Anti-Infinity

Great Debates
1

A nonreligious, nonpolitical debate.

I present two formal proofs:

Proof 1

Universal Set: the set of real numbers.

Hypothesis: .9… equals 1.

Definition: .9… = .9[n].

Axiom 1: A = A.

Axiom 2: If A = B and B = C, then A = C and C = A.

Axiom 3: If A = B, then AX = BX.

Axiom 4: If A = B, then A - X = B - X.

Axiom 5: The property of subtraction for the set of real numbers is an operation in the field of arithmetic.

Axiom 6: If A = B, then A / C = B / C and C does not equal 0.

Axiom 7: The property of division for the set of real numbers is an operation in the field of arithmetic so long as the divisor does not equal 0.

Premise 1: Let X = .9[n] (by Axiom 1)

Premise 2: 10X = 9.9[n] (by Axiom 3)

Premise 3: 10X - X = 9.9[n] - .9[n] (by Axiom 4 and Premise 1)

Premise 4: 9X = 9 (by Axiom 5)

Premise 5: 9X / 9 = 9 / 9 (by Axiom 6)

Premise 6: X = 1 (by Axiom 7)

Premise 7: .9[n] = 1 (by Axiom 2)

QED

Proof 2

Universal set: the set of real numbers

Hypothesis: .9… does not equal 1.

Definition: .9… = .9[n].

Axiom 1: A = A

Axiom 2: If A = B, then B - A = 0.

Axiom 3: The property of subtraction for the set of real numbers is an operation in the field of arithmetic.

Axiom 4: If the rightmost digit in a decimal expression is greater than zero, then it is significant.

Axiom 5: The rightmost digit in a decimal expression decrements by exactly one order of magnitude for place inserted to the left of it.

Premise 1: 0 is not greater than 0 (by Axiom 1)

Premise 2: 10 - .9 = .1 (by Axiom 3)

Premise 3: 10 - .99 = .01 (by Axiom 3)

Premise 4: 10 - .999 - .001 (by Axiom 3)

Premise 5: 10 - .9999 = .0001 (by Axiom 3)

Premise 6: 10 - .99999 = .00001 (by Axiom 3)

Premise 7: 10 - .9[n] = .0[n-1]1 (by Axiom 5)

Premise 8: .0[n-1]1 > 0 (by Axiom 4 and Premise 1)

Premise 9: .9[n] <> 1 (by Axiom 2)

QED

One of these two is wrong because they contradict.

I’d like you to take out your logic magnifying glass, and review each of them for any flaw. I know which one I think is flawed and where the flaw is, but I thought it might make for a great debate to pick at it together first. For one thing, I’m afraid that if I told you right away what I think, you’d call me crazy and run the other way. So, let’s debate first just the logic in these two proofs.

But after a little while, I expect there to unfold a really great debate that will test whether we can see things in a whole new way. And that debate will unfold right neatly along the lines of who buys proof number one and who buys proof number two.

We will end up in a debate over the very nature of quantity and magnitude themselves, and dealing with the beautiful and esoteric things found only in the very bowels of number theory. We will divide into Infinitists versus Anti-infinitists, almost certainly many more of the former than the latter.

If you will, please hold off on introducing other proofs right now. Hold those for the ensuing debate over infinity. Concentrate right now on the two proofs given here.

And remember, infinity is not a number.

(Spiritus: I trust you’re chompin’ at the bit to go at this.)

“It is lucky for rulers that men do not think.” — Adolf Hitler

2

And no, typos don’t count.

Axiom 5 of Proof 2 should have read:

Axiom 5: The rightmost digit in a decimal expression decrements by exactly one order of magnitude for each place inserted to the left of it.

“It is lucky for rulers that men do not think.” — Adolf Hitler

3

Let’s do this the easy way, o.k.?
Infinity=Un-ending Time
Time=The measurement of a sequence of events.
Therefore, as long as events continue to occur, we have Infinity.
Now, if you can postulate a scenario where no events occur, you can disprove Infinity. Good luck.

4

Lib - what do you mean by .9[n]? How does it differ from .9[n-1]?

.999… , with the 9’s going out to infinity, equals 1, by the simple fact that there’s no number in between .999… and 1. But the way you use your notation suggests .9[n] involves a finite number of 9’s (n of them, for any given positive integer n, I would expect). This could be a problem, but it won’t be much of a debate. :slight_smile:

5

Well, I’m not a mathematician, but you lost me right here:

Axiom 3 being:

Defining .9[n] as X leads to a valid re-phrasing of Premise 2 as 10X = 9X. Since 10 (A) clearly does not equal 9 (B), the proof goes “poof!”

Am I missing something here?

6

slythe:

I have no idea what you’re talking about. This is for number theory, not philosophy.

RTFirefly:

Obviously, you side with Proof 1.

Let me give you, but just for the moment for argument’s sake, your point. Then, feel free to recast Proof 1 using .9… notation (without Definition 1). Please explain how Proof 1 manages to pull off Premise 2 if there are no fewer places to the right of the decimal than there are in Premise 1.

If you say there are an infinite number of places, and that Infinity - 1 = Infinity, then you imply that Infinity is a number because you are doing operations with it in the field of arithmetic.

Besides, the universal set is the set of real numbers, of which infinity is not an element. If infinity were a real number, then it could be decremented. And if you say that the universal set is infinitely large, then you are undermining your claim that nothing is between .9… and 1.

In order to draw your conclusion (which I disagree with) from your premise (which I do agree with), you have made an a priori assumption, namely that if .9… does not equal 1, there ought to be something between them.

The challenge is to find a flaw in either Proof 1 or Proof 2.

If you agree that .9… = 1, then point out a flaw in Proof 2. If you say that the flaw is in Definition 1 of Proof 2, then you may remove it just as above and use substitute notation. Definitions aren’t flaws; they’re just definitions.

“It is lucky for rulers that men do not think.” — Adolf Hitler

7

Quixotic:

Well, you are taking .9[n] as a sort of string literal. In other words, 10 times .9[n] is given as 9.9[n], or 9 + .9[n]. See, it’s 9 + X, and not 9X.

“It is lucky for rulers that men do not think.” — Adolf Hitler

8

Hi Lib,

I agree with proof one. There are, in fact, other ways of deriving the same result. I would attack proof 2 on premise 7:

Premise 7: 10 - .9[n] = .0[n-1]1 (by Axiom 5)

The value .0[n-1]1 is undefined. The notation you used to define .9[n] denotes a sequence of digits that repeats without end. Such a sequence cannot have a “1” stuck at the end, because there is no end to the sequence of zeros.

I am also not certain why you felt the need to invoke premise 1 to get from premise 7 to premise 8, but that is not really germane to my complaint. Premise 8, of course, uses the same undefined value introduced in premise 7, so it also cannot be accepted.

The best lack all conviction
The worst are full of passionate intensity.
*

9

Lib, I know your first proof is right, so the second one must be flawed somehow. I don’t know where the flaw is.

The first one is right, .9bar (or .999… or .9[n]) is equal to one. If you were to subtract .99999… from 1, what would the difference be? The difference between two numbers is the distance between them on a number line. There are no numbers between the infinitely long string .999… and one, there is no distance between them, the difference is zero. If A-B=0, then A=B.

10

Oh, and how about this.

Lib, you noted that 9.9[n] equals 9 + X.

Well, if you say 10X = 9 + X, like you said there, and solve for X, you get 1. So 9.9[n] is the equivalent of one, a separate proof.

11

Oh, and they’re the same by law of concatentative assemblage (things equal to the same are equal to eachother), or if A=C and B=C then A=B.

12

Well first of all they’re not proofs. You can’t prove anything through circular logic. You assumed something and using that assumption proved it to be true.

So taking the theorems as premises both 1 and 2 are right (but using different premises for each). Either .999… = 1 or it doesn’t. If it does then 1 is right, if it doesn’t then 2 is right. Working from the same premise only one of them can be right.

13

I’m not a mathematician, but I’d have to agree with Spiritus here. “The value .0[n-1]1 is undefined.” Actually, I’m unclear as to what “[n-1]” itself means in this notation.

14

I knew I probably should have avoided this thread. It’s probably making me look stupid, but:

But Axiom 3 references multiplication, not addition. And 10 times X does not equal 9 plus X; it equals (9 times X) plus X.

In fact, the statement 10X = 9 + X is only true if X = 1, whereas 10X = 9X + 1 is true for all values of X.

I’ll have to leave, now.

<font size=1>Still not getting it . . . </font>

15

Dammit, of course that should be:

10X = 9X + X

16

Konrad:

The hypotheses were never used in the proofs. What happened was that the conclusions matched the hypotheses, which indicated that the proofs were finished.

I assure you they are (fairly) strict formal proofs, and neither of them is circular.

Surgoshan:

Actually, that’s a lemma, and just another way to state the identity.

Well, yes, but the transitive law, cited in Axiom 2 of Proof 1 covers it nicely. There is no need to introduce some other entity, Ockhamly speaking.

Actually, I see a glaring flaw in Proof 1, but can find no nontrivial flaw in Proof 2.

Greetings. When you get mad, please remember we are friends. :smiley:

I thought you might.

Oh yeah, I know. But they have the same flaw as the one in the OP. I appreciate your withholding them, though, as the OP requested.

I thought that might be the jugular you would go for. :wink:

Well, no it isn’t. It is simply very small. The value X / 0, that is undefined, because its operation is undefined.

But in .0[n-1]1, there are 1 fewer zeros than there are nines in .9[n], so there is room for the 1 stuck at the end.

Citing Premise 1 avoided the error of setting .0[n-1]1 equal to 0.

How long, I wonder, before an anti-infinitist wanders in?

“It is lucky for rulers that men do not think.” — Adolf Hitler

17

Actually you didn’t end up using the hypothesis directly so it isn’t really circular logic. The problem is that in one case you’re using the n in .9[n] as a finite number and in the other as an infinite number.

For 9.9[n] - .9[n] = 9 in between premise 3 and 4 n must be infinite otherwise you will have an infintesmal difference.

In Premise 7 of part 2 you wrote:
10 - .9[n] = .0[n-1]1
This is only true if n is finite. You can’t perform those operations on infinite numbers. If n is infinite then it’s limit goes to infinity therefore .0[n-1]1 = 0.

So like I said before, you have to agree on a premise. Either n is infinite or it isn’t. What you decide changes the actual defintion of the problem. (.999… = .9[n]). So both proofs are correct if in 1 the definition is different from the definition in 2.

18

Gilligan:

Well, it’s just hard to do subnominal notation in this medium, so I used [n] to indicate the number of times a digit repeats. [n - 1] is one fewer time.

Quixotic:

Thar ye go, lad. Be ye an anti-infinitist?

“It is lucky for rulers that men do not think.” — Adolf Hitler

19

Konrad:

Now, you’re zeroing in on the flaw in Proof 1. (It’s in Premise 2!)

“It is lucky for rulers that men do not think.” — Adolf Hitler

20

Lib: That is the “flaw”. Premise 2 is just a result of that. For premise 2 to work n must be infinite.

It’s not really a flaw as much as a word game. It’s an ambigious definition.