The Mystery of the Endless Decimals: Can .9 Recurring Equal 1?

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In summary, the conversation discusses the mathematical concept of whether 0.999... is equal to 1. The consensus among the participants is that by definition and derivation, 0.999... is indeed equal to 1. The discussion also touches on the concept of real numbers and their completeness, which is beyond high school mathematics. It is suggested that a standard answer on this topic be created for the PF library.
  • #1
Ed Aboud
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Hi all.
I can't remember where I saw this but it really confuses me.

If

x = .999999999...
10x = 9.9999999...
10x - x = 9x

x = 1?

So does that mean if that .9 recurring is 1 exactly?
Thanks for any help.
 
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  • #2
Yes, by definition and derivation. You got it.
 
  • #3
Yes it does mean that. .99999... is an alternate representation for the number 1.

Another way of looking at it: Suppose a and b are real numbers with b>a. There there exists another number, c, such that a < c < b. Suppose we have the hypothesis that .9999... is not equal to 1. Can you find a number between .999999... and 1?
 
  • #4
I am so glad someone finally posted about this, I have never seen it here in the forum before.
 
  • #5
diffy said:
i Am So Glad Someone Finally Posted About This, I Have Never Seen It Here In The Forum Before.

Sarcasm alert!
 
  • #6
Assuming .99999... is actually represents a number, it can't be bigger than 1. It also can't be smaller than 1 because of Nicksauce's reasoning. Therefore the only option is for .9999.. to be 1.

The question of whether .999999... is even a real number at all - the question if the sequence .9 + .09 + .009 + ... converges - is a bit more complicated and requires some basic real analysis.
 
  • #7
maze said:
Assuming .99999... is actually represents a number, it can't be bigger than 1. It also can't be smaller than 1 because of Nicksauce's reasoning. Therefore the only option is for .9999.. to be 1.

The question of whether .999999... is even a real number at all - the question if the sequence .9 + .09 + .009 + ... converges - is a bit more complicated and requires some basic real analysis.

Really?

From what I know, a usual axiom appended to the real number field is that every bounded, increasing sequence converges..
Not too complicated to show that the condition holds in this case..
 
  • #8
arildno said:
Sarcasm alert!

:smile:
 
  • #9
Ok, thanks very much for the help guys.
And I am sorry if it was posted before, I'm relatively new to PF.
 
  • #11
arildno said:
Really?

From what I know, a usual axiom appended to the real number field is that every bounded, increasing sequence converges..
Not too complicated to show that the condition holds in this case..

Yeah, I mean it's not really too complicated, its just probably a little beyond a standard high school level.

The possible "problem" would be if the real numbers weren't complete (they are complete, so it's not a problem). The analogy would be, the rational numbers are not complete, so the rational sequence 3, 3.1, 3.14, 3.1415, ..., a bounded increasing sequence, doesn't actually converge in the rational numbers. If the real numbers were not complete, .9, .99, .999, ... might converge to some non-real "infintessimal" sort of number, in the same way that 3.14, 3.141, ... converges to a non-rational number.
 
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  • #12
arildno said:
Really?

From what I know, a usual axiom appended to the real number field is that every bounded, increasing sequence converges..
Not too complicated to show that the condition holds in this case..

The form of that axiom that I've always seen is that the real numbers are complete (i.e. every Cauchy sequence of real numbers has a limit in the real numbers). A bounded, increasing sequence is a Cauchy sequence though, so it has a limit.
 
  • #13
LukeD said:
The form of that axiom that I've always seen is that the real numbers are complete (i.e. every Cauchy sequence of real numbers has a limit in the real numbers).
This is true, but be fair to the young'uns. This is far beyond high school mathematics. It is even beyond the typical Calc 1/2/3/... sequence required of all science and engineering undergraduates in most colleges. Science majors do study advanced math, but it is stuff like Green's theorem, complex analysis, numerical methods, ... That [tex]0.\bar{9}=1[/tex] is even beyond a lot of high school math teachers. I know, because long, long ago I saw the simple 9*0.111... "proof" and pointed it out to a high school math teacher who insisted I was wrong. I didn't learn that I was right until I went to college and took a real algebra course. Bottom line: We should go a bit easy on the kind of people who typically ask this question.

What we need is a standard answer to this question, and now we have the vehicle for this answer: the PF library. Unfortunately, I will be very busy for the next few weeks. Any takers on writing this up? Then when the question comes up again we can simply point to the library entry.
 
  • #14
D H said:
What we need is a standard answer to this question, and now we have the vehicle for this answer: the PF library. Unfortunately, I will be very busy for the next few weeks. Any takers on writing this up? Then when the question comes up again we can simply point to the library entry.

That's a good idea that perhaps someone could undertake.

There are also hundreds of threads on this topic that one can find by searching. I'm thus locking this thread.
 

1. Can .9 recurring truly equal 1?

Yes, .9 recurring (or .999...) is equivalent to the decimal representation of the number 1. This can be proven mathematically through various methods, such as using infinite geometric series or converting to fractions.

2. How is it possible for .9 recurring to equal 1?

This concept can be difficult to grasp, but it essentially comes down to the fact that the decimal system has limitations in accurately representing certain numbers. Just like how 1/3 in decimal form is represented as .333..., 1 in decimal form is represented as .999...

3. Is .9 recurring just an approximation of 1?

No, .9 recurring is not an approximation of 1. It is the exact representation of 1 in the decimal system. It may seem like an approximation because our minds have been conditioned to think of numbers with finite decimal representations, but mathematically, it is equivalent to 1.

4. Can you provide a real-life example of .9 recurring equaling 1?

Sure, one example is the concept of repeating decimals in the metric system. For instance, 1 meter is equivalent to 1.000... meters, but we typically round it to 1 meter for simplicity. This is similar to how .9 recurring is equivalent to 1, but we often round it to 1 for ease of understanding.

5. Why is this concept important in mathematics?

The concept of .9 recurring equaling 1 may seem trivial, but it has significant implications in fields such as calculus and number theory. It also challenges our understanding of numbers and the limitations of the decimal system. Additionally, it serves as a reminder to always question and critically think about mathematical concepts.

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