S

Does .9999.... equal 1?

13 Answers
I look at this as more of a philosophy question than math.
F

It does equal 1, which means that any rational number other than 0 that has a terminating decimal expansion can also be expressed as a nonterminating decimal. For instance,

1/4 = 0.25 = 0.24999...

Why does 0.9999... equal 1?

It is a geometric series.

If we add the terms of a geometric sequence

{a, ar, ar^2, ar^3, ...}

we obtain the geometric series

a + ar + ar^2 + ar^3 + ar^4 ... = a(1 + r + r^2 + r^3 + ...)

The sum of the geometric series is the limit of its partial sums. The partial sums of the geometric series are

s_1 = a

s_2 = a + ar = a(1 + r)

s_3 = a + ar + ar^2 = a(1 + r + r^2)

s_4 = a + ar + ar^2 + ar^3 = a(1 + r + r^2 + r^3)

In general, the nth partial sum is

s_n = a + ar + ar^2 + ar^3 + ... + ar^{n - 1} = a(1 + r + r^2 + r^3 + ... + r^{n - 1})

If we multiply 1 + r + r^2 + r^3 + ... + r^{n - 1} by 1 - r, we obtain

(1 - r)(1 + r + r^2 + r^3 + ... + r^{n - 1})
= 1 + r + r^2 + r^3 + ... + r^{n - 1} - r - r^2 - r^3 - ... - r^{n - 1} - r^n
= 1 - r^n

Thus, if r does not equal 1, then

1 + r + r^2 + r^3 + ... + r^{n - 1} = (1 - r^n)/(1 - r)

Hence, if r does not equal 1, then the nth partial sum of the geometric series is

s_n = a(1 + r + r^2 + r^3 + ... + r^{n - 1}) = a(1 - r^n)/(1 - r)

When |r| < 1, the term r^n approaches 0 as n approaches infinity. Hence, the limit of the geometric series is

S = a * 1/(1 - r) = a/(1 - r)

when |r| < 1.

0.9999...
= 9/10 + 9/100 + 9/1000 + 9/10000 + ...
= 9/10(1 + 1/10 + 1/100 + 1/1000 + ...)
= 9/10(1 + 1/10 + (1/10)^2 + (1/10)^3 + ...)

Hence, 0.9999... is a geometric series with a = 9/10 and r = 1/10. Thus, the sum of the geometric series is

0.9999...
= 9/10(1 + 1/10 + (1/10)^2 + (1/10)^3 + ...)
= 9/10[1/(1 - 1/10)]
= 9/10[1/(9/10)]
= 9/10(10/9)
= 1

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N

Hey, I used to drive my prof's crazy too, challenging there statements and often times leaving them wondering if they were wrong or right all along! They'd try to mark one of my problems wrong on a test and I'd almost always figure a way to argue my way out of it. It was funny that the class would get quite and listen as best as they could, waiting in suspense... then they'd hear, "Ok problem number seven has been proven to be unanswerable, so if you got it wrong bring your test to me and you'll get credit for it!" You could hear the whole class give a collective sigh of relief and a little "Yes!" under their voice. What fun it was to mess with them!

Back to our question, I personally have a hard time fully accepting infinity as a natural concept. If the universe were infinite in size or time, then this would be a different argument, but it's believed to be finite. So I find it hard to find a place for the concept in reality. Similarly, infinitesimals are difficult to make sense of in a truly quantized universe! They can be useful in theory, and allow problems that are otherwise unsolvable, possible to work through to their answer. So infinites might be said to be a necessity "behind the curtains" of a working universe, or just abstractions of nature that make it much easier to describe - I tend to favor the latter, and it's fun to make such a proclamation in a room full of mathematicians!

In reality it doesn't really matter. 0.99... will equal 1 whether it truly is possible to have a real world analog or not. it can't mean any other value, so even if infinity is merely a product of a manmade system of mathematics it's not the solution, it only helps us arrive at the solution. Generally, when an infinite is a solution to a problem describing the real-world, it's telling us that the theory is flawed or your using it wrong or the phenomenon being described isn't possible. You will see it come up as "singularities" in physics, it comes up if you try to predict what happens when a massive particle travels at the speed of light... energy => infinity, can't do it! Try to use general relativity to predict the center of a black hole... density => infinity, the theory breaks down. Try merging classical gravity with quantum theory, gravitational attraction => infinity, your using the theories incorrectly (they speak different languages.)

But if you apply infinity and/or infinitesimals to a problem that spits out a usable solution, then it translates easily to real-world meanings or properties or whatever. So, you can see why I (at least) find it difficult to say they are "natural". Then again, I could argue against myself by saying that we can use infinites in our minds, they exist there in concept, and what are we if we're not the universe itself? So technically I could say that infinity does indeed exist in the universe, and even say so objectively!

Finally, here's some extremely straightforward proof of the original problem (1 = 0.999...):

If you take any real number x, and add it to itself (i.e., double it), then subtract the original number, you're left with x. This is about as simple as I think it can be explained: z = x + x; y = z - x; then y = x. You can clearly see by substitution of z, that y = (x + x) - x = 2x - x = x. So now apply this to the question your having trouble with:

x = 0.999...

z = x + x = 0.999... + 0.999... = 1.9999...

y = z - x = 1.999... - 0.999... = 1

Now y = x:

1 = 0.999...


And to dispute that is to say that NO fractions can have infinitely repeating numerals or patterns, like 10/9 = 1.111..., or 1/3 = 3.333..., or 7/6 = 1.1666...! If you work with them in different forms you will understand it better, like this:

3(1/3) = 1/3 + 1/3 + 1/3 = 3/3 = 1/1 = 1

3(0.333...) = 0.333... + 0.333... + 0.333... = 0.999... = 1

3(0.333) = 0.333 + 0.333 + 0.333 = 0.999 =/= 1 (because there are not infinite repetitions; i.e. 0.999 =/= 0.999...)

or

10/9 - 1/9 = 9/9 = 1/1 = 1

Alternatively: ( 10/9 - 1/9 = 1&1/9 - 1/9 = 1 )

1.111... - 0.111... = 1


10/9 - 1 = 10/9 - 9/9 = 1/9

Alternatively: ( 10/9 - 1 = 1&1/9 - 1 = 1/9 )

1.111... - 1 = 0.111... ( = 1/9 )


1/9 * 9 = 9/9 = 1/1 = 1

0.111... * 9 = 0.999... ( = 1 )


You can't just accept some and not others in such a definite, self-consistent science such as math. If you accept: 1/9 = 0.111..., and 9 = 9, and 9/9 = 1/1 = 1, and you accept that multiplication applies the same operation on all numbers, then 1 = 0.999... becomes logically undisputable.

So there you have it, from my point of view.

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F

The fact that 0.9999... = 1 is an artifact of our numeration system. Since it is not possible to create as many symbols as there are numbers, we find it convenient to use a base 10 numeration system. When we write 125, we mean

1 x 100 + 2 x 10 + 5 x 1

The cost of doing this is that not every number has a unique decimal representation. In particular, any rational number except 0 that has a terminating decimal expansion also has a representation as a repeating decimal that ends in infinitely many 9's. Since the geometric series

0.9999... = 9/10 + 9/100 + 9/1000 + 9/1000 + ... = 9/10[1/10 + (1/10)^2 + (1/10)^3 + ...] = 1

the number

125
= 1 x 100 + 2 x 10 + 5 x 1
= 1 x 100 + 2 x 10 + 4 x 1 + 1
= 1 x 100 + 2 x 10 + 4 x 1 + 9 x 1/10 + 9 x 1/100 + 9 x 1/1000 + 9 x 1/10000 + ...
= 124.9999....

If we were working in a different base, we would encounter the same issue. The binary (base 2) number

0.11111... = 1

since the binary number

0.11111...
= 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + ...
= 1/2[1 + 1/2 + (1/2)^2 + (1/2)^3 + (1/2)^4 + ...]
= 1/2 * 1/(1 - 1/2)
= 1/2 * 1/(1/2)
= 1/2 * 2
= 1

By representing a number in base b, where b is 10 (decimal) or 2 (binary) or something else, we create a system in which the number 1 has two representations, the terminating decimal expansion 1 and the nonterminating decimal 0.aaa..., where a = b - 1.

This is how Georg Cantor showed that the real numbers are uncountable. He considered the numbers in the open interval (0, 1). If the numbers in that set were countable, we could write down a list of each real number in this interval. To guarantee a unique representation, we choose the nonterminating decimal expansion of numbers with terminating decimal expansions (for instance, 1/2 = 0.5 = 0.49999...). This would yield a sequence {a_1, a_2, a_3, ...} consisting of all the real numbers in the interval (0, 1). However, as Cantor argued, we could form a new number by changing the first digit of a_1, the second digit of a_2, the third digit of a_3, and, in general, the nth digit of a_n. Since that number differs from a_1 in the first decimal place, a_2 in the second decimal place, a_3 in the third decimal place, and so forth, it could not have been on our original list. Hence, we have found a real number that was not on our list, so the set of real numbers in the interval (0, 1) is uncountable.

Note that the set of real numbers in the interval (0, 1) is uncountable even though we gave each number in the interval a unique decimal representation. The fact that the real numbers are uncountable does not depend on our choice of numeration system. The fact that rational numbers with terminating decimal representations do not have a unique decimal representation is an artifact of our numeration system.

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N
Lynden90 - Just because a value exists, does not make it measurable. This is too narrow of a scope to answer an question involving any type of infinite values. The value 0.99... when applied to distance, exceeds the resolution of Plank's constant, and therefore you cannot say it never reaches one definitively. Within the framework of mathematics it does, so we must draw the abstract conclusion that the most reasonable answer in a real world situation would be that it, too, equals one. If the two were not equal, then we find a paradox that presents itself. We must pass the point measuring 0.999... in the given unit system before we can possibly reach point 1 when walking across an arbitrary distance. So how could we ever reach distance 1 before passing 0.999.... if the latter is not a real meaningful value in reality?
Passio25 - that's already been stated (probably several times) as "0.999 =/= 0.999...". See answer 10/12 by me [ Newibr81 ] for one such instance.


I realize it [0.999...] appears to imply the existence of an asymptote at 1 to the average student, and this would be true if it were not for the fact that it becomes infinitely closer to one. One can theoretically NEVER reach infinity - by definition, this is why it is described as a "concept", not a number, and it's easier to talk about in terms of limits. But speaking "outside of the mathematical box" as it were, if and when you DID reach infinity (which is what the 'repeating' symbol '...' indicates,) 0.99999... WOULD finally EQUAL 1. And that is precisely what is occurring here.

"Numbers don't' exist in nature" is a pretty difficult hypothesis to defend! Sure the physical symbols or other manmade methods of representing them is not how they exist in nature, but numbers are mere symbolic and/or mental representations of an analogous concept in the real-world... "a number by any other name..." Mathematics may be a 'black-box theory', but nevertheless it gives the same, correct answer 99.999.....% of the time (I hope you caught that one!) Aside from Godel's incompleteness theorem, there has been nothing that suggests an inconsistency between numerical representations nor the formulae that we use to give accurate theoretical representations of real-world phenomena and that which we determine the results to actually be, i.e. empirical = theoretical, within the mathematically expected degree of precision.

So, ultimately, the answer to the question is simply, "Yes, 1 = 0.999...!"

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N

Hi Scooby48

By definition, no two numbers exist that do not have a number between them. You have effectively answered your own question in your previous post. Since there are no numbers that exist between 1 and 0.9999., then they cannot be unique numbers. Therefore they are equal by definition, so you could write a triple bar equal sign between 1 and 0.999.., and interchange them in any part of any equation. Consider them two representative symbols of the identical number, namely the number one (1).

To fully appreciate this concept, I suggest you don't take my answer or even that of our resident expert, Forbie89 at face value. Go and get a pen and paper, a math book and any material you can locate on the subject and develop a real deep understanding and appreciation of this basic number theory concept as well as that of infinitesimals by working it through yourself, inside and out. These two concepts relate to one another beautifully in this nearly perfect relationship for studying these concepts at the most fundamental level possible.

You unwillingness to accept what you don't understand, even when such a superb set of proofs has been presented to you is truly a virtue. It's really quite uncommon to find this trait among scientists/mathematicians and many consider it quite annoying, but it's important that you learn these concepts and become fluent with them in your mind, not because someone showed you an elegant proof. That's not learning, and your entire understanding of mathematics will become a fragile hollow shell if you were to take such concepts as factual and move on just because they are proven to you. Open a book and a notebook, and finally open your mind. Work it forwards and backwards ten different ways, then teach someone else the relationship, no, convince someone else that doesn't believe you when you first meet. THEN and only then will you speak this language of mathematics as fluently as you speak your native language.

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F

No such number exists.

Let x < 1. Then the number

(x + 1)/2

satisfies the inequalities

x < (x + 1)/2 < 1

Therefore, given any real number x < 1, it is always possible to find a larger real number that is also less than 1, namely its average with 1.

In case you are interested, here is a proof that

x < (x + 1)/2 < 1

We must show that

(x + 1)/2 - x > 0

This inequality is equivalent to the statement that

x + 1 - 2x > 0

which we obtain by multiplying both sides of the inequality by 2. Simplifying yields

1 - x > 0

Adding x to each side of the inequality yields the equivalent inequality

1 > x

which is true by assumption. Hence, the assertion that

x < (x + 1)/2

is equivalent to our hypothesis that x < 1. It remains to show that

(x + 1)/2 < 1

This inequality is equivalent to the statement that

x + 1 < 2

Subtracting 1 from each side of the inequality yields the equivalent statement that

x < 1

which was our hypothesis. Hence, if x < 1, then

x < (x + 1)/2 < 1

as claimed.

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S

Thanks for your reply Newibr81
Forbie89 has been very helpful (and patient) on this post as well as others and I am truly appreciative for that. I hope I don't get too annoying by prying into the subject matter deeper. I am pretty sure I frustrate some of my professors to no end with questions occasionally.

I believe I am understanding the nature of the situation better. If .999. and 1 were distinct numbers it would imply that the set of real numbers is countably infinite, which it certainly isn't. I am not sure if that is correct but that seems to make sense to me.

Initially I took the question to be more of a metaphysical question then a question of mathematics. Now I am starting to see that my question is rooted deeply in the foundation of mathematics and really allows for no debate.

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L

The number .999 repeating exists in theory, but it is a number without any practical value as it is impossible to measure something and get this value as a result. This is because to measure something is to limit and to limit means that .9 cannot repeat infinitely.

Within the framework of mathematics .999 repeating does equal 1. But in practice, .999 repeating would never come up as a value. The value measured would always be 1 in this case. So to answer your question regarding whether or not .999 repeating exists in nature I would say that it does not because numbers don't exist in nature, as I see it, rather they exist in our minds to help us make sense of existence, or at least the physical aspects of existence.

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P

.999 with a line above, meaning an infinitely repeating .9999999999999 essentially means 1. If, however, the decimal is finite, such as 0.999990000000000 (repeating 0's forever), this number by no means is equal to 1.

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C

Yes, .999... (repeating) equals one. A simple way to prove this to divide one by three, then multiply that by three.

1 / 3 = 0.333...

0.333... * 3 = 0.999...

1 = 0.999...

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G

Assuming you mean .9999.... as in repeated, no it does not equal one. It's like being 99.9999999... % to completion, but you're still not complete.

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S

For those of you who responded saying they are equal, how would you respond to someone asking "What is the number directly less then 1?"

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C

Yes these are equal. There are no possible numbers between the two and so they must be the same.

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