Math Question

[Master of Cards]

.9 repeating is not equal to one. I'm trying to get through to this idiot and he wants math to prove it. I tried to do logic allegorys like lightspeed but it didnt get through to him, it devoled into a screaming match, we made a bet to see who can get proof.
My question is can someone show me a way to do math to show .9 repeating is not infact 1? or am i wrong and hes right?

[Fire Fly]

I don't have a definitive answer for you, but the other math-versed members have discussed this prior.

[TheLemur]

.9 repeating is one. There are several proofs here.

[SeeingRed]

.9 repeating IS in fact, equal to 1.

There are many ways to prove this, but here's a simple one I can think of offhand:

consider n = .999... [1]
then 10*n=9.999... [2]
subtracting [1] from [2]: 9n=9
and we obtain: n=1.

I can give you others if you'd like, but this should suffice for most purposes I should think.

[Sriad]

There is no number X such that
1-x=.999...
therefore 1=.999...

No, 0.000...1 ISN'T a number. It's a smart-ass way of writing 0. And A+0=A.

[Sriad]

OR:

1/3=.333...
multiply both sides by 3:
3/3=.999...=1

If you can't accept that A/A=1 we'll need to take away your math license.

[Andrew_Fireborn]

Or, for the average person:

It's one because it repeats infinately, meaning that it would reach a point at which there is no practical or decernable difference between it and 1.

And really, it's just an artifact of our 10 based system being incompatible with the number 3... (not that I really see a 12 or whatever other number system being any better in any respect.)

[Nova Andromeda]

It's one because it repeats infinately, meaning that it would reach a point at which there is no practical or decernable difference between it and 1.

-That is exactly wrong and probably what confuses most people. An infinite quantity DOESN'T ever "reach a point" (i.e., end).
-The whole question is basically answered depending on how you define things. Under the common definitions .9999... is 1- 1/x where x is infinity and 1/infinity = 0 NOT 1/infinity is approximately/nearly equal to 0: http://mathworld.wolfram.com/Infinity.html.
-The trick with anything involving infinity is to determine exactly how it is defined.

[Andrew_Fireborn]

Functionally, a number is a measure of something.

The point it would reach to being rediculous, would be at which the numbers are representing the difference between electrons. Having reached it's practical limit several orders of magnitude prior...

Therefore, my explination makes the most sense to myself. (although I suppose that precludes the average person tag I gave it...)


Also, I'd contest the thought of reaching something being the terminus of whatever reached it, rather than being another manner of measurement most applicable to time, but I'm not in the mood, or particularly well armed for such a contest.

[The Spartan]

A variation on what Sriad did:

1/3=0.33333...
2/3=0.66666...
the sum of 1/3 and 2/3 is 3/3=0.99999...

therefore 0.99999...=1

QED

[Ariphaos]

Proving the first two is an entirely different topic, but anyway:

1: An infinite amount of rational numbers exist between any two irrational numbers.
2: An infinite amount of irrational numbers exist between any two rational numbers.
3: Therefore, there exists no pair of numbers wherein another number cannot fit between them.
4: No number exists between .999... and 1
5: Therefore, .999... and 1 are not a pair.
6: Since they are in fact numbers, they must represent the same number.

[Surlethe]

What you have to keep in mind is that decimals are simply sequences, and are governed by the rules of convergent sequences. In this case, 0.9999... converges to 1; we therefore consider it equal to 1.

[petesampras]

Is 0.999 recurring an actual number?

The various types of numbers and their representations are defined, but I'm not sure any of them contain the means to express 0.999 recurring as a number.

In which case asking if 0.999 recurring equals 1 is a meaningless question, the answer would be undefined. Like asking if 'w' equals 1.

[Gaidin]

Is 0.999 recurring an actual number?

The various types of numbers and their representations are defined, but I'm not sure any of them contain the means to express 0.999 recurring as a number.

In which case asking if 0.999 recurring equals 1 is a meaningless question, the answer would be undefined. Like asking if 'w' equals 1.

The Reals?

The odd part is that with the 3/3=.999... idea is that I'm not sure I could tell you off the top of my head whether its rational or irrational...my mind goes both ways on that for what I hope are apparent reasons.

[petesampras]

Is 0.999 recurring an actual number?

The various types of numbers and their representations are defined, but I'm not sure any of them contain the means to express 0.999 recurring as a number.

In which case asking if 0.999 recurring equals 1 is a meaningless question, the answer would be undefined. Like asking if 'w' equals 1.

The Reals?

The odd part is that with the 3/3=.999... idea is that I'm not sure I could tell you off the top of my head whether its rational or irrational...my mind goes both ways on that for what I hope are apparent reasons.

How would you write down 0.999 recurring as a real number? Before you can decide whether a number belongs to the set of real numbers, you must have a means of formaly defining it. What I'm asking for is 0.9999 recurring expressed according to one of the formal methods for constructing numbers.

[Wyrm]

Is 0.999 recurring an actual number?

Yes.

1 and 0.9999... are both representations of real numbers. As it so happens, they are the representation of the same real number, because the real number represented by 0.9999... is defined as the limit of the sequence S=(9/10, 99/100, 999/1000, ...), which has the limit of 1, and therefore the decimal 0.9999... = 1. It's not difficult to show this.

Remember, just because the representation looks different doesn't mean that they don't name the same real number. sin(π/2) looks different from both 1 and 0.9999..., yet all of them are equal to each other.

The odd part is that with the 3/3=.999... idea is that I'm not sure I could tell you off the top of my head whether its rational or irrational...my mind goes both ways on that for what I hope are apparent reasons.

1 = 3/3 = 0.999... is manifestly a rational number. Rationality (and for that matter, irrationality) is a property of real numbers, not representations. A rational number is by definition equal to a ratio of two whole numbers. Since the notation '0.9999...' represents the real number 1, and 1 is equal to a/a for any integer a, 1 is obviously rational.

π, on the other hand, is irrational, not because it has an infinite decimal representation, but because it is not equal to any ratio between two whole numbers. That π cannot be equal to any ratio of whole numbers automatically precludes it from being represented by anything but an infinite, nonrepeating decimal.

[Ariphaos]

Is 0.999 recurring an actual number?

The various types of numbers and their representations are defined, but I'm not sure any of them contain the means to express 0.999 recurring as a number.

You just did, what's the problem?

There is no rule that declares a given number must be assigned a single, specific representation. Only convention. In the case of that, it's

. .___
0.999

Or 1, which it equals.

(ignore the periods above the zero)

In which case asking if 0.999 recurring equals 1 is a meaningless question, the answer would be undefined. Like asking if 'w' equals 1.

You don't get an undefined answer to a boolean question. You have true, false, or indeterminable.

[petesampras]

Is 0.999 recurring an actual number?

The various types of numbers and their representations are defined, but I'm not sure any of them contain the means to express 0.999 recurring as a number.

You just did, what's the problem?

There is no rule that declares a given number must be assigned a single, specific representation. Only convention. In the case of that, it's

That's a bit of a strawman, I'm not saying that a number must have only way means of representation. 1 and 4/4 both represent the same number, both are validly represented. What I'm asking is how 0.9999 recurring is represented via one of the formal methods for contructing numbers.

. .___
0.999

Or 1, which it equals.

(ignore the periods above the zero)

In which case asking if 0.999 recurring equals 1 is a meaningless question, the answer would be undefined. Like asking if 'w' equals 1.

You don't get an undefined answer to a boolean question. You have true, false, or indeterminable.

'Equals' is a function. A function is only defined for certain types of argument, in this case numbers. Give it something which is not a number and it is undefined, this is not the same as indeterminable.

[Wyrm]

How would you write down 0.999 recurring as a real number?

I can write the number represented by '0.999...' more simply as '1', since the two representations name the same real number. However if you want the number itself, then you can't for any real number period. In that case, I guess math is built on sand.

Before you can decide whether a number belongs to the set of real numbers, you must have a means of formaly defining it. What I'm asking for is 0.9999 recurring expressed according to one of the formal methods for constructing numbers.

Bullocks. There are plenty of numbers we don't have a finite representation for. Indeed, most numbers do not have any way of expressing them at all in a finite way.

Think about it: the set of all finite sequence of operations is manifestly countable. If each sequence of operations represents a number, then the set of all numbers specified by finite sequences of operations must also be countable. But the set of all real numbers is uncountable (by Cantor's diagonalization proof). Therefore, most real numbers will be unrepresentable by a finite sequence of operations.

[petesampras]

. .___
0.999

Ooops, sent the reply before responding to this bit.

Do you have a link to this means of expressing numbers?

[Wyrm]

That's a bit of a strawman, I'm not saying that a number must have only way means of representation. 1 and 4/4 both represent the same number, both are validly represented. What I'm asking is how 0.9999 recurring is represented via one of the formal methods for contructing numbers.

Yes, 0.9999... is a perfectly legitimate way of representing a real number. It is the limit of the sequence S = (9/10, 99/100, 999/1000, ...), where S_n (the n'th member of sequence S) is ∑_{i=1}^n 9 (1/10)^i. Limits do not need the representation 0.999..., because 0.999... is defined as a kind of limit. And limits are real numbers.

I can give you the full proof that 1 = 0.999... if you ask.

'Equals' is a function.

Wrong. 'Equals' is a logical assertion: x = y is defined such that, for any predicate P(∙), P(x) if and only if P(y).

A function is only defined for certain types of argument, in this case numbers. Give it something which is not a number and it is undefined, this is not the same as indeterminable.

There are plenty of mathematical objects that are 'not a number' yet are very much defined. A square of side 9 is perfectly defined, yet is not a number. And a square (meaning a geometrical shape) is certanly not equal to any real number. For instance, a real number certainly is not a four-sided polygon, because it isn't a polygon, whereas a square is a four-sided polygon. Therefore, there is one predicate P(∙) for which the statement P(a square) iff P(a real number) is false. Therefore, the two are not equal.

[Ariphaos]

That's a bit of a strawman, I'm not saying that a number must have only way means of representation. 1 and 4/4 both represent the same number, both are validly represented. What I'm asking is how 0.9999 recurring is represented via one of the formal methods for contructing numbers.

Where horizontal bars and other means of representation are not possible or feasible, an ellipses works, and conveys the idea properly.

This is more formal, of course, but for some reason HTML does not support an Overline tag.

'Equals' is a function. A function is only defined for certain types of argument, in this case numbers. Give it something which is not a number and it is undefined, this is not the same as indeterminable.

Undefined means that something does not have meaning, the typical posit being 0/0.

A true or false question has a limited set of answers, which are all meaningful, as far as I'm aware. It can be true, false, or unprovable. Certain wonky logic allows for both true and false to apply, but we're not dealing with anything remotely like that - this thread's seen about a dozen different proofs that .999... equals 1, it's not even up for debate.

[drachefly]

Another way of solving this is to change from normal base 10 to base 10 with the digits
-4 -3 -2 -1 0 +1 +2 +3 +4 +5

With the effect that 76 would be represented +1 -2 -4

in this case, 0.999 repeating takes on the decimal representation
1.0000 with a -1 digit pushed off to infinity.

[Surlethe]

How would you write down 0.999 recurring as a real number? Before you can decide whether a number belongs to the set of real numbers, you must have a means of formaly defining it. What I'm asking for is 0.9999 recurring expressed according to one of the formal methods for constructing numbers.

Real numbers can be defined as Cauchy sequences of rational numbers (there was a nice little discussion on this very subject about a year and a half ago; if you want to find it, it's in the HoS, IIRC). Since 0.999... is simply shorthand for the sequence {0, 0.9, 0.99, 0.999, 0.9999, 0.99999, ...}, which is Cauchy, as you can easily verify, is in the same equivalence class as 1, we say 1 = 0.999 ... .

What you're really doing is looking at the definition of equality: two real numbers {a_j} and {b_k} are equal iff for any ε>0 there exists an N s.t., for n>N, |a_n - b_n| < ε. It's not difficult to check that 0.999 ... and 1 are equal by this definition*. Then we partition the reals based on equality just like we partition the rationals based on equality (1/1 = 2/2 = 3/3 = ..., e.g.).

*Say 1 = {1,1,1,1,1,1,1,...} = {1^j} and 0.999... = {0,0.9,0.99,0.999,...} = {1-10^-k}, as above. Then (WLOG, since ε tends to 0) let 1 > ε > 0 be given. Set N = [-log_10{ε}] (i.e., the smallest integer greater than -log_10{ε}). Then, for n > N > -log_10[ε}, and |(1^n) - (1-10^-n)| = |10^-n| = 10^-n < 10^-(-log_10{ε}) = ε, and we have equality.

EDIT: Asterisk, clarity, and a_n - b_n is in modulus.

[fnord]

Another way that you could look at it would be from the viewpoint of a geometric sequence.

Namely

0.99999.... = 0.9 + 0.09 + 0.009 + 0.0009 + ....

Which looks suspiciously like a geometric series with a (initial term) being 0.9 and r (common ratio of terms) being 0.1.

From there, proceed to the sum of the geometric series

= 0.9 / (1 - 0.1) (a / (1-r) )

= 0.9 / 0.9

= 1