0.9999... != 1

Status
Not open for further replies.
The proof is flawed in that it fails to provide a limit step. All it succeeds in showing is that no finite string of 9s after a decimal point can represent 1. But that's uncontroversial.

You either need to provide a limit step or switch to strong induction. Unfortunately, you can't do either without begging the very question you've set out to prove.
 
It comes down to the simple fact that they are both representaions of the same number.

and james want trying to prove .999=1 , he was proving that .999... is not irrational.
 
It seems to me that induction breaks down at n=infinity. Consider the following proof that 1+1+1+... is finite:

If x is finite, then x+1 is finite. 1 is finite, therefore 1+1+1+... is finite.

But then, I'm no expert, and it's been a long time since I've seen the inside of a math class.
 
Weak induction works fine for infinity, both countable and uncountable. It's just that it requires a limit step. Strong induction works without limit steps, but it tends to be harder to prove the inductive step.
 
drnihili:

Can you please explain what you mean by "weaK" and "strong" induction?
 
Weak induction starts with a base case, typically 0 or 1, and proves that the base case has the property of the induction. Then it shows that if n has the property of the induction, so does the successor of n. If you want the proof to apply to infinity you also have to prove the limit case, that if m is not a successor but all cases below it have the property, so does m.

Strong induction shows simply that if everything below n has the property, so does n. Notice that doing this automatically encompasses the base and limit cases and also the inductive step.

Weak induction is typically a bit easier to do, but only works for certain kinds of cases. You can't use it to prove things about the rationals in their normal ordering for example because there are no successor rationals.

Strong induction tends to be a bit trickier, but if you can do it, it has wider application.

Strong and weak induction both work on infinite collections of any cardinality, but of course you absolutely must have a limit step if you want to use weak induction in non-finite cases.
 
0.9999...... is equal to 1.

Ok......

You have two real numbers, a and b. If a and b have different values then a - b is not equal to zero.

You could try to prove this for a = 0.99999 and b = 1 using reducio ad absurdum, induction relies on a few things that are a pain to use here.
 
Last edited:
Ok, Case one.

a = 1
b = 0.9

a - b = 1 - 0.9 = 0.1

I will put it in this form to show what I am doing:
Code:
  a = 1
  b = 0.9
a-b = 0.1


Case two.

a = 1
b = 0.99

a - b = 1 - 0.99 = 0.01

Code:
  a = 1
  b = 0.99
a-b = 0.01

Case three.

a = 1
b = 0.999999....... where b is zero followed by a decimal place followed by a never-ending series of nines.

a - b = 1 - 0.9999999...... = 0

Code:
  a = 1
  b = 0.9999999999999999999....
a-b = 0.0000000000000000000....

From the earlier post, you can see that if a - b is not equal to zero, then a and b are different. (a and b are real numbers).

In this case, a-b is zero followed by a never-ending series of zeroes. The next part is to prove that 0.0000...... is equal to zero.




Now, we have 0.00...... where 0.00..... is the number zero followed by a decimal point followed by a never-ending series of zeroes. Let's call this c.

Now, if we have a real non-zero number r then 2r is not equal to r.

Now, let r = c:
Code:
 r = 0.000....
2r = 0.000....

where 2r is the number zero followed by a decimal point followed by a never-ending series of zeroes. 2r = r, so r is not a non-zero number. We can define it as the number zero followed by a decimal point followed by a never-ending series of zeroes, so r is not undefined, therefore it is defined, therefore r = 0.

r = c, therefore c = 0.

c = a - b, therefore a = b - c, therefore a = b.

Therefore

0.999.... = 1

where 0.999.... is the number zero followed by a decimal point followed by a never-ending series of nines.
 
Last edited:
Note: My proof relies on this law of logic.

You have two real numbers, a and b. If a and b have different values then a - b is not equal to zero.

Let the statement P be "a and b have different values" and the statement Q be "a - b is not equal to zero".


Now because P implies Q then ~Q implies ~P.

so the following statement is also true:

" If a - b is equal to zero then a and b do not have different values."

I just realised it relied on this.
 
Last edited:
Originally posted by James R
StrangeDays:

<i>Proofs by induction are valid only for finite n.</i>

No. They are valid for countable infinities, too.

No, James, strangedays is correct. induction shows that something is true for an infinite countable set, let s say for the purposes of this thread, N, the set of natural numbers.

induction shows that a property holds for all n in N. is infinity in N? NO! every element in N is a finite number, even though the set itself is infinite.

Originally posted by StrangeDays
It seems to me that induction breaks down at n=infinity. Consider the following proof that 1+1+1+... is finite:

If x is finite, then x+1 is finite. 1 is finite, therefore 1+1+1+... is finite.

But then, I'm no expert, and it's been a long time since I've seen the inside of a math class.

this is a good example. induction proves that every finite number is finite. it is ridiculous to extrapolate that infinite numbers are also finite. induction only holds for finite n. again i will say: it holds for an infinite number of n, but n is still in N, still finite.

Originally posted by drnihili
Weak induction works fine for infinity, both countable and uncountable. It's just that it requires a limit step. Strong induction works without limit steps, but it tends to be harder to prove the inductive step.

whoa!!?! induction can work for any well ordered set. by the Axiom of Choice, the reals can be well ordered, so i suppose in principle one could use induction on a set of cardinatlity aleph_1, but i have never seen this done in practice. remember that this ordering can never be constructed. the axiom of choice is nonconstructive. so i don t know if this would be useful at all.

so i would say, let s keep our feet on the ground here, and only apply induction to countable sets.

next thing: limits??? what limit? this is not analysis class, this is set theory. there is no limit in induction. and i reiterate that a mathematical induction proof holds for the infinite number of elements of our countable set. unless you include infinity in your well ordered set (which might be possible, but certainly isn t the case for sets like N or Z or R), then induction gives no result for infinity. it only holds for finite n.

lastly, the strong and weak forms of mathematical induction are equivalent. one may be harder to use in some contexts, but it doens t matter which one you use, you get the same result either way.
 
Originally posted by lethe


whoa!!?! induction can work for any well ordered set. by the Axiom of Choice, the reals can be well ordered, so i suppose in principle one could use induction on a set of cardinatlity aleph_1, but i have never seen this done in practice. remember that this ordering can never be constructed. the axiom of choice is nonconstructive. so i don t know if this would be useful at all.

so i would say, let s keep our feet on the ground here, and only apply induction to countable sets.

Induction gets applied to all sorts of cases that aren't countable. It's pretty common actually. Most of transfinite arithmetic would evaporate without it. Every set theory book I've ever taught from or learned from does it. How do you propose to develop the theory of ordinals by only appealing to countable sets? And that just at the basic level.

As for the axiom of choice, I assume we're talking ZFC here unless otherwise specified. Sure choice is non-constructive and we can have a debate about whether it's true and such. But this isn't a thread about constructivist approaches to mathematics. So unless otherwise stated it's pretty safe to assume that the underlying set theory is something akin to ZFC.

next thing: limits??? what limit? this is not analysis class, this is set theory. there is no limit in induction.

Slow down bucko, nobody said anything about limits in the sense of analysis. I'm talking about limit ordinals here.

and i reiterate that a mathematical induction proof holds for the infinite number of elements of our countable set. unless you include infinity in your well ordered set (which might be possible, but certainly isn t the case for sets like N or Z or R), then induction gives no result for infinity. it only holds for finite n.

If you're only using /omega as the ordinal underlying the induction (i.e. if you have no limit stage) then induction only aplies to finite n. It applies to infinitely many of them, but each of them are finite. That's my point about having proven only that no finite string of 9s after a decimal represents the same number as 1. However, unless you a finitist at heart it would be pretty odd to insist on only using /omega sequences.

lastly, the strong and weak forms of mathematical induction are equivalent. one may be harder to use in some contexts, but it doens t matter which one you use, you get the same result either way.

Yup, anything that can be proven one way can be proven the other. That's why I said he could switch his proof to strong induction, but that he'd still get hung up. However, despite the equivalence of results, the methods are somewhat different. Specifically, weak indcution requires a limit step to prove anything transfinite, strong doesn't because it's built in to the indictive step.
 
Ok. I'm learning something about transfinite sets here (I think).

The question is, to either drnihili or lethe: can you prove that 0.999... = 1? i.e. do you believe that a rigorous proof is possible, given what you have said? If so, can you show me?
 
Originally posted by James R
Ok. I'm learning something about transfinite sets here (I think).

The question is, to either drnihili or lethe: can you prove that 0.999... = 1? i.e. do you believe that a rigorous proof is possible, given what you have said? If so, can you show me?

OK, well firstly, drnhill, i guess you re right, when you re doing transfinite arithmetic, you do deal with sets which contain infinities. i m not very familiar with that, so i didn t consider that when i wrote my response.

however, strangedays comment still stands. BenGodding was doing induction over the integers, and so his result can only hold for finite n. it is still true that when you re doing induction over the integers, you cannot extrapolate to infinity.

now, on to your question James. drnhills interesting comments on induction don t effect the fact that the statement 0.9999... =1 practically doesn t even require proof. it is built in to the definition of the reals. there are basically three choices for defining the reals:

1. dedekind cuts. my favorite. a real number is a set of rationals with an upper bound. associate that set with its least upper bound, which is guaranteed to exist by the continuity axiom.

2. cauchy sequences of rationals.

3. equivalence classes of decimal representations.

all three definitions are equivalent.

if you view 0.9999.... as a dedekind cut, it is easy to see that the least upper bound of all rationals less than 0.9999..... is 1. for example, proof by contradiction. assume the least upper bound is not one. by the dichotomy axiom, it must be less than one or greater than one. it obviously cannot be greater than 1, suppose it is 1+1/n, for some large n. then 1+1/2n is a rational, less than the LUB. 1+1/2n is obviously greater than 0.999... (assuming we have defined "greater than" for decimal representations. just compare digit by digit.). therefore our least upper bound must be less than one. say, 1-1/n, for some large n.

now we just have to pick a long enough (finite) sequence of 9s in our expansion, and show that this will eventually be greater than 1-1/n, no matter how large n is. because any finite sequence of 9s is obviously in the dedekind cut. but this is easy. for the finite sequence, 1-0.9999 is 1/10^m, where m is the number of 9s. so we just have to choose m > log n 9s, and this will be satisfied. this is our contradiction, and we have our proof. 0.9999999.... =1.

or, perhaps, the proof would be more natural using the decimal representation definition, since 0.9999.... is really a relic of that system anyway. and here is what i mean when i say it hardly requires proof. two decimal representations represent the same real number iff they converge to the same real number as a limit. then, just write down the geometric series, and you re done.
 
Good stuff drnihili and lethe, this is a really invigorating discussion.

Cjwinnit proof is good enough for me, and so I have learnt something today

0.999...=1

I am a physicist, not a mathematician, so I regard this proof to be rigorous enough. Continue though as Like James, I think I am learning something here.
 
Originally posted by drnihili
Induction gets applied to all sorts of cases that aren't countable. It's pretty common actually. Most of transfinite arithmetic would evaporate without it. Every set theory book I've ever taught from or learned from does it. How do you propose to develop the theory of ordinals by only appealing to countable sets? And that just at the basic level.

As for the axiom of choice, I assume we're talking ZFC here unless otherwise specified. Sure choice is non-constructive and we can have a debate about whether it's true and such. But this isn't a thread about constructivist approaches to mathematics. So unless otherwise stated it's pretty safe to assume that the underlying set theory is something akin to ZFC.
OK, so you can use induction on any well ordered set, and by the AC, any set can be well ordered. i brought up the AC, not because i wanted to suggest that we shouldn t work in ZFC, but just because i thought that since the ordering can t be constructed, i thought it would be very difficult to actually prove anything that way. when you write a proof by induction over the integers, you have to actually calculate the succesor of your nth term, and with an ordering that you cannot construct, i don t see how you could write a proof by induction over other sets.

however, just because i haven t seen such a proof, doesn t mean they don t exist. so i will defer to you on this.

could you say a little bit about how such an inductive proof would look?


Slow down bucko, nobody said anything about limits in the sense of analysis. I'm talking about limit ordinals here.
i have never seen limits of ordinals. the only limits i have ever seen are those found in analysis. but that doesn t mean that it can t be done, so again i defer.


If you're only using /omega as the ordinal underlying the induction (i.e. if you have no limit stage) then induction only aplies to finite n. It applies to infinitely many of them, but each of them are finite. That's my point about having proven only that no finite string of 9s after a decimal represents the same number as 1. However, unless you a finitist at heart it would be pretty odd to insist on only using /omega sequences.
OK. yes, well it was my contention that a proof by induction makes no sense to try and prove that 0.999...=1. it is not true for any finite n, and induction over the integers (\omega?) will be totally useless. i have a hard time imagining that it would get you anywhere even with bigger ordinals....

so whether or not i m a finitist, doesn t seem like it is relevant here.

my point above is that, the way the real numbers are defined, induction is not necessary. 0.999... as a real number has to be 1.

and my other point is this:
BenGooding said:
1. 0.999 is not 1 for any finite number of 9s.
2. by induction, let n be infinity. therefore 0.999... repeating is not 1.

to which StrangeDays replied:
"induction over finite n cannot be extrapolated to n=\infty"

(i paraphrase)

and i agree with StrangeDays. induction shows that 0.999 is not 1 for any finite number, and it is not valid to induct up to infinite number of 9s. i would be interested to hear how you apply transfinite induction to this argument, and i sincerely hope that you re not going to claim that it allows you to prove that 0.999... repeating does not equal 1.

but like i said, you can make a completely rigorous argument, just using the definition of a real number, without any recourse to induction, that 0.999...=1.
 
Originally posted by lethe
you can make a completely rigorous argument, just using the definition of a real number, without any recourse to induction, that 0.999...=1.

That's what I tried.

The problem is when you have "an infinte number of nines" or "it's infinitesemmly close to nine". There are better ways of describing it.
 
As long as we're talking about Real Numbers, 0.999... = 1. There are a number of ways to prove it. Lethe has already provided one of them. The proofs won't generally involve induction. Instead they proceed by reductio. The assumption that 0.999... does not equal 1 is inconsistent with the basic properties of real numbers. You can of course talk about 0.999... as representing something other than a Real, but then it's up to you to give a complete definition of what you're talking about. As long as you pick one of the standard definitions of the Reals, 0.999...=1.

Lethe and I are in agreement that the original proof only shows that no finite string of 9s after a decimal will be sufficient to represent 1. I think we're also in agreement that induction isn't the way to deal with this particular issue. My point was that the original proof simply is not a correct application of induction. It can't prove it's conclusion. It fails because it omits one case.

There are three types of ordinals: zero, successor ordinals, and limit ordinals. /omega is the first limit ordinal. When doing proofs by weak induction, you have to do a separate case for each type. Of course if you only care about the finite cases you can omit the limit ordinals since all limit ordinals are infinite (Note I did not say that all infinite ordinals are limits.)

Since I get the impression that induction on the ordinals isn't very familiar to people, I'll try to write a snippet about it. But that will go in a different post.
 
First let's talk about cardinals. We can construct the finite cardinals in the following way. There are alternate ways, but this is as good as any.

0 = {} = the null set
1 = {0} = {{}} = (the singleton set with the null set as it's only member.)
2 = {0, 1} = {{}, {{}}}
.
.
.
n+1 = {0, 1, 2, ..., n}

So each cardinal is the set with all smaller cardinals as it's members. We get infinite cardinals in the same way, but they end up being a touch different. \aleph_0 is the smallest infinite cardinal. It is the set of all finite cardinals.

\aleph_0 = {0, 1, 2, ..., n, n+1, .... }

Notice that \aleph_0 has no greatest member, but all the finite cardinals do. Now, you might think that the next cardinal would be {0, 1, 2, ..., n, n+1, ..., \aleph_0 }. Alas, you would be wrong. By definition, there is no bijective function between distinct cardinals. But there is a bijective function between \aleph_0 and this new set. Thus the new set is not a distinct cardinal. However you do get a new cardinal if you take the power set of \aleph_0, the set of all subsets of \aleph_0. In fact, it can be proven that the powerset of any cardinal \aleph_n is bigger than \aleph_n. There may be cardinals in between \aleph_0 and it's powerset, but the powerset is always bigger.

So, with infinite cardinals, X+1=X. In fact X+X=X.

Now, the ordinals are built up in a very similar fashion. The main difference is that the ordinals are ordered sets, not mere sets. Here we go:

0 = <> = the null sequence.
1 = <0> = <<>> = the sequence with the null sequence as it's only member.
2 = <0, 1> = <<>, <<>>>
.
.
.
n+1 = <0, 1, 2, ..., n>

The smallest infinite ordinal is \omega which is just the sequence of finite ordinals in their standard ordering. I.e.

\omega = <0, 1, 2, ..., n, n+1, ...>

and \omega sequence is just a sequence that is equally as long as \omega. The natural numbers in their standard ordering are a good example. Two ordered sets have the same ordinality if there is a bijective function pairing the nth member of one set with the nth member of the other.

But notice that when we add \omega to the end of itself, we get a new ordinal. We get

\omega+1 = <0, 1, 2, ..., n, n+1, ... \omega>

\omega and \omega+1 have the same cardinality, but different ordinality. Ordinal arithmetic is importantly different from cardinal arithmetic in this respect.

An important point is that \omega is not the successor of any other ordinal. This follows from the fact that it has no last member. This is why induction whithout limit steps cannot prove anything about infinite. If you prove that 0 has a property, and that if a number has the property, so does it's successor, it's still an open question whether \omega has the property. You have to explicitly prove that limit ordinals have the property.

This is where the original proof failed. <0.9, 0.99, 0.999, ...> is an omega sequence. But each member of the sequence is a finite string. Since the proof wants to prove soemthing about the infinite string 0.9999... The proof must rest on and \omega+1 sequence with 0.999... as the \omega+1 member. But in this case you're now doing indcution on transfinite ordinals and you have to include the limit step.

*hopes that was relatively clear*
 
Status
Not open for further replies.
Back
Top