Is 0.999... Really Equal to 1? Exploring the Mathematical Proofs

[Mentat]

I'm sure most of you already know this. The real point of the thread is finding different ways of approaching it. You see, I have a friend who refuses to accept what seems to me to be so obvious: .9 repeating (infinite 9s after the decimal) is exactly equal to the whole number 1.

Here are the proofs that I've used so far:

1) 1/3 is precisely equal to .333... (to which he agreed). 3*.333... equals .999... and 3*1/3 equals 1. Therefore, since I multiplied equal numbers by the same number, they should still be equal...ergo 1 = .999...

2) If 1 is greater than .999..., then by how much, exactly, is it greater? After all, 1 is greater than .9 by .1. It is greater than .99 by .01. It is greater than .99999999999999999999 by .00000000000000000001. And so on, and so on. However, if the number of nines is infinite, then the number of zeros preceding the 1 will be infinite, and 0.000... is obviously equal to 0, which means that 1 is greater than .999... by exactly 0.

3) After this point, my friend revealed that he doesn't really believe in actual infinites. He thinks you will eventually reach the 1, at the end of the infinite series of 0s. Ridiculous as that may sound, I respected that as his opinion and posed this next point: If there are no infinites then the term equal to 1/3, which is usually considered .333..., would actually (eventually) terminate in a 4. I asked him if that was right. After some thought, he said "yes". So, I asked, if we multiplied this number by 2, would you logically expect to get the same number 6s as you had 3s, and then terminate in an 8 (since 4+4 is 8)? He said, "yes", not realizing that he had just implied that the infinitely repeating number .666..., which is equal to 2/3 must now terminate in an 8, instead of a 7 (as his reasoning would naturaly have assumed).


Are there any other points I can use? I wasn't going to ask for help on this, until he talked to his math teacher, and the teacher agreed with him! Now he's got this quaint little ipse dixe argument, "well, the math teacher said I'm right".

Any new points or comments on my previous ones are appreciated.

 

[Mentat]

btw, I don't know if this was the right place to have posted this thread, but I couldn't think of a better place for it.

 

[confutatis]

The real point of the thread is finding different ways of approaching it. You see, I have a friend who refuses to accept what seems to me to be so obvious: .9 repeating (infinite 9s after the decimal) is exactly equal to the whole number 1.

You have provided your friends some arguments as to why .9999... must be the same as 1, otherwise math would not be perfect. Your friend is saying math is not perfect. Countering his argument with an unsubstantiated claim that math is perfect, end of discussion, doesn't seem to do much for him.

But you can still get him to think about the issue from a different perspective. If he doesn't accept that numbers can be infinitely small, then he must agree that there exists a minimum quantity which cannot be divided. For instance, if he thinks the smallest number possible is N, then you can ask him, how much is N/3? Is it N or is it 0? There is no known answer to that question, but if the problem is real then an answer must necessarily exist. And that means the answer can be determined with probabilities. The probability of N/3 being N is 1/3, and the probability of it being 0 is 2/3.

So the correct answer to his problem is that the result of 3 * 0.3333...3 must also be determined by probabilities, and 1, not 0.9999...9, is by far the most likely answer.

 

[NateTG]

There is a fundamental question that is generally skipped in the discussions of $$0.\bar{9} = 1$$, which is the assumption that $$0.\bar{9}$$ is a real (or rational) number.

A second shortcoming is that 'bar' notation is generally not well explored. If your friend is comfortable with bar notation, the following might be interesting for him:

$$1$$
$$=\frac{1}{9} + \frac{8}{9}$$
$$=0.\bar{1}+0.\bar{8}$$
$$=0.\bar{9}$$

Although this is, of course, identical to the $$3 \times \frac{1}{3}$$ example that you gave.

The traditional elementary school method is:
$$9 \times 0.\bar{9} = 10 \times 0.\bar{9} - (1)0.\bar{9} = 9.\bar{9} - 0.\bar{9} = 9$$
so
$$9 \times 0.\bar{9} = 9$$
so
$$0.\bar{9}=1$$

You can also talk about infinite sums, but getting satisfactory answers with sums would (eventually) involve invoking the completeness of the real numbers - which is not necessarily an accessible subject.
$$0.\bar{9} = \sum_{i=1}^\infty \frac{9}{10^i}$$

You can ask your friend what:
$$\frac{1}{1 - 0.\bar{9}}$$
is equal to. Since $$0.\bar{9} \neq 1$$ the denominator is non-zero right?

 

[Integral]

http://home.comcast.net/~rossgr1/Math/one.PDF

 

[Dissident Dan]

$$x = 0.\bar{9}$$
$$9x = 10x - x$$
$$9x = 9.\bar{9} - 0.\bar{9}$$
$$9x = 9$$
$$x = 1$$

Of course, I tend to think that the idea of a value of anything repeating is illogial--after all, it's just a defect of the number system (decimal, aka base-10). The proof that I just listed relies on the fact that:
infinity - 1 = infinity

0.\bar{9} cannot be viewed as anything other than a defect of the number system. It is not a value, as it involves infinity.

 

[Integral]

0.\bar{9} cannot be viewed as anything other than a defect of the number system. It is not a value, as it involves infinity.

I don't see it that way. Just the opposite if we were UNABLE to assign a value to any infinitely repeating number that would be a defect in the number system.
This is NOT a facet of DECIMAL numbers, it is a facet of the density of the Real number line. No matter how you choose to represent the points of the real number line there will always be numbers which require an infinite number of digits to be specified.

The Base does not matter. for example

$$.1_{10} = .00011001100... _2$$

So in binary .1 has an infinitely repeating representation which MUST be truncated in all computations. This fact was first made apparent to me years ago when I did the following loop in Applesoft Basic
10 x= 0
20 For i = 1 to 10
30 x= x+ .1
40 Print x
50 next i

The final number would not be 1 but .999999, in only 10 steps the round off error would become apparent.

(Round off error has nothing to do with the fact that 1=.999...)

 

[Tango_dark]

If we allow .0...1 (sorry not sure how to code for the bar notation) to have a non-zero value, then it must be the smallest positive number. If we allow for numbers to have a minimum value then the number line ceases to be continuous as any 2 numbers no longer have an infinite range of values between them. Now we have x, x+.0...1, x+.0...2, etc as our number line. There can be no number between these values because then there would be a smaller number then .0...1 Since this causes a contradiction, we can determine that .0...1 cannot have a non-zero positive value.

 

[Hurkyl]

If we allow .0...1 (sorry not sure how to code for the bar notation) to have a non-zero value, then it must be the smallest positive number.

If we allow this, then why is .0...01 not a smaller number? After all, if we're going to allow ourselves "decimals" with infinity + 1 digits, then why not decimals with infinity + 2 digits?

Before we ask "what is .0...1?" we should ask "does the notation .0...1 even make sense?"

Are there any other points I can use?

Try asking him how many digits 1/3 has.

 

[Tango_dark]

Since there are already an infinite number of zeroes before the one, adding one more will not make a difference (property of infinity). Thats part of the problem. Giving the 1 at the end of infinity a value means a) allowing the infinity to end and b) since the value of the one is equal to 1/10^(infinity+1) we are allowing 1/infinity to have a non-zero defined positive value. Neither of these is consistent with the way which math works. The number .0...1 cannot possibly exist in any form which would allow it to be added to .99... to equal 1 (and hence prove that the two are not equal).

 

[hypnagogue]

Mentat, you have your friend stuck on a pair of horns. If 1/3 = 0.333...4, then 3 * 1/3 = 1.000...2, contradicting 3 * 1/3 = 1. So if your friend continues to hold that 1/3 = 0.333...4, then he must believe that

a) 1 = 1.000...2, which is just as bad (in his eyes) as the original position he was seeking to refute, or
b) 1/3 * 3 = 1 is false, or
c) that mathematics is not internally consistent, which essentially means that statements like 1 = 2 can be shown to be true.

If this forces him to reject that 1/3 = 0.333...4, then you can return to the conventional arguments. If not, then your friend should have no faith in mathematics whatsoever.

Another plan of attack would be to convert the numbers to a base where we don't get infinitely repeating decimals. For instance, $$3_{10} = 10_3$$, and $$1/3_{10} = 0.1_3$$. So if he accepts that 1/3 = 0.333..., and therefore that 3 * 1/3 = 0.999..., then he must accept that 0.999... = 1, since in base 3, $$10_3 * 0.1_3 = 1_3$$, and $$1_3 = 1_{10}$$.

 

[Mentat]

Thanks for all the replies, guys . All very useful stuff (even those who seem to agree with my friend, as I can use my own rebuttals to their reasoning as added arguments against his).

Hypna, you are absolutely right. He should indeed have no faith in math whatsoever. The problem is, his math teacher agreed with him...which means I'm going to have to talk to the math teacher first, thus removing his ability to argue from authority.

 

[Dissident Dan]

I don't see it that way. Just the opposite if we were UNABLE to assign a value to any infinitely repeating number that would be a defect in the number system.
This is NOT a facet of DECIMAL numbers, it is a facet of the density of the Real number line. No matter how you choose to represent the points of the real number line there will always be numbers which require an infinite number of digits to be specified.

The Base does not matter. for example

$$.1_{10} = .00011001100... _2$$
QUOTE]

Well, yes, all number systems have their own flaws. I didn't say that flaws were unique to decimal.

The invention of repeating numbers is a result of the fact that some values cannot be exactly represented using certain bases. A repeating number is just an attempt to express a number using partitions that do not allow for exact representation.

 

[leto]

This has bothered me for years. My own logic tells me they aren't equal, while math tells me they are. I haven't decided yet if my logic is flawed or the number system is. I never got anywhere with it when I ask a math head because they are stuck on the definitions they are taught and seem to be oblivious to anything else. I personally think it is as close to 1 as you can get without ever being 1. In any pratical use of math it can be called 1, so it really doesn't matter anyway.

 

[Hurkyl]

I personally think it is as close to 1 as you can get without ever being 1.

Suppose that's true. Then how does $(0.\bar{9} + 1)/2$ compare?

 

[Jeebus]

Suppose that's true. Then how does $(0.\bar{9} + 1)/2$ compare?

D00d, you got served! Just kidding, of course. I just had to say that.

 

[Tango_dark]

Also, if we assume that they are not equal and that it is "as close to one as possible without being one". Then we define must define a smallest number which represents their difference. If we assume a smallest possible number, bad things happen. Like the number line ceases to be continuous, distance ceases to exist, and other bad stuff.

 

[Integral]

Consider this inequality.

$$1 - .1^n < x < 1+ .1^n$$

It seems clear that there is only a single number for which this is true for ALL values of n>0, x= 1 . This is a simple statement that any number added to x results in something greater then 1, or any number subtracted from x results in something less then one. I doubt that you will find many people who will argue with the truth of the statement, x=1.

Now, in the link I posted above, I show that using simple arithmetic, involving only valid rational numbers, one can construct this inequality.

$$1- .1^n < .999... < 1+.1^n$$

notice that is the exact inequality as above, thus we have x=.999... If the original statement is correct that the inequality can only be satisfied by 1 you must be lead to the conclusion that
1=.999...

 

[honestrosewater]

What if you make a distinction between 1.000... and .999..., i.e. so that they are different numbers, and then figure out "how infinite" this set of numbers is. If this new set is "more infinite" than the set of real numbers, case closed. That is, if your friend doesn't object to 1-1 correspondence, countablilty, and the various ways of comparing infinite sets ;)

I'm not sure how to make such a distinction. Perhaps the sets of all arrangements, where an "arrangement" is an infinite string of digits and one decimal point, ex.
1010= ...0001010.000...
98765.4321= ...00098765.4321000...
1= ...0001.000...
.999...= ...000.999...

The set of all arrangements should include all real numbers, plus two strings for each terminating decimal, but may not be more infinite than the reals. Just a thought. The answer is probably already known. Though now I find it interesting in itself :)

The whole idea is that 1=.999... implies that there must be more representations (numerals) than numbers. This may be something your friend has not realized. And it may be a way to get him/her to take another look at the axioms and theorems. Of course, "more" becomes vague when dealing with the infinite, but...
Happy thoughts
Rachel
EDIT- Nevermind, I see my mistake.
But I would like to know more about the sets of all numerals.

 

[Imparcticle]

Does your friend deny the fact that .9999... when rounded is 1? Is that the problem he/she seems to have difficulty with?

 

[Math Is Hard]

Does your friend deny the fact that .9999... when rounded is 1? Is that the problem he/she seems to have difficulty with?

when rounded, yes, but, I don't think rounding enters into this. This is .9999... with the decimal places stretching out to infinity. No rounding allowed- we're going all the way to the bitter end*.
My personal visualization of this is that 1 is reached only at the infinite decimal place and not anywhere before that. (But let me add the caveat that I am a million times lower on the rungs of mathematics than the majority of posters to these types of threads - so I may be completely full of baloney here.)

 

[Polly]

And I am definitely worse than Math Is Hard and just shooting some hot air here. But I am thinking, assuming 1 to be infinity, will taking a particle from it make it less than infinity? Probably not.

 

[pig]

What many people don't understand is that decimal representation is just that, a method for representation of quantities. That's why they think that any possible combination of numbers, dots, etc, should represent some quantity, and they invent meaningless stuff like "0.000...0001". What it means is important, not what it looks like when written down. We don't "allow" 0.000...0001 to have a non-zero value, or a zero value - it is simply not a number.

Polly - taking a finite amount out of something infinite doesn't reduce its size. This is the most important difference between infinite and finite sets. This is why the 0.999... * 10 = 9.999... demonstration makes sense - you moved all the nines one place to the left, but there is still the same number of them after the decimal point.

 

[quddusaliquddus]

Can Fuzzy Logic help here? - or am I confusing the topic?

 

[NateTG]

Can Fuzzy Logic help here? - or am I confusing the topic?

The technical term 'fuzzy logic' refers to systems of logic, for example electronics, that have more than 2 states, or many inputs. It is not at all apropos.

If you mean fuzzy logic in the sense of what mathematicians refer to as handwaving -- that is, saying that something makes sense, and leaving it there -- then it's also not appropriate for a discussion of this nature.

 

[quddusaliquddus]

Lol - not hand writing. I once read a book on it (quiet a while ago), and from what I think I understood - fuzzy logic allows one to have in-between states like 80% true. It probably doesn't help though ... :D

 

[Imparcticle]

This is why the 0.999... * 10 = 9.999... demonstration makes sense - you moved all the nines one place to the left, but there is still the same number of them after the decimal point.

No, you move one nine to the left.

We don't "allow" 0.000...0001 to have a non-zero value, or a zero value - it is simply not a number.

What is a number? .000...0001 is a possible measurement for some subatomic entity such as an atom or something.

 

[Hurkyl]

No, you move one nine to the left.

Which 9?

.000...0001 is a possible measurement for some subatomic entity such as an atom or something.

How do you figure?

What is a number?

What kind of number? I presume you mean "decimal number".

A decimal number is (leaving out some details), a function from the integers into the set {0, 1, 2, ..., 9}. The intuition is that the function takes as input a position and it returns the digit in that position.

0.00...001 ("infinite" zeroes) whatever it means, cannot be a decimal number. There has to be some integer which says where the '1' occurs, but that would mean there are only a finite number of zeroes between the '1' and the decimal point.

 

[Integral]

No, you move one nine to the left.


What is a number? .000...0001 is a possible measurement for some subatomic entity such as an atom or something.

A real number is a very well defined "thing". Numeric representations of real numbers consist of an infinite sum, you MUST be able to assign a integer to each digit. What is the integer assigned to the above 1? To say an infinite number of zeros then a 1, is an self contradictory statement.

Physical objects do not enter into the definition of the real numbers. Even if it did Planck's length, the theoretic smallest physical length, is not infinitely small.

Edit: a few words for clarity.

 

[Sariaht]

It's true if all pieces are shareable. I guess they are.

 

[HallsofIvy]

This has bothered me for years. My own logic tells me they aren't equal, while math tells me they are. I haven't decided yet if my logic is flawed or the number system is.

I know that feeling. Just the other day I was driving down a one-way street and all the other cars were going the other way! For the life of me, I just could decide whether I was going the wrong way or everyone else was!

I never got anywhere with it when I ask a math head because they are stuck on the definitions they are taught and seem to be oblivious to anything else.

Exactly! That blasted cop was so stuck on HIS definition of "one-way" he just wouldn't listen to me! (I was, after all, only going one way!)

 

[Math Is Hard]

Exactly! That blasted cop was so stuck on HIS definition of "one-way" he just wouldn't listen to me! (I was, after all, only going one way!)

I tried to apply similar logic. "Officer, the sign said 'No, U turn'." So I did.

 

[Math Is Hard]

from Marilyn Vos Savant's Parade Magazine column

Dear Marilyn,

If it is a fact that 1/3 = .333 repeating, why is it true that 1/3 + 1/3 + 1/3 = 1?

-Daniel

Dear Daniel,

"... no matter how far you extend the .333 repeating, the part you didn't write down (the repeating part) totals 1/3 of the next decimal place. And no matter how far you extend a .666 repeating, the part you didn't write down totals 2/3 of the next decimal place. And no matter how far you extend a .999 repeating, the part you didn't write down totals 3/3 of the next decimal place. This means that you can stop at any point and add it to what you have already written, and the result will be 1.000."

 

[leto]

I know that feeling. Just the other day I was driving down a one-way street and all the other cars were going the other way! For the life of me, I just could decide whether I was going the wrong way or everyone else was!

Your sarcastic one-way street analogy is interesting enough, but it doesn't have any value aside from your apparent need to mock me. We both know what appeal to popularity is. Traffic rules require no justification beyond keeping order. My problem with .999 repeating = 1, is simple - There is NEVER a point where you can stop the repetition and say that number = 1. I understand that the repetitions never stop, however, the amount of places from the decimal point also never stop. You are infinitely taking smaller and smaller steps closer to one. If infinity at some point arbitrarily fills the gap then why can't .999 repeating = 2, or any other number? Certainly the repetition must continue to go on even after 1 is approached. I can not give a value between .999 reapeating and 1, I understand this - This is because I cannot stop the repetition; just as you cannot stop the repetition and have 1.

I don't doubt I may be wrong. What I will not do is mindlessly accept what someone tells me unless I understand the logic behind it.

 

[Hurkyl]

0.999~ is a (decimal) number. It doesn't change, it doesn't approach anything. It doesn't have any steps, nor repetitions. It simply has a 9 in all of the positions to the right of the decimal point.