Does 0.9999 repeating 9 equal 1?

[Larry]

I think I've just realized what the problem is.

If you'll notice, the ones in this thread that refuse to accept that .999 repeating equals one and refuse to accept that .333 repeating equals one-third - all these members are YOUNG. They're kids! Look at their ages! They're like nineteen. :doh:

These kids have done math on calculators all their lives! And because of that, they are limited in their thinking and in their minds. They are limited to eight place marks or sixteen place marks or however many their respective computer/calculator goes to. That's why they don't understand the concept of, for example, .333 repeating equals EXACTLY one-third. Their calculator only tells them it's close. But not exact. And they believe it!!! Their calculators have been giving them bad information for their whole lives in respect to infinitely repeating numbers. And these kids have never known anything different.

Those of use who grew up doing math with a pencil and a sheet of white paper can think outside 16 place marks. That's why we understand that .333 repeating is EXACTLY one-third.

And I bet none of them ever watched Captain Kirk tell a computer to calculate pi to the last digit, either.

[Mark The Homer]

Or the older generation has not been exposed to the higher level of math actually required to understand the concept of limits and infinity. I will qualify myself here by stating I am currently getting my graduate degree in statistics at GW right now and have spent a lot of time with higher level math courses and especially limits. Infinity for example is not a tangible number. Infinity only exists conceptually. .9999999.... APPROACHES 1 as the number of digits reaches infinity. It never gets to 1. NEVER! Its an asymptote to 1 like many others have said. The reason for limits is to approximate values as they become extremely large to turn a continuous value into something discrete so it can be worked with. Its a manipulation in order for mathematicians to be able to work with functions. .99999.... is 1 for all practical purposes meaning if we were to use it in some way but only because we have manipulated it. The true value is and never will be 1.

Okay. Then how come you guys can't see what, to us, is clear as day?

We're not talking about infinity. We're talking about a value expressed with infinite trailing numbers. It's apples and oranges.

I'm going to ask you to unclutter your mind, remove all that crap about limits and differentials, clear it all out, and take a fresh look at the following.

Ready? Here we go.

Look at the relationship between these two values expressed in fractions:

one-half and one-third.

Now, look at the relationship between these two values expressed in decimals:

.5 and .33333333333333 (infinitely repeating 3's)

What can we conclude from this? Put your pride aside for one moment and give us an honest answer.

[jnhay]

I think I've just realized what the problem is.

If you'll notice, the ones in this thread that refuse to accept that .999 repeating equals one and refuse to accept that .333 repeating equals one-third - all these members are YOUNG. They're kids! Look at their ages! They're like nineteen. :doh:

These kids have done math on calculators all their lives! And because of that, they are limited in their thinking and in their minds. They are limited to eight place marks or sixteen place marks or however many their respective computer/calculator goes to. That's why they don't understand the concept of, for example, .333 repeating equals EXACTLY one-third. Their calculator only tells them it's close. But not exact. And they believe it!!! Their calculators have been giving them bad information for their whole lives in respect to infinitely repeating numbers. And these kids have never known anything different.

Those of use who grew up doing math with a pencil and a sheet of white paper can think outside 16 place marks. That's why we understand that .333 repeating is EXACTLY one-third.

So you think we learned math with calculators? Maybe you should ask an elementary school teacher about that?

I understand that no other fraction can be used as .3 repeating other than 1/3, but that still doesn't convince me that we should say that .3 repeating is the EXACT answer.

[Mark The Homer]

So you think we learned math with calculators? Maybe you should ask an elementary school teacher about that?

I understand that no other fraction can be used as .3 repeating other than 1/3, but that still doesn't convince me that we should say that .3 repeating is the EXACT answer.

You shouldn't have to be convinced. You should be able to see it without the need to be convinced.

[PeterMP]

So you think we learned math with calculators? Maybe you should ask an elementary school teacher about that?

I understand that no other fraction can be used as .3 repeating other than 1/3, but that still doesn't convince me that we should say that .3 repeating is the EXACT answer.

Seriously, don't take our word for it. At least some of you that doubt it are still in school. Go find a Calc teacher that you trust and ask them.

***EDIT***

And remeber we aren't talking about the limit as the number of 9s goes to infinity, but the actual number with an infinite number of 9s.

[jnhay]

You shouldn't have to be convinced. You should be able to see it without the need to be convinced.

That doesn't make any sense.

[JMUGator19]

I think my two favorite proofs are

finding a number between .999... and 1

and

1/3 = .333... X 3 = .999... therefore 1 = .999...

[Mark The Homer]

For the naysayers:

http://wiki.answers.com/Q/What_is_one_third_as_a_decimal

http://mathforum.org/library/drmath/view/55845.html

http://math.wikia.com/wiki/Proof:The_Decimal_0.999..._is_Equivalent_to_1

http://uk.answers.yahoo.com/question/index?qid=20090225122725AApnPuX

http://www.ehow.com/how_2239655_change-fraction-decimal.html

Found this on usenet:

http://groups.google.com/group/rec.puzzles/browse_thread/thread/7e94aa5aea335146/b7e0f1d98afeb4dc?hl=en&ie=UTF-8&q=decimal+one+third

We're laboring under the assumption that when you write "0.333..."

you mean "decimal point followed by repeated 3s forever." A formal

_

notation for this is 0.3 (horizontal bar over the 3).

If this is not what you mean, if instead you mean "decimal point

followed by a lot of 3s, for example maybe a million or two," then

let's stop right now, because such a number is not equal to 1/3.

If you really mean "3s forever" then it is obvious that you cannot

write all the digits of this number. If you were to attempt to

write down all the digits, you would fail: you would either use up

all the paper and ink in the world or die first.

All the division algorithm can do for you is to generate the digits

of 0.333... one at a time. If after each step you still have a non-zero

remainder, well, you haven't finished the division yet!

>Besides since 1/3 * 3 = 1... shouldn't 0.333... * 3 be 1 as well then?

Of course it is, because 0.999... = 1. But that begs the question

of what an infinitely long decimal means in the first place, so I

don't particularly like this as a "proof" that 0.999... = 1.

>Is 1/3 = 0.333... or is it just a convention that is used because it is

>otherwise impossible to feed 1/3 into a computer (or for another reason,

>point is, is it a convention)?

It certainly has nothing to do with computers, since any real computer

is completely incapable of storing an infinite number of decimal digits.

I suppose you might regard the notation as a "convention." I prefer

to call it an idealization; it is mathematics, not mere arithmetic.

The long division algorithm is arithmetic and will never actually

generate more than a finite number of digits.

In mathematics, however, we can define infinite sets and manipulate

them according to well-defined, consistent rules, so we can handle

infinite decimals provided that we make appropriate definitions of

what we want to do with them. It turns out that there is a very

useful set of definitions that identifies each infinite decimal with a

real number. This makes it easy to prove that there are more real

numbers than rational numbers (by the Cantor diagonalization proof).

According to these definitions, we would write 0.333... = 1/3 and

0.999... = 1. A disadvantage of these definitions is that there are a

few real numbers (anything of the form n*10^(-m) for n, m integers)

that end up with two representations, e.g. 0.999... = 1.000...

and 0.124999... = 0.125000... (the latter example, IIRC, is from

Tom Apostol's _Calculus_), but this is a relatively minor problem.

>third question: should I ask this in sci.math?

I would recommend reading the sci.math FAQ first, which will give

another explanation of why 0.999... = 1.

David A. Karr

I googled for the sci math faq...

Here's the proof within - two proofs actually, one using limits... references included...

http://faqs.cs.uu.nl/na-dir/sci-math-faq/0.999999.html

In conclusion, the answer to the question in the thread title in an unequivocal YES.

/thread

[edgun88]

Any number repeating is not a real number. It goes on for an infinite number of decimal places. It is not exactly equal to anything. 1 is a real number. Only a real number can equal a real number.

[Spec138]

Any number repeating is not a real number. It goes on for an infinite number of decimal places. It is not exactly equal to anything. 1 is a real number. Only a real number can equal a real number.

Ughhh :doh::doh::doh:

[Thinking Skins]

Any number repeating is not a real number. It goes on for an infinite number of decimal places. It is not exactly equal to anything. 1 is a real number. Only a real number can equal a real number.

Can you state whether thats opinion or fact?

And if you call it a fact, then whats that based off of?

Can you state the definition of a real number?

Can you state the definition of a rational number?

Can you state how the decimal representation of a rational number relates to the number itself?

I don't feel like getting back into the deep end of this argument, but here's a quote from wikipedia:

http://en.wikipedia.org/wiki/Rational_number

The decimal expansion of a rational number always either terminates after finitely many digits or begins to repeat the same sequence of digits over and over. Moreover, any repeating or terminating decimal represents a rational number.

[Mark The Homer]

Any number repeating is not a real number. It goes on for an infinite number of decimal places. It is not exactly equal to anything. 1 is a real number. Only a real number can equal a real number.

:doh:

[Spec138]

It's sort of funny that people will have blind faith in so many things, but whenever something is spelled out and proven they won't.

Note: .999... is just another way to say 1.

[Thinking Skins]

Thanks MTH. I just learned something new:

http://faqs.cs.uu.nl/na-dir/sci-math-faq/0.999999.html

Another informal argument is to notice that all periodic numbers such

as 0.46464646... are equal to the period divided over the same number

of 9s. Thus 0.46464646... = 46/99. Applying the same argument to

0.9999... = 9/9 = 1.

but once again, we have the same proof that people in this thread have been repeating:

In modern mathematics, the string of symbols 0.9999... is understood

to be a shorthand for ``the infinite sum 9/10 + 9/100 + 9/1000 +

...''. This in turn is shorthand for ``the limit of the sequence of

real numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ...''. Using

the well-known epsilon-delta definition of the limit (you can find it

in any of the given references on analysis), one can easily show that

this limit is 1. The statement that 0.9999... = 1 is simply an

abbreviation of this fact.

And this is from sci.math.

[Spec138]

Thanks MTH. I just learned something new:

http://faqs.cs.uu.nl/na-dir/sci-math-faq/0.999999.html

That's actually a really simple way to prove it.

Though you have to accept the part about the repeating part over that many 9s. I don't know how you prove that.

[Mark The Homer]

Thanks MTH. I just learned something new:

http://faqs.cs.uu.nl/na-dir/sci-math-faq/0.999999.html

Did you know this?

Put your hands on the table.

To calculate 9x whatever....

Fold the appropriate finger and count the fingers on either side.

ex. 9 x 7 .... fold the 7th finger ... answer 63 ...

...and so on

Here's something else.

11 x a two digit number...

add the two digits and stick it in the middle .... that's your answer

ex 11 x 27 .... Add the two and the seven ... answer is 297.

...and so on...

(this is what we had to do before calculators were invented

[Spec138]

That's neat, though for 9x something I've always just multiplied by 10 and subtracted whatever I was multiplying by 9.

[Larry]

(this is what we had to do before calculators were invented

Were you going to school in West Virginia, or something? We used pencil and paper.

(There. That ought to hijack this thread.)

[Thinking Skins]

.3 repeating is not exactly one third. It's just the closest number to 1/3 that we use for simplicity.

In the real numbers there's no such thing as a 'closest number to 1/3'. If .3333... was not 1/3, then they would be two distinct numbers. So then by the density property of the real numbers there would be a number between them, which would make the number between them closer to 1/3 than .3333...

So what you're saying contradicts the density property of real numbers. Are you saying you don't believe that either?

Or are you saying that .33333... is not a real number?

Lets take that to task. Is .3333... a real number?

Well .3333... is shorthand for writing the series

.3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity.

This is an infinite series. Does this series converge?

Well, lets look at this sequence.

{.3, .33, .333, .3333, ...}

Question 1: Is this a bounded sequence? A sequence is bounded above is there is a real number M such that a_n <= M for all n.

Well, we know that a_n < 1/3 for every term in this sequence, so this sequence is clearly bounded above.

A sequence is bounded below if there is a real number N such that N <= a_n for all n. Well since 3 > 0 and 10^n > 0, we know that 3/10^(n) > 0. This implies that each term a_n > 0. So 0 serves as a lower bound.

A sequence is a bounded sequence if it is bounded above and below. Then we see that our sequence is a bounded sequence.

Question 2: Is this a monotonic sequence?

A sequence {a_n} is monotonic if its terms are nondecreasing or nonincreasing:

a_1 <= a_2 <= a_3 <= a_4 <= ...

In this series we have a_{i+1} = a_i + (3/10^(i+1)) and since 3 > 0 and 10^(n) > 0 for all n, we know that (3/10^(n)) > 0 for all n. So this is a monotonic (nondecreasing) sequence.

We know that if a sequence has an upper bound, then it has a least upper bound - an upper bound that is smaller than all other upper bounds for the sequence.

We see then that 1/3 is the least upper bound for this sequence and we can thus use the following theorem from calculus

Then from calculus we have a theorem: If a sequence {a_n} is bounded and monotonic (nondecreasing), then it converges (to its least upper bound).

Getting back to the original sum

.3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity.

what we can see is that the sequence, {.3, .33, .333, .3333, ...}, is special too.

Each term in the sequence {.3, .33, .333, .3333, ...} is called the nth partial sum, S_n of the series above.

The definition of convergence of a series is that if the sequence of parital sums {S_n} converges to some number M, then the series converges to that number M.

So since we just showed that the sequece {.3, .33, .333, .3333, ...} converges to 1/3, we get that the series .3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity converges to 1/3 as well.

This is written as 1/3 = .3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity

[jnhay]

In the real numbers there's no such thing as a 'closest number to 1/3'. If .3333... was not 1/3, then they would be two distinct numbers. So then by the density property of the real numbers there would be a number between them, which would make the number between them closer to 1/3 than .3333...

So what you're saying contradicts the density property of real numbers. Are you saying you don't believe that either?

Or are you saying that .33333... is not a real number?

Lets take that to task. Is .3333... a real number?

Well .3333... is shorthand for writing the series

.3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity.

This is an infinite series. Does this series converge?

Well, lets look at this series

{.3, .33, .333, .3333, ...}

Question 1: Is this a bounded sequence? A sequence is bounded above is there is a real number M such that a_n <= M for all n.

Well, we know that a_n < 1/3 for every term in this sequence, so this sequence is clearly bounded above.

A sequence is bounded below if there is a real number N such that N <= a_n for all n. Well since 3 > 0 and 10^n > 0, we know that 3/10^(n) > 0. This implies that each term a_n > 0. So 0 serves as a lower bound.

A sequence is a bounded sequence if it is bounded above and below. Then we see that our sequence is a bounded sequence.

Question 2: Is this a monotonic sequence?

A sequence {a_n} is monotonic if its terms are nondecreasing or nonincreasing:

a_1 <= a_2 <= a_3 <= a_4 <= ...

In this series we have a_{i+1} = a_i + (3/10^(i+1)) and since 3 > 0 and 10^(n) > 0 for all n, we know that (3/10^(n)) > 0 for all n. So this is a monotonic (nondecreasing) sequence.

We know that if a sequence has an upper bound, then it has a least upper bound - an upper bound that is smaller than all other upper bounds for the sequence.

We see then that 1/3 is the least upper bound for this sequence and we can thus use the following theorem from calculus

Then from calculus we have a theorem: If a sequence {a_n} is bounded and monotonic (nondecreasing), then it converges (to its least upper bound).

Getting back to the original sum

.3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity.

what we can see is that the sequence, {.3, .33, .333, .3333, ...}, is special too.

Each term in the sequence {.3, .33, .333, .3333, ...} is called the nth partial sum, S_n of the series above.

The definition of convergence of a series is that if the sequence of parital sums {S_n} converges to some number M, then the series converges to that number M.

So since we just showed that the sequece {.3, .33, .333, .3333, ...} converges to 1/3, we get that the series .3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity converges to 1/3 as well.

This is written as 1/3 = .3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity

Ow. Okay, you win.:allhail:

[Thinking Skins]

Here's another proof that

.3 + .03 + .003 + .0003 + ... + 3/10^n + ... as n approaches infinity is 1/3.

We can write this as

sum_{n = 0 to infinity} (.3 * 10^n)

We can see this as a geometric series, where the first term is .3 and the common ratio is 1/10.

Then the theorem on the convergence of a geometric series says that the series converges if the absolute value of the ratio is between 0 and 1 and the formula to calculate the sum is

a / (1 - r),

where a is the initial term and r is the radius.

In this case, the ratio's absolute value is 1/10, which satisfies these conditions, so this series converges.

What does it converge to?

a / (1 - r) = .3 / (1 - (1/10)) = .3 / (9/10) = .3 / .9 = 1/3.

What about the series .9 + .09 + .009 + .0009 + ... + 9/10^n + ... as n approaches infinity?

Here a = .9 and r = 1/10.

a / (1 - r) = .9 / (1 - (1/10)) = .9 / (9/10) = .9 / .9 = 1.

Q.E.D.

[Thinking Skins]

Ow. Okay, you win.:allhail:

can the congregation say Amen!!!

[Spec138]

Ow. Okay, you win.:allhail:

The math...

It hurts.

[abdcskins]

8+3= 12. At least I know that.

[Thinking Skins]

Another informal argument is to notice that all periodic numbers such

as 0.46464646... are equal to the period divided over the same number

of 9s. Thus 0.46464646... = 46/99. Applying the same argument to

0.9999... = 9/9 = 1.

I can also prove this using the geometric series argument that I just stated above. We'd replace the initial term by whats being repeated and the ratio by 1/10^k, where k is the number of digits that are being repeated.

So if we have 0.464646464646...., then a = .46 and r = 1/100

a / (1 - r) = .46 / (1 - 1/100) = .46 / .99 = 46 / 99, as stated in the quote.

This is cool stuff!