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

  • Context: High School 
  • Thread starter Thread starter Ed Aboud
  • Start date Start date
  • Tags Tags
    Mystery
Click For Summary

Discussion Overview

The discussion revolves around the mathematical concept of whether the repeating decimal .999999... is equal to the integer 1. Participants explore various arguments, definitions, and implications related to this topic, touching on aspects of real numbers, sequences, and convergence.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a derivation suggesting that if x = .999999..., then 10x - x = 9x leads to x = 1, implying .999999... equals 1.
  • Several participants agree with the assertion that .999999... is an alternate representation of the number 1.
  • Another perspective is introduced questioning if there exists a number between .999999... and 1, challenging the hypothesis that they are not equal.
  • Some participants discuss the implications of .999999... being a real number and the convergence of the sequence .9 + .09 + .009 + ... as a more complex issue requiring real analysis.
  • There is mention of the completeness of real numbers and how bounded, increasing sequences converge, with some participants noting that this concept may be beyond standard high school mathematics.
  • A participant reflects on the need for a standard answer to the question of .999999... equaling 1, suggesting the creation of a library entry for future reference.

Areas of Agreement / Disagreement

Participants express a mix of agreement and differing viewpoints, with some asserting that .999999... equals 1 while others raise questions about the implications and understanding of real numbers and convergence. The discussion remains unresolved with multiple competing views present.

Contextual Notes

Some participants note that the mathematical concepts discussed may be complex and not typically covered in high school curricula, indicating a potential gap in understanding for some individuals.

Ed Aboud
Messages
200
Reaction score
0
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.
 
Mathematics news on Phys.org
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?
 
I am so glad someone finally posted about this, I have never seen it here in the forum before.
 
diffy said:
i Am So Glad Someone Finally Posted About This, I Have Never Seen It Here In The Forum Before.

Sarcasm alert![/size]
 
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.
 
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..
 
arildno said:
Sarcasm alert!

:smile:
 
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.
 
Last edited:
  • #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 0.\bar{9}=1 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.
 

Similar threads

High School Simple maths but i don't understand why (about recurring decimals or repeating)
  • · Replies 27 ·
Replies
27
Views
5K
High School Fractional/decimal powers
  • · Replies 10 ·
Replies
10
Views
3K
High School Difference between powers of numbers equaling 1
  • · Replies 11 ·
Replies
11
Views
2K
High School Is 0.9 Recurring Truly Considered Equal to 1 in Mathematics?
  • · Replies 34 ·
2
Replies
34
Views
17K
High School Can a log have multiple bases?
  • · Replies 44 ·
2
Replies
44
Views
5K
MHB Can GCD of m and n be forced to equal 1 for given equations and conditions?
  • · Replies 3 ·
Replies
3
Views
4K
High School What is the link between proportion and multiplication?
  • · Replies 3 ·
Replies
3
Views
2K
MHB How Can One Demonstrate the Ratio Between the Hypotenuses of Two Triangles?
  • · Replies 2 ·
Replies
2
Views
2K
High School What type of equation is this?
  • · Replies 6 ·
Replies
6
Views
1K
Undergrad I would like to argue about .999
  • · Replies 109 ·
4
Replies
109
Views
21K