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Author Topic:   0.99999~ = 1 ?
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 34 of 237 (543226)
01-16-2010 11:27 AM
Reply to: Message 13 by PaulK
01-15-2010 5:51 PM


Re: Hmm, equal?
That is, two numbers a and b are equal if there is no number, e such that
|a - b| < e
Read through that again.
You meant to say "if for every (positive) number e, |a - b| < e".
Or possibly you meant to say "if there is no (positive) number e such that |a - b| > e".
Edited by Dr Adequate, : No reason given.

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Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 35 of 237 (543228)
01-16-2010 12:04 PM
Reply to: Message 16 by Straggler
01-15-2010 7:26 PM


Re: 1 and NOT 1
Surely the difference between 0.999R and 1 is practically non-existant and philosophically massive?
It is the difference between claiming complete certainty (for example) and always allowing for the possibility of that which is unexpected (no matter how likely or unlikely). It is the difference between an obtainable destination and that which can never exist or be obtained in reality.
I am no mathematician. But surely the difference between 1 and NOT 1 is as significant as ever. No matter what the NOT 1 may be?
It's funny, no-one has any difficulty with grasping that 1/2 and 2/4 are the same thing.
Well, here we have a similar situation. We happen to use a system of notation in which sometimes there is more than one way to write exactly the same number.
Edited by Dr Adequate, : No reason given.

This message is a reply to:
 Message 16 by Straggler, posted 01-15-2010 7:26 PM Straggler has replied

Replies to this message:
 Message 38 by Straggler, posted 01-17-2010 3:20 PM Dr Adequate has replied

  
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 55 of 237 (543418)
01-17-2010 10:03 PM
Reply to: Message 38 by Straggler
01-17-2010 3:20 PM


Re: 1 and NOT 1
In which case I stand corrected. And in which case I will need to find a new nomenclature for expressing all but certain without the philosophical possibility of complete and absolute certainty.
How about "a probability of 1 - ε"?
But just to be clear is it false to say that 0.999R < 1?
False as false can be.
If the former was smaller than the latter, there would have to be some quantity by which it was smaller. But there isn't.
Edited by Dr Adequate, : No reason given.

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 Message 38 by Straggler, posted 01-17-2010 3:20 PM Straggler has replied

Replies to this message:
 Message 57 by Straggler, posted 01-18-2010 6:17 AM Dr Adequate has replied

  
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 56 of 237 (543423)
01-17-2010 11:42 PM
Reply to: Message 50 by Straggler
01-17-2010 4:31 PM


Re: 1 and NOT 1
Screwed if I know. But then degrees of infinity is no less mad ...
It's manifestly sane.
Suppose we have a set S. We can consider the set of all subsets of S --- call this P(S).
So for example if S = {1,2} then P(S) = {{},{1},{2},{1,2}}.
Now, the size of P(S) must be strictly greater than the size of S. This is obviously true in the finite case. Less obviously, it must also be true if S is infinite.
This means that there must be lots of different sizes of infinity. If S is the set of natural numbers, which is infinite, then P(S) must be bigger than that, and P(P(S)) bigger than that, and P(P(P(S))) bigger than that, and so forth.
---
I shall give the proof of the statement above, because it's one of my favorite proofs in mathematics.
Proof that P(S) always has more members than S:
Note that P(S) always has at least as many members as S, because for every member s of S, P(S) contains the element {s}. So if they're different sizes, P(S) must be the larger of the two.
This we shall prove by contradiction. Suppose that they were the same size. Then we could put the members of S and the members of P(S) into one-to-one correspondence, since that's what it means for two sets to be the same size. Hence there would be a function f from S to P(S) such that for every y in P(S) there is an x in S such that f(x) = y.
Now, for every z in S, z either is or is not a member of the set f(z). So, under our assumption that the function f exists, we can form the set T = {All z in S such that z is not in f(z)}.
Now, T is a subset of S. Therefore T is a member of P(S). Therefore, there is some member t of S such that f(t) = T.
Now, here's the clever bit. Is t a member of T?. Well, if it is, then it isn't, and if it isn't, then it is --- by definition of T.
So we have produced an absurdity, therefore the function f can't exist, therefore P(S) is larger than S.
Edited by Dr Adequate, : No reason given.
Edited by Dr Adequate, : No reason given.

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Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 61 of 237 (543454)
01-18-2010 7:58 AM
Reply to: Message 57 by Straggler
01-18-2010 6:17 AM


Re: Is 0.999R a Whole Number?
Well I accept what you say. And when you guys explain it does all kinda make sense. But there is still something that seems intuitively wrong about the whole thing. For example is it true to say that 0.999R is a whole number?
Yes. Specifically, it's the number 1.
Edited by Dr Adequate, : No reason given.

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Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 88 of 237 (544109)
01-23-2010 7:51 PM
Reply to: Message 87 by Jon
01-23-2010 7:27 PM


Re: Totally right!
Your logic goes like this:
P "All real numbers have properties X"
P ".9999| does not have properties X"
C ".9999| is = 1 and it is false that .9999| ≠ 1"
No, it goes like this:
P "Any pair of distinct real numbers has property X."
P "The pair of real numbers .9999| and 1 does not have property X."
C ".9999| and 1 are not a pair of distinct real numbers."
Edited by Dr Adequate, : No reason given.

This message is a reply to:
 Message 87 by Jon, posted 01-23-2010 7:27 PM Jon has replied

Replies to this message:
 Message 89 by Jon, posted 01-23-2010 9:41 PM Dr Adequate has replied

  
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 92 of 237 (544123)
01-23-2010 10:59 PM
Reply to: Message 89 by Jon
01-23-2010 9:41 PM


Re: Totally right!
LOL. Again, you base your conclusion on the premisethough, I give you extra credit for more cleverly disguising it this timethat .9999| is a real number.
Which it is by definition.
BTW, I have obviously not attempted to disguise the fact that we are reasoning about real numbers and not about omelets or geraniums. Your admiration of how "clever" I am in doing so, though flattering, is therefore misplaced.
Your flow can just as easily be reworded as follows ...
No it can't.
{ABE: I noticed you edited your post after I started typing my reply; worry not, though; italicizing your conclusion doesn't affect its truth value}
That was not the change I made.
If you will look at my edit and compare it to the original, you will see what change I made. The purpose of it was to prevent you from reading it as though being "distinct" could be the property of one number, or of two numbers separately, rather than of a pair of numbers.
Edited by Dr Adequate, : No reason given.

This message is a reply to:
 Message 89 by Jon, posted 01-23-2010 9:41 PM Jon has replied

Replies to this message:
 Message 95 by Jon, posted 01-23-2010 11:36 PM Dr Adequate has replied

  
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 93 of 237 (544125)
01-23-2010 11:11 PM
Reply to: Message 89 by Jon
01-23-2010 9:41 PM


Re: Totally right!
DP.
Edited by Dr Adequate, : No reason given.

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Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 96 of 237 (544130)
01-23-2010 11:52 PM
Reply to: Message 91 by Jon
01-23-2010 10:55 PM


Re: Totally right!
I guess my prayers were not answered. Can you provide a definition of 'real number' so that I do not have to look for one myself?
Well, try the following definition:
---
Let R be the set of Cauchy sequences of rational numbers. That is, sequences x1,x2,x3,... of rational numbers such that for every rational ε > 0, there exists an integer N such that for all natural numbers m,n > N, |xm-xn|<ε. Here the vertical bars denote the absolute value.
Cauchy sequences (x) and (y) can be added, multiplied and compared as follows:
* (xn) + (yn) = (xn + yn)
* (xn) (yn) = (xn × yn)
* (xn) ≥ (yn) if and only if for every rational ε > 0, there exists an integer N such that xn ≥ yn - ε for all n > N.
Two Cauchy sequences are called equivalent if and only if for every rational ε > 0, there exists an integer N such that |xn -yn|<ε for all n > N.
This does indeed define an equivalence relation, it is compatible with the operations defined above, and the set R of all equivalence classes can be shown to satisfy all the usual axioms of the real numbers.
---
Note that the sequence 0.9, 0.99, 0.999, 0.9999 ... is indeed a Cauchy sequence, since this is a sequence of rational numbers and since for any rational ε > 0 there exists some N for which ε > 10-N.
Indeed, one can also immediately prove from the definition of the equivalence relation that 0.999999... = 1.
Edited by Dr Adequate, : No reason given.

This message is a reply to:
 Message 91 by Jon, posted 01-23-2010 10:55 PM Jon has replied

Replies to this message:
 Message 101 by Jon, posted 01-24-2010 12:26 AM Dr Adequate has replied

  
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 97 of 237 (544131)
01-23-2010 11:53 PM
Reply to: Message 95 by Jon
01-23-2010 11:36 PM


Re: Totally right!
LOL. I'm too busy for malarkey.
Your posts belie this claim.
My debate is with lyx2no, now.
This declaration will not, of course, prevent me from pointing out that you are wrong.

This message is a reply to:
 Message 95 by Jon, posted 01-23-2010 11:36 PM Jon has replied

Replies to this message:
 Message 98 by Jon, posted 01-24-2010 12:14 AM Dr Adequate has replied

  
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 100 of 237 (544134)
01-24-2010 12:24 AM
Reply to: Message 98 by Jon
01-24-2010 12:14 AM


Re: Totally right!
LOL. Yes; too busy for malarkey, but I can make time for bullshit.
So I see.
If you wish to be obtuse, care to point out where in my rewording the meaning of your original was lost?
But what if I do not wish to be obtuse?

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Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 102 of 237 (544136)
01-24-2010 12:28 AM
Reply to: Message 99 by Jon
01-24-2010 12:22 AM


My math is a little rusty; can you explain this one to me? What does it mean for numbers to converge upon e?
A sequence of numbers x1, x2, x3 ... converges on e if for every ε > 0 there is some N such that for all i > N, |xi - e| < ε.

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Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 103 of 237 (544137)
01-24-2010 12:32 AM
Reply to: Message 101 by Jon
01-24-2010 12:26 AM


Re: Totally right!
I did not think 'rational' and 'real' numbers to be the same.
You're right. Why do you mention it?
Which are...?
A model for the real number system consists of a set R, two distinct elements 0 and 1 of R, two binary operations + and * on R (called addition and multiplication, respectively), a binary relation ≤ on R, satisfying the following properties.
1. (R, +, *) forms a field. In other words,
* For all x, y, and z in R, x + (y + z) = (x + y) + z and x * (y * z) = (x * y) * z. (associativity of addition and multiplication)
* For all x and y in R, x + y = y + x and x * y = y * x. (commutativity of addition and multiplication)
* For all x, y, and z in R, x * (y + z) = (x * y) + (x * z). (distributivity of multiplication over addition)
* For all x in R, x + 0 = x. (existence of additive identity)
* 0 is not equal to 1, and for all x in R, x * 1 = x. (existence of multiplicative identity)
* For every x in R, there exists an element −x in R, such that x + (−x) = 0. (existence of additive inverses)
* For every x ≠ 0 in R, there exists an element x-1 in R, such that x * x-1 = 1. (existence of multiplicative inverses)
2. (R, ≤ ) forms a totally ordered set. In other words,
* For all x in R, x ≤ x. (reflexivity)
* For all x and y in R, if x ≤ y and y ≤ x, then x = y. (antisymmetry)
* For all x, y, and z in R, if x ≤ y and y ≤ z, then x ≤ z. (transitivity)
* For all x and y in R, x ≤ y or y ≤ x. (totalness)
3. The field operations + and * on R are compatible with the order ≤. In other words,
* For all x, y and z in R, if x ≤ y, then x + z ≤ y + z. (preservation of order under addition)
* For all x and y in R, if 0 ≤ x and 0 ≤ y, then 0 ≤ x * y (preservation of order under multiplication)
4. The order ≤ is complete in the following sense: every non-empty subset of R bounded above has a least upper bound. In other words,
* If A is a non-empty subset of R, and if A has an upper bound, then A has a least upper bound u, such that for every upper bound v of A, u ≤ v.
Edited by Dr Adequate, : No reason given.
Edited by Dr Adequate, : No reason given.

This message is a reply to:
 Message 101 by Jon, posted 01-24-2010 12:26 AM Jon has replied

Replies to this message:
 Message 104 by Jon, posted 01-24-2010 2:35 AM Dr Adequate has replied

  
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 105 of 237 (544142)
01-24-2010 2:50 AM
Reply to: Message 104 by Jon
01-24-2010 2:35 AM


Re: Totally right!
How is this different from saying:
Real numbers are numbers that meet definition X
Clearly the only way to define the real numbers is to supply a definition of the real numbers.
What purpose does the Real/Non-real distinction serve? Does it actually differentiate, or is the distinction merely made up?
Clearly there are structures which are not the real numbers. Such as the complex numbers. Or quaternions. Or, for that matter, asparagus.

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Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 114 of 237 (544191)
01-24-2010 4:51 PM
Reply to: Message 113 by Jon
01-24-2010 4:32 PM


Re: proof or begging the question?
I was under the impression that infinite numbers (which is what you get if you subtract 0.9999| from 1, no?)
No. You get 0. 0 is finite.
When we stop it, we are admittedly not working the the number proposed, which was infinite
No. It's equal to 1. 1 is finite.
In other words, there never seems to be a way to link the mathematical 0.9999| with anything in reality that would make it meaningful
Your meaning is obscure.
So, is there any reason to assert the existence of 0.9999| other than to dazzle the Kindergartners?
Yes. Our system of decimal numbers allows us to write it.
Would you agree that there is a use for the expression 0.3333|, namely to write 1/3 in decimal notation? And a use for 0.6666|, namely to write 2/3? Very well, what do you get if you add them together using the standard algorithm for the addition of decimal numbers?
Oh yes, you get 0.9999|.
If we declare that expression meaningless, we have a problem.

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