equal one (using your logic, and evidence you have given) But in one sense it will always be less then 1. So yes and no.

 

 

 

sort of; since you cant actually write out an infinite amount of 9's any form of this you write out excluding the .999... as infinity notation will be less then one. With an infinite amount of 9's it does however=1

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The first proof is terrible. I'd reword it to connect the bits about 1/9 and 1/3 because right now they're meaningless and the last statement "so 0.9999...=1" seemingly comes from nowhere. In otherwords, it isn't proven with that proof. You just proved "3*0.333....=1" and "9*0.111...=9*(1/9)=(9*1)/9" not "0.9999...=1".

 

 

 

It's basically

 

1=9/9(fractional laws)

 

=9*(1/9)(because 9*1=9, thus 9/9)

 

=9*0.1111...(because 1/9=0.111....)

 

1=0.999.....(because 9*0.111...=0.999....)

 

 

 

and you're correct; I left out a couple of lines; I'll just replace the first one with this one #-o

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The first proof is terrible. I'd reword it to connect the bits about 1/9 and 1/3 because right now they're meaningless and the last statement "so 0.9999...=1" seemingly comes from nowhere. In otherwords, it isn't proven with that proof. You just proved "3*0.333....=1" and "9*0.111...=9*(1/9)=(9*1)/9" not "0.9999...=1".

 

 

 

It's basically

 

1=9/9(fractional laws)

 

=9*(1/9)(because 9*1=9, thus 9/9)

 

=9*0.1111...(because 1/9=0.111....)

 

1=0.999.....(because 9*0.111...=0.999....)

 

 

 

and you're correct; I left out a couple of lines; I'll just replace the first one with this one #-o

 

 

 

ah, that proof makes a lot more sense now, confused me the first time I read it.

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The proof is interesting, but I think it's just another way of looking at it. Infinity, essentially a nonreal concept cannot be compared to real numbers on our decimal system; it can easily be said that .999... is infinity increasing but never actually reaching 1. It's the same concept as approaching the event horizon of a black hole; to observers, it goes infinity slowly and will never reach the actual horizon.

 

 

 

Basically, yes, it is increasing - but by an infinitely small amount. Your proof represents another way of looking at it, but it isn't accurate to treat infinity as a number when doing comparisons. Essentially, 1/9 and .999... are numbers that cannot be compared in the standard decimal system, as they represent a concept instead of an actual value.

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I believe that 0.999... = 1.

 

 

 

My proof is that if you were to infinitely divide a number, you would get 0. However, if you look at in a different way, 1/(2^n), you would see that there must be a numerical value greater than 0, since everything (Excluding 0), divided by anything (Again, excluding 0), must have a value greater than 0.

 

 

 

So, would you not agree that 0.999...9 + 0.000...1 = 1?

 

 

 

Since 0.000...1 equals zero...

 

 

 

0.999... + 0 = 1

 

0.999... = 1

 

 

 

I doubt this isn't flawed though, but it makes sense to me.

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It's a simple matter of logic.

 

 

 

0.999... Will ALWAYS be 0.00...1 away from 1, no matter how many zeros you put in.

 

 

 

 

 

 

 

~ Captainkidd

 

But 1/3 = .3333...

 

 

 

And 1/3*3 equals 3/3 which equals 1

 

 

 

So therefore .333...*3 = 1.

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It's a simple matter of logic.

 

 

 

0.999... Will ALWAYS be 0.00...1 away from 1, no matter how many zeros you put in.

 

 

 

 

 

 

 

~ Captainkidd

 

 

 

yes, but with an infinite amount of nines the difference between 1 and .999... is a 1 an infinite amount of places to the right of the decimal. Since you cant go an infinite amount of places to the right you will only be able to read 0.00000000000.... forever; therefore, .999... is 1

 

 

 

edit--and what riku said.

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Since you cant go an infinite amount of places to the right you will only be able to read 0.00000000000.... forever; therefore, .999... is 1

 

 

 

1.00000000->

 

-0.99999999->

 

--------------

 

0.<-00000001

 

 

 

The infinity goes in the direction of the decimal point forever, but there is still always going to be that 1 on the far right.

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I never understood this concept, even as my last course being AP Calc. If I walk precisely 99.999...% am I touching the wall? I assumed that I would begin walking infinitely closer to the wall, without ever touching it. Assuming that based on your proof that 99.999... also equals a value equal to that of 100. I understand the proof, don't get me wrong, I don't understand the application.

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I'm not the best math student, but I make it easy for myself.

 

 

 

"Find me a number between .999... and 1!"

 

 

 

"Okay! .0000000000000000 . . . 0 . . . 0 . . . oh . . ."

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