Does 2.9999999~9 = 3

[LASTGUY2GETPS2]

Originally posted by: Taejin
Here's a proof.

x = .999999~9
10x = 9.9999999~9

10x - x = 9.99999~9 - .99999~9
9x = 9
x = 1.

So, 2 + .9999~9 = 2 + 1 = 3.

I think this is actually kind of a paradox. 2.9999~9 does not *actually* equal 3, since it is approaching 3, but using simple math you can argue it does equal 3. I'm not EXACTLY sure, but I think this is just a flaw within the decimal number system humanity uses. It's been too long ago since I read this in a book (I was browsing) so I can't recall.

This is also an interesting POV of the same problem.

1/9 = .1111~1
2/9 = .2222~2
....3/9 4/9 5/9 etc.
8/9 = .8888~8
9/9 = .9999~9

Except we also know that 9/9 = 1

beat me to it

 

[sharkeeper]

I think this is actually kind of a paradox. 2.9999~9 does not *actually* equal 3, since it is approaching 3, but using simple math you can argue it does equal 3. I'm not EXACTLY sure, but I think this is just a flaw within the decimal number system humanity uses. It's been too long ago since I read this in a book (I was browsing) so I can't recall.

It's asymptotic.

Cheers!

 

[91TTZ]

Originally posted by: Taejin
Here's a proof.

x = .999999~9
10x = 9.9999999~9

I'm not EXACTLY sure, but I think this is just a flaw within the decimal number system humanity uses.

No, it's just a rounding error. The fact that it's called 2.99999whatever tells you that it is ever so slightly less than three. If it was equal to three, they would call it three.

 

[CrackRabbit]

:beer: = 0.:wine::wine::wine::wine::wine::wine::wine::wine::wine:... ?

 

[spidey07]

Originally posted by: 91TTZ

Originally posted by: Taejin
Here's a proof.

x = .999999~9
10x = 9.9999999~9

I'm not EXACTLY sure, but I think this is just a flaw within the decimal number system humanity uses.

No, it's just a rounding error. The fact that it's called 2.99999whatever tells you that it is ever so slightly less than three. If it was equal to three, they would call it three.

and they do. 2.999... = 3.

 

[91TTZ]

Originally posted by: spidey07

and they do. 2.999... = 3.

They don't. The ones saying that 2.999 = 3 are incorrect. They are making a rounding error. Everything thereafter builds upon this rounding error.

I can make a formula with the premise that 2+2= 7, but that doesn't make anything else that follows correct.

 

[sharkeeper]

No, it's just a rounding error. The fact that it's called 2.99999whatever tells you that it is ever so slightly less than three. If it was equal to three, they would call it three.

Depends on your requirements of precision. Some people use 3 for pi, for example.

Cheers!

 

[91TTZ]

Originally posted by: sharkeeper


Depends on your requirements of precision. Some people use 3 for pi, for example.

When you do math, one of the fundamental things to remember is to make sure that rounding errors do not ruin the desired results. You need to decide how many significant figures you need. If you were trying to get an answer that comes within 10, it would be perfectly acceptable to use precision that delivers an answer that's within 1 or 2. But when you need to be accurate to within .1, it isn't acceptable to do the math in a way that only gives you an answer within 1 or 2.

In the case of the OP's problem, to prove or disprove that equation you'd need infinite precision to do the problem using math. But you can do it using logic, and state that since 2.9999.. is a different number than 3, that therefore 2.9999.. does not equal 3.

 

[spidey07]

Originally posted by: 91TTZ

Originally posted by: spidey07

and they do. 2.999... = 3.

They don't. The ones saying that 2.999 = 3 are incorrect. They are making a rounding error. Everything thereafter builds upon this rounding error.

I can make a formula with the premise that 2+2= 7, but that doesn't make anything else that follows correct.

alright, you're on. show me 2+2=7

2.999... does indeed equal 3, it isn't rounding. They are equal. They are one and the same.

 

[91TTZ]

Originally posted by: spidey07
[
alright, you're on. show me 2+2=7

2.999... does indeed equal 3, it isn't rounding. They are equal. They are one and the same.

2+2 doesn't equal 7. I'm illustrating that you cannot use an invalid premise as the basis for a mathematical problem.

2.9999.... does not equal 3.

2.999.. signifies that there are an infinite amount of 9's following the decimal point. This means that it is less than three by an infinitely small margin.

 

[ohtwell]

No! Now go away!


: ) Amanda

 

[User1001]

Originally posted by: virtualgames0

Originally posted by: Supa
Does 0.000000 .... 00001 = 0 ?


----

Are you dumb?
0.0000.. forever wouldn't let you have a 1 at the end, because there IS NO END.

I think he meant an infintesimal.

 

[91TTZ]

I think in the context that this is written, this is more of a logic problem than a math problem.

.9999...9 is as close as you can get to one without actually being one. It is infinitely close, just coming up short by an infinitely small margin.

 

[MrDudeMan]

Originally posted by: 91TTZ
I think in the context that this is written, this is more of a logic problem than a math problem.

.9999...9 is as close as you can get to one without actually being one. It is infinitely close, just coming up short by an infinitely small margin.

you are wrong. if you claim .999... is not = to 1 then you dont accept calculus. this is a simple proof that is readily available on google.

.999... = 1 and anyone that knows calculus should know this.

 

[xirtam]

If it's truly infinitely close, then there's no finite distinction between the two. They've become joined. "Infinitely close" is equal to "the same as," and truly "infinitely small" is the same as none, since you can't have any less than none, and you also can't have any less than "infinitely small."

So 2.999... = 3.

See the thread, "Does .999... = 1", and I hate you for the respost in disguise.

 

[91TTZ]

As a followup to my previous post, if you did it as a logic problem it is easier to solve.

.9999...9 < 1

therefore

2.9999...9 < 3

therefore

2.9999...9 != 3

 

[spidey07]

Originally posted by: 91TTZ
As a followup to my previous post, if you did it as a logic problem it is easier to solve.

.9999...9 < 1

therefore

2.9999...9 < 3

therefore

2.9999...9 != 3

Nice try, you are terminating the decimal, hence no infinity

But the fact remains that 2.999... = 3.

 

[91TTZ]

Originally posted by: spidey07


Nice try, you are terminating the decimal, hence no infinity

I can't find the proper key on the keyboard that I'm looking for. The figure that I want would be a 9 with a little line on top of it, indicating that the 9's repeat forever.

 

[MrDudeMan]

Originally posted by: 91TTZ
As a followup to my previous post, if you did it as a logic problem it is easier to solve.

.9999...9 < 1

therefore

2.9999...9 < 3

therefore

2.9999...9 != 3

dude you are wrong. read above and search google for the calculus proof. its real, it works, and 2.999... = 3.

 

[ActuaryTm]

91TTZ has been here only a few short days, and is undoubtedly unfamiliar with this thread.

After reading the above, please refer to the already posted depiction of the decease equine.

 

[yhelothar]

Originally posted by: 91TTZ
As a followup to my previous post, if you did it as a logic problem it is easier to solve.

.9999...9 < 1

therefore

2.9999...9 < 3

therefore

2.9999...9 != 3

Can't believe people are still arguing this. You really are dumb. Just stop posting. Anyone with half a brain would understand the logic that .9999... = 1.

 

[91TTZ]

Originally posted by: ActuaryTm
91TTZ has been here only a few short days, and is undoubtedly unfamiliar with this thread.

Holy crap. If they couldn't come to an agreement in 260 pages we're not about to solve it here on page 3, lol

 

[91TTZ]

I'm not going to waste my time with this. I could be posting all weekend and still not come to an agreement.

 

[So]

Also, do you people not get basic math... if .999... = 1 then x + .9999 = x + 1

It's the same reason they invented i (j for use EE's :roll: )

 

[arcenite]

damn you kyteland... damn you...