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0.9999 = 1, true or false and why?

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Anaxagoras

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Oct 25, 2007
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y= 0.9999999
10y=9.9999999
9y=10y - y
9y=9.9999999 - 0.9999999
9y=9
y=1
1=0.9999999

Where did I go wrong?
 

flsh

Banned
Aug 19, 2004
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You have one 9 too many on that 9.999..
Though 0.9999... (to infinity) does equal 1.
 

StuBurns

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Jan 9, 2008
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flsh said:
You have one 9 too many on that 9.999..
Though 0.9999... (to infinity) does equal 1.
This.

Although the OP said it strangely, I ususally see it like this:

1/3=0.333*
0.333*x3=0.999* so...

1=0.999*
 

Earthstrike

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Dec 9, 2005
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0.9999... is exactly equal to one.
You went wrong nowhere. This is just a consequence of the definition of a repeating decimal as setup by mathematics. A lot of people tend to have a knee jerk reaction to reject for some reason.

A much simply way to think about this is 1/3 + 2/3 = 3/3
1/3 = .333...
2/3 = .6666...

There's several easy ways to rationalize this.
The first is to consider the difference between 0.999... and 1.
You would have 0.000... with a hypothetical 1 as the inifitieth digit. Consequently, you would have a number that is infinitely small. If you have a number that's infinitely small, then it must be the smallest number (absolute value of course). a la zero.


Suffice it to say though, the math will show through.
 

RevoDS

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Apr 21, 2007
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StuBurns said:
This.

Although the OP said it strangely, I ususally see it like this:

1/3=0.333*
0.333*x0.333*=0.999* so...

1=0.999*

I believe you meant to say 0.333* x 3 = 0.999*. Because 0.333* x 0.333* = 0.110889.

As for the OP's question, it all depends on the point of view. If you see things technically, then no, 0.999999 does not = 1 because no matter what you do with it, how many 9s there are, there is still an infinitely small difference which keeps shrinking as you add 9s but never completely goes away. In practical use however, 0.999999 = 1.

The fraction argument is a fallacy, because the conversion of 1/3 to decimals relies on a small amount of rounding which introduces a margin of error big enough to make up for the actual difference between both numbers.
 

tokkun

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Jan 29, 2007
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Yes, .9 repeating is equal to 1. We just had a thread about this too.

Here's another one:

1/9 = .1111111111111111...
2/9 = .2222222222222222...
3/9 = .3333333333333333...
4/9 = .4444444444444444...
...
8/9 = .8888888888888888...
9/9 = 1
 

Kipz

massive bear, tiny salmon
Aug 29, 2007
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Because .9999... is an imaginary number that is infinitely close to 1 and therefore for all intents and purposes is 1?
 
Nov 7, 2007
1,982
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1,215
72/9 = 8.0
73/9 = 8.111111
74/9 = 8.222222
75/9 = 8.333333
76/9 = 8.444444
77/9 = 8.555555
78/9 = 8.666666
79/9 = 8.777777
80/9 = 8.888888

Following this same pattern, the next answer should naturally be 8.999999. However:

81/ 9 = 9.0
 

StuBurns

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Jan 9, 2008
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RevoDS said:
I believe you meant to say 0.333* x 3 = 0.999*. Because 0.333* x 0.333* = 0.110889.
I did indeed, I edited before you posted, but you caught me. Moronic mistake.
 

BananaBomb

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Kipz said:
Because .9999... is an imaginary number that is infinitely close to 1 and therefore for all intents and purposes is 1?

no it's a real number that is equal to one.
 

tokkun

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Kipz said:
Because .9999... is an imaginary number that is infinitely close to 1 and therefore for all intents and purposes is 1?

The term 'imaginary number' means something in math.

.999.... is not an imaginary number.
 

bachikarn

Member
Jun 7, 2004
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RevoDS said:
As for the OP's question, it all depends on the point of view. If you see things technically, then no, 0.999999 does not = 1. In practical use however, 0.999999 = 1.

This is wrong (no offense). .9999... = 1 in theory and in "practice."

The easiest way to think about it is that numbers (real numbers specifically for this problem) are abstract quantities. We represent them with "numerals". So when we write "5," we are writing down a representation of the abstract quantity that is the real number 5. It turns out because of how our number system was set up, the numeral ".999...." and the numeral "1" represent the same real number. Hence, we say .9999... = 1.


Edit: PS, there is no fallacy in the fraction argument.
 

SmokeMaxX

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Apr 3, 2009
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Haven't seen one of these threads on here in a while. They occurred so often in a short period of time and then suddenly stopped. I thought they were bannable at some point in time.

Anyway, 0.9999 repeating is 1. Numbers are weird, but that's just how it is. You can rationalize it in many ways or you can just acknowledge it as truth. The latter is easier. While there are many ways to prove that 0.99999... = 1, there is no way to confirm the opposite.

I've always thought a better question was this:
If you had a hat with an infinite amount of slips of paper consisting of an infinite amount of numbers, the probability of pulling any random number is zero (1/infinity = 0). In fact, the probability of pulling any real subset of numbers is zero (i.e. 10000/infinity = 0). However, if you actually reach down in there and pull out a number, the probability obviously isn't zero because you just pulled out a number. Why is that?

I had my Cal professor explain the answer to this one for me, but I totally forgot what he said haha.
 

Cruceh

Banned
Dec 26, 2006
2,998
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It's 1. Learned this in HS math. Something about limits or wahtever. Forgot everything already.
 

Evlar

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Dec 22, 2006
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0.9999999 != 1 (does not equal)

0.9999... = 1 (the ellipsis means the previous series continues perpetually)
 

Earthstrike

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Dec 9, 2005
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RevoDS said:
I believe you meant to say 0.333* x 3 = 0.999*. Because 0.333* x 0.333* = 0.110889.

As for the OP's question, it all depends on the point of view. If you see things technically, then no, 0.999999 does not = 1 because no matter what you do with it, how many 9s there are, there is still an infinitely small difference which keeps shrinking as you add 9s but never completely goes away. In practical use however, 0.999999 = 1.

The fraction argument is a fallacy, because the conversion of 1/3 to decimals relies on a small amount of rounding which introduces a margin of error big enough to make up for the actual difference between both numbers.


There's not rounding in the fractional representation of 1/3. Also, answers to mathematical questions do not depend on "points of view".

Plain and simple mathematics is the evaluation of the set of consequences and relations amongst a given set of definitions. By the definitions in our number system 0.99999 repeating is exactly equal to one.

You're argument contains a fallacy however because you say "there is still an inifinitely small difference that never goes away". Actually it HAS to go away by definition.

Proof: Lets say there is some infinitely small difference that is greater than zero. Divide that by 10. It is now smaller, meaning it wasn't an infinitely small difference. The only infinitely small difference is zero.
 

Tntnnbltn

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Jul 12, 2007
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RevoDS said:
As for the OP's question, it all depends on the point of view. If you see things technically, then no, 0.999999 does not = 1 because no matter what you do with it, how many 9s there are, there is still an infinitely small difference which keeps shrinking as you add 9s but never completely goes away. In practical use however, 0.999999 = 1.
No, it doens't depend on your point of view. They are different representations of the same number. Technically, practically, and in every way.

"4/4" looks different to "1", but people easily accept that these are different representations of an identical number. The exact same deal applies to 0.999... = 1
 

RevoDS

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Earthstrike said:
There's not rounding in the fractional representation of 1/3. Also, answers to mathematical questions do not depend on "points of view".

Plain and simple mathematics is the evaluation of the set of consequences and relations amongst a given set of definitions. By the definitions in our number system 0.99999 repeating is exactly equal to one.

You're argument contains a fallacy however because you say "there is still an inifinitely small difference that never goes away". Actually it HAS to go away by definition.

Proof: Lets say there is some infinitely small difference that is greater than zero. Divide that by 10. It is now smaller, meaning it wasn't an infinitely small difference. The only infinitely small difference is zero.

By "infinitely small", I meant that it was equal to 1x10^-n, where n is the amount of 9s after the 0. As the 0.9999.... grows ad infinitum towards 1 without ever actually reaching it, the difference between 1 and that number shrinks at the same rate.

There IS rounding, for the very same reason there is a 0.0000....1 difference between 0.9999....9 and 1

In practice, we simply assume 0.999999 = 1 for convenience and because the 0.000001 difference is mathematically insignificant. But it is still there no matter what, thus 0.999999 and 1 are not the same number.

Assuming 0.999... is 1 is the very same thing as assuming 99 is 100, only on a smaller scale.

99/100 =9.9/10
9.9/10=0.99/1
 

Evlar

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Dec 22, 2006
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RevoDS said:
By "infinitely small", I meant that it was equal to 1x10^-n, where n is the amount of 9s after the 0. As the 0.9999.... grows ad infinitum towards 1 without ever actually reaching it, the difference between 1 and that number shrinks at the same rate.

There IS rounding, for the very same reason there is a 0.0000....1 difference between 0.9999....9 and 1

In practice, we simply assume 0.999999 = 1 for convenience and because the 0.000001 difference is mathematically insignificant. But it is still there no matter what, thus 0.999999 and 1 are not the same number.
You are correct that 0.9999...9 does not equal 1. However, we aren't talking about 0.9999...9.

Or, to put it another way,

0.9999...9 != 1
0.9999...9 != 0.9999...
RevoDS said:
Assuming 0.999... is 1 is the very same thing as assuming 99 is 100, only on a smaller scale.

99/100 =9.9/10
9.9/10=0.99/1
Specious reasoning. You are offering one pair in a convergence series. The correct observation is that the difference between the 0.9...9 series and 1 approaches 0 as the series is extended.
 

Enkidu

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RevoDS said:
There IS rounding, for the very same reason there is a 0.0000....1 difference between 0.9999....9 and 1
This magical 0.000...1 number doesn't exist. You can't have an infinite amount of 0's and then add a 1 at the end, because if you can add a 1 at the end then you obviously could have added more 0's instead hence it wasn't an infinite amount. Therefore, as you say, the difference between 0.999... and 1 is 0.000... which is simply another way of writing 0.
 

Tntnnbltn

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RevoDS said:
By "infinitely small", I meant that it was equal to 1x10^-n, where n is the amount of 9s after the 0. As the 0.9999.... grows ad infinitum towards 1 without ever actually reaching it, the difference between 1 and that number shrinks at the same rate.

There IS rounding, for the very same reason there is a 0.0000....1 difference between 0.9999....9 and 1

In practice, we simply assume 0.999999 = 1 for convenience and because the 0.000001 difference is mathematically insignificant. But it is still there no matter what, thus 0.999999 and 1 are not the same number.
But there is no 0.000000.........1 difference (for the proper 1-0.999...= argument). No matter how far you go, that tailing one doesn't exist. And because there is no difference between the numbers, they are the same.

The entire mathematic world agrees that they are equal to each other, and not just due to "mathematical insignificance" like you claim.

Assuming 0.999... is 1 is the very same thing as assuming 99 is 100, only on a smaller scale.

99/100 =9.9/10
9.9/10=0.99/1
No it doesn't, because the second it stops being an infinite series it stops being equal.
 

ratcliffja

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tokkun said:
Yes, .9 repeating is equal to 1. We just had a thread about this too.

Here's another one:

1/9 = .1111111111111111...
2/9 = .2222222222222222...
3/9 = .3333333333333333...
4/9 = .4444444444444444...
...
8/9 = .8888888888888888...
9/9 = 1

This is the one I always use. It seems weird, but it's actually true. The entire concept of repeating decimals is to express a ratio or fraction in decimal form. Repeating numbers can always be expressed as a number out of a series of 9s. For instance, .123123123... = 123/999.
 

Earthstrike

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RevoDS said:
Assuming 0.999... is 1 is the very same thing as assuming 99 is 100, only on a smaller scale.

You do not make assumptions in math. We're not assuming 0.999... is equal to one. We have proven it with nothing more than addition, subtraction, division, multiplication, and algebraic substitution.

Math is not felt out or worked out by analogy to thought processes. You have the rules defined within the system and the definitions of the system. With these you can demonstrate the 0.999... = 1, ergo, 0.999... is exactly equal to one.
 

Ashes

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Dec 11, 2008
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Earthstrike said:
You do not make assumptions in math. We're not assuming 0.999... is equal to one. We have proven it with nothing more than addition, subtraction, division, multiplication, and algebraic substitution.

Math is not felt out or worked out by analogy to thought processes. You have the rules defined within the system and the definitions of the system. With these you can demonstrate the 0.999... = 1, ergo, 0.999... is exactly equal to one.

Excellent argument against a false analogy. Although I would argue against using the word exact. It is 'theoretically equal' within 'our' number system.
 

Zaptruder

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Ok. Accepting that 0.9recurring = 1, then how to you represent the number that is smaller than 1 by the smallest margin?
 

ccbfan

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I think I never got about the proof is that

.3333333333333333333333333~ * 3 = .99999999999999999999999~

Is there any proof that says you can even multiply this sort of number like real numbers.

its similar to trying to say

3! * 3 = 9!
 

BananaBomb

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Jul 1, 2009
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Zaptruder said:
Ok. Accepting that 0.9recurring = 1, then how to you represent the number that is smaller than 1 by the smallest margin?

What does the smallest margin even mean? There isn't such a thing. How do you represent the number one less than infinity?

And to the guy above me, unless I am in grievous error, 3! and 9! are both real numbers...
 

Stat Flow

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Oct 27, 2005
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If you had 99 cents and what you were trying to buy was a dollar, you wouldn't be getting shit.

Nah, but most people have already said it, 0.999... = 1, I prefer to go with the 1/3 method of proving it.
 

Evlar

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Zaptruder said:
Ok. Accepting that 0.9recurring = 1, then how to you represent the number that is smaller than 1 by the smallest margin?
The "smallest margin" is 0. In the real number system there is no number that is the "smallest possible size but larger than 0". Number systems that include such numbers (called infinitesimals) include hyperreal numbers; however, for analysis purposes these are considered non-standard.
 

Slavik81

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Earthstrike said:
You do not make assumptions in math.
Wikipedia said:
Logical axioms are usually statements that are taken to be universally true (e.g., A and B implies A), while non-logical axioms (e.g., a + b = b + a) are actually defining properties for the domain of a specific mathematical theory (such as arithmetic). When used in the latter sense, "axiom," "postulate", and "assumption" may be used interchangeably. (source)
There are assumptions made in math, but they are fundamental assumptions.

Zaptruder said:
Ok. Accepting that 0.9recurring = 1, then how to you represent the number that is smaller than 1 by the smallest margin?
I'd guess:
y = 1 - ε

I think I never got about the proof is that

.3333333333333333333333333~ * 3 = .99999999999999999999999~

Is there any proof that says you can even multiply this sort of number like real numbers.

its similar to trying to say

3! * 3 = 9!
Code:
 12.8
+ 6.1
-----
 18.9

The sum of each column can be taken to find
the total sum.

  0.333333...
+ 0.333333...
+ 0.333333...
-------------
  0.999999...
 

Socreges

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BananaBomb

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Since I feel like there is more controversy in this thread than is acceptable, let's just seal the issue with a simple (and mathematically sound) proof.

Sum of a geometric series with a = 9 and r = 1/10 is 1. End of story.
 

Tntnnbltn

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ccbfan said:
I think I never got about the proof is that

.3333333333333333333333333~ * 3 = .99999999999999999999999~

Is there any proof that says you can even multiply this sort of number like real numbers.
There probably is, but I'm not a mathmetician.

If it's just the multiplication bit you don't like, try...


Otherwise, there are a whole bunch of the proofs at http://en.wikipedia.org/wiki/0.999...#Proofs don't use decimal representation of a fractions at all.
 
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