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Proving 0.999... Is Equal To 1
eldavojohn writes "Some of the juiciest parts of mathematics are the really simple statements that cause one to immediately pause and exclaim 'that can't be right!' But a recent 28 page paper in The Montana Mathematics Enthusiast (PDF) spends a great deal of time fielding questions by researchers who have explored this in depth and this seemingly impossibility is further explored in a brief history by Dev Gualtieri who presents the digit manipulation proof: Let a = 0.999... then we can multiply both sides by ten yielding 10a = 9.999... then subtracting a (which is 0.999...) from both sides we get 10a — a = 9.999... — 0.999... which reduces to 9a = 9 and thus a = 1. Mathematicians as far back as Euler have used various means to prove 0.999... = 1."
0.999... = 1 is second place to the Monty Hall Problem on the list of things that people have difficulty understanding and accepting the proof of. It is second place because the only department where I do not see graduate students giving me a confused look is the math department; with the Monty Hall problem, I will sometimes get a confused look even from people in the math department.
The other reason I put it in second place is that most people have difficult understanding the problem at all, whereas very few people have trouble understand what the Monty Hall problem is asking.
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And seriously... is this really front page material? The simplest proof is to say "express 1/9" as a decimal. Now multiply both sides by 9. I remember this in elementary school algebra.
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Small numbers usually win; express 1/3 as a decimal, and multiply by 3. The problem with that, though, is that people have trouble accepting that there was nothing wrong with what they did -- a lot of people have this implicit assumption that if a few simple steps bring them to a result that doesn't look like it makes sense, then they did something wrong. If you get them a more complicated proof (assuming they can follow it), they are more willing to accept the result.
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1/3 = 0.3333...
2/3 = 0.6666...
0.3333.... + 0.6666.... = 0.9999....
1/3 + 2/3 = 1 = 0.9999.....
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Humans are used to natural numbers because they're simple. But do natural numbers even exist in the real world? For the vast majority of practical purposes, 0.99999 can be thought of as one. But "one" itself is usually just a construct in the real world. There is no such thing as the perfect one of anything. The more precise we get, the more "one" becomes more of a mathematical ideal than a reality. So we spend our entire lives rounding off, because that's practical. We teach kids to count 1, 2, 3, 4... We can't very well teach them to count 0.000001, 0.00001, 0.0001, 0.001... (or any of the infinite variations of "counting" without resorting to natural numbers).
Proving that 0.99999 = 1 is an interesting intellectual exercise. But in the real world, we do it every minute of every day.
In other words--eh, close enough.
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Actually, if you define 0.999... as having an infinite number of decimal points, then it is true. And that's how that ellipsis is defined! It means exactly infinite repeating decimals.
You've demonstrated the first hurdle that this problem raises in people's brains: they start thinking about adding "one more" decimal point to the expression, meaning they're thinking of a large but finite number of decimal points. And the second hurdle: people find it hard to believe that you can do mathematics with "infinity" as a meaningful quantity.
No kidding!!! What do you say at this point?
Decimal numbers are just names for points on the real number line (relative to a chosen point we call "0"). Thus one reason 0.999... is equal to 1 is that if they were referring to two different point on the number line, there would have to be a point (acutally infinitely many points) between them. Since every point on the real line can be written as a decimal (this is called the completeness property of the reals), and there is clearly no decimal greater than 0.999... and less than 1, then 0.999... and 1 must be the same point on the real line: the same number.
If pressed, many logicians will admit that the modern foundation of mathematics (ZFC) is probably inconsistent.
See this article:
http://www.math.princeton.edu/~nelson/papers/warn.pdf
The author discusses an informal survey he took among loogicians on page three.
If someone ever discovers a paradox, we can simply scale back to some other system and keep most of what we know, but still...
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0.999.... might be equal to 1, but 0.999 is not equal to 1:
x=0.999
10x = 9.99
10x-x = 9.99 - 0.999
9x = 8.991
x = 0.999
This is true whether there's three 9s or a hundred 9s, so I can see the confusion.
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The series is infinite, you don't lose one.
Just because you can not show the number as a whole does not mean
you can not perform operations using it.
i.e. Think of pi.
I'm not even close to a mathematician, so forgive this possibly very stupid question:
.999... by anything at all? If the sequence is infinity, then any application of task or step can never be completed as it would take infinite time to perform the calculation.
How can you multiply
Conceptually, 0.999... keeps getting closer and closer to 1, as you add more decimal places. It approaches 1. This limit is how all calculus works. Any series that approaches another number as you flesh out the series further and further, will be that number once you have taken the series to infinity.
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a lot of people have this implicit assumption that if a few simple steps bring them to a result that doesn't look like it makes sense, then they did something wrong.
It should be noted that this is not a bad thing. Indeed, it is one of the first things that a good math teacher will teach to the class - all answers should go through a 'does this make sense?' filter before you consider the problem done. It is only very rarely that it causes problems, and it is exceedingly common that it prevents them.
Suppose you have 3 numbers, a, b and c such that c = b - a.
Multiply each side by (b - a) to get:
c(b - a) = (b - a)(b - a) => Or....
cb - ca = b^2 - 2ba + a^2 => Now add (ab - a^2 - cb) to both sides
ab - ca - a^2 = b^2 - cb - ba => Or....
a(b - c - a) = b(b - c - a) => Divide both sides by (b - c - a) and.....
a = b
There you go! Proof that any two numbers (such as 0 and 1) are equal.
(Yes, I know there's a flaw in there. Let's see who'll spot it first.)
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From my sophomore algebra class: .333333... .666666... .999999...
1/3 =
2/3 =
1 =
We sort of had the can't-be-right disbelief the summary expresses until our teacher pointed out that the decimal representations were really limits.
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Okay, but this isn't a problem with the foundation of math being inconsistent, this is a problem with people not knowing how to write the number normally known as "1" in a different way. Most people would grasp "3/3" as being the same as 1, but this *looks* different because they're unused to seeing it.
The fact that the fractions 1/3 (known in decimal notation as .3...) and 2/3 (known in decimal as .6...) have a sum that can be written funny doesn't mean that they don't still add up to 1.
A mathematical amusement causes people confusion and consternation. It's like asking someone why they appear reversed left-to-right in a mirror, but not top-to-bottom, and saying there's an inconsistency in the foundation of physics.
The problem is that partial understanding of a subject and an associated problem in that subject makes things *appear* inconsistent when they are not.
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And how do you do multiplication of an infinite series of digits? I'm guessing that you don't start from the right-hand side.... but beyond that your approach seems to be simple because it is incomplete. Kind of the point of the article really.
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There is no point in trying to convince me: I am well aware that 0.999... = 1. It is a very basic fact about decimals that I was taught in school at a young age.
I said nothing about multiplication by 10. I know that is trivial, but if you read the GP you will see that he claimed it is easy to multiply an infinite series of digits by a number that was not the base. It's not trivial and it requires some definitions far deeper than what he was alluding to. This is why his "simple" proof can only be proven correct using more complex methods.
Try reading the posts that you respond to. It does help.
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Your statement that 1=0! is actually true but your proof is wrong. The factorial of 0 is 1 so 1=0! or 0!=1
The problem with that, though, is that people have trouble accepting that there was nothing wrong with what they did -- a lot of people have this implicit assumption that if a few simple steps bring them to a result that doesn't look like it makes sense, then they did something wrong.
Nope, the problem is that the people who discuss this question are lousy teachers. They set it up deliberately to create a block in other people's minds that makes it unnecessarily difficult for them to understand what is being claimed and why it is true.
If instead they said, "It is possible to represent numbers in different ways. We all know this, and it's completely uninteresting, but I'm going to bore you with it anyway. You know you can represent 1/3 as 0.3333... right? No big deal. Now curiously that also means you can represent 1/1 = (3*1/3) as 3*0.3333... or 0.99999... It's just a different representation of exactly the same value. You can of course also represent 1 as 5*1/5 1/2+1/2 and all kinds of other awkward and unintersting ways, too."
I'm not sure why people insist on presenting this result in the most counter-intuitive way possible and then wasting vast amounts of time trying to undo the damage they've inflicted with their incompetent introduction of the problem. My guess is that they are simply not very smart, as anyone who isn't fairly dumb would see that there is an obvious pedagogical problem at play here, and correct their presentation accordingly, rather than blindly and stupidly repeating the rote "0.9999... = 1" introduction to the remarkably dull fact that you can represent the same value in different ways.
Of course, in an insanely strictly typed language with infinite precision 0.999... would not quite be the same as 1, as the former is a real and the latter is an integer, so despite having the same value their different types would mean they could not be used identically in all circumstances.
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The problem with the argument you present is that people who don't believe 0.999...=1 also don't believe that 0.333...=1/3. They can't quite wrap their heads around the concept of infinity, so in their minds 0.333... continually comes closer to 1/3, but never quite reaches it because they can only imagine a finite number of digits. They honestly think of infinity as being a really large finite number, so they believe that no matter how many digits you add to 0.333..., it never quite reaches 1/3.
Another part of the problem is that many people simply can't wrap their heads around is that they don't separate the idea of a number and the symbols used to represent numbers, thus they cannot grasp that some numbers can be represented in more than one way by our number system.
0.999... is a number. It's not a series or a limit, it's a static and constant number. There's no "as you add on more 9s," because the number is the number. Just because it may take you time (maybe an infinite amount of time) to write it out doesn't mean it's a different number as you write it. If I write out 1,000,000,000, it's 1 billion. And it's always 1 billion. It's not some limit that starts at 1, then goes to 10, then 100, then 1000, until we reach 1,000,000. It's just 1 billion. And 0.999... is just 0.999..., which is also just 1.
As soon as you get to "You know you can represent 1/3 as 0.333... right?", you hit a brick wall. People who believe that 0.999... does not equal one also believe that 0.333... does not equal 1/3, and for many of the same reasons. Taking your approach, you simply shift from arguing about whether or not 0.999... equals one to arguing about whether or not 0.333... equals 1/3. You have to get at the root of the problem of why they refuse to believe those numbers are equal before you can get anywhere.
No, the problem is different. Moreover, it isn't the case that .999... = 1. Here is the problem with the proof in the OP: Take a = .999... = 1 - (1/infinity). So now you multiply a by 10 and you get 10a = 9.999... = 10 - (10 / infinity). Now you want to subtract out "a", so you end up with (10 - (10 / infinity)) - (1 - (1 / infinity)) = 9 - (9 / infinity). Which divided by 9 is not equal to 1.
The floor operation arguably doesn't make much sense for infinite decimal places. I don't know if it's ever used for anything in mathematics except finite-precision numeric methods.
(also, it was a joke. Laugh.)
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