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  #31 (permalink)  
Old 16-March-2005, 12:50 PM
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First line: "let n = 0.99999999..."
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  #32 (permalink)  
Old 16-March-2005, 12:50 PM
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It's just a label we assign to the number denoted by "0.999...", to make the steps in the proof clearer. You could do the same without using a variable.
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  #33 (permalink)  
Old 16-March-2005, 12:50 PM
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Quote:
Originally Posted by Nicolas
First line: "let n = 0.99999999..."
Dear me. ops: I better go get that second cup of coffee now.

It's always the obvious stuff that trips me up.
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  #34 (permalink)  
Old 16-March-2005, 12:56 PM
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Quote:
Originally Posted by farmerjumperdon
Is the first version (the simple one, the non-calculus one), correct?
I see nothing wrong with it. Of course, you could say that it "begs the question", in a sense. It assumes that the expression "0.999..." represents a unique real number, and that it is defined in such a way that the usual properties of real numbers that we all know and love remain valid. But then, why define it in any other way?

If we're very picky, then the first proof also assumes some knowledge of calculus, since infinite decimals like 0.333... can only be rigorously defined as limits. However, most people accept such expressions intuitively without protest (at least until we tell them that 0.999...=1 :wink.
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  #35 (permalink)  
Old 16-March-2005, 01:30 PM
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Quote:
Originally Posted by Frog march
RULE 6 ? 8-[
There is NO Rule 6!
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  #36 (permalink)  
Old 16-March-2005, 01:57 PM
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Perhaps this is where we are going wrong? :wink:
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Old 16-March-2005, 03:21 PM
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Quote:
Originally Posted by farmerjumperdon
Is the first version (the simple one, the non-calculus one), correct?
I don't think it's a rigorous proof. How do we know that 10x0.9999.... = 9.9999.... ? Intuitively it looks right, but how would you actually do the calculation?

10xn means n+n+n+n+n+n+n+n+n+n. So 10x0.9999.... means:

0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
__________
????????????

Since there's no right-hand column in which to start your addition, you can't even begin to do the sum.
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  #38 (permalink)  
Old 16-March-2005, 03:26 PM
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ToSeek: I beat you!

In the summation part, one must be careful how one simplifies the series:
  • 9 * ((1/10) / (1 - 1/10))
  • 9 * ((1/10) / (9/10))
  • 9 * 1/9
At this point you can either say
  • 9/9 = 1
which is cool,

or
  • 9 * 0.111111111... = 0.999999999...
and you're right back where you started! ](*,)

Fred
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  #39 (permalink)  
Old 16-March-2005, 03:28 PM
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Quote:
Originally Posted by Eroica
Quote:
Originally Posted by farmerjumperdon
Is the first version (the simple one, the non-calculus one), correct?
I don't think it's a rigorous proof. How do we know that 10x0.9999.... = 9.9999.... ? Intuitively it looks right, but how would you actually do the calculation?

10xn means n+n+n+n+n+n+n+n+n+n. So 10x0.9999.... means:

0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
0.9999....
__________
????????????

Since there's no right-hand column in which to start your addition, you can't even begin to do the sum.
You don't have to do addition from right to left. That's just a convenience. You can go in any order you choose because of the associative property.

Also, I'm not sure that multiplication needs to be defined as a series of additions.

That said, it's not a rigorous proof, just a demonstration.
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  #40 (permalink)  
Old 16-March-2005, 03:30 PM
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Eroica, multiplying by ten moves each digit one place to the left (it's one of those nice "properties of the real numbers we all know and love" :wink. [Edit: See Frog march's post below.]

Edited to simplify and correct.
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  #41 (permalink)  
Old 16-March-2005, 03:33 PM
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The 9 in 0.9 is in the tenths column and so represents 9/10(nine tenths) and so when multiplied by 10 becomes 9.

The 9 in 0.09 is in the hundredths column and so represents 9/100(nine hundredths) and so when multiplied by 10 becomes 9/10 ie 0.9

.
.
.
.
.
etc


every thing just moves left by one place.
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  #42 (permalink)  
Old 16-March-2005, 03:34 PM
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Quote:
Originally Posted by Nowhere Man
ToSeek: I beat you!

In the summation part, one must be careful how one simplifies the series:
  • 9 * ((1/10) / (1 - 1/10))
  • 9 * ((1/10) / (9/10))
  • 9 * 1/9
At this point you can either say
  • 9/9 = 1
which is cool,

or
  • 9 * 0.111111111... = 0.999999999...
and you're right back where you started! ](*,)

Fred
But that in itself implies that .999...=1.

What these explanations really show is that .999... must be 1 as long as .111...=1/9, .333...=1/3, etc.
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  #43 (permalink)  
Old 16-March-2005, 03:40 PM
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Actually, Nowhere Man, you just cleared it up for me.

Isn't it true that 9 * 1/9 = 1? and isn't 1/9 = .111111...... ?

Well, there you go. I already figured it equalled 1, but sometimes you got that little voice in your head that says : wait, you know its wrong!

Like my test I just took yesterday, how I got the right answer but then thought on it for the remainder of the test time, changed the answer, and ended up getting it wrong.....

kinda sucks.

Edit: ugg, Tomba you beat me to a reply to Fred's comment......
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  #44 (permalink)  
Old 16-March-2005, 04:20 PM
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Quote:
Originally Posted by W.F. Tomba
But that in itself implies that .999...=1.
Which is the question that started the whole thing.

Quote:
Originally Posted by W.F. Tomba
What these explanations really show is that .999... must be 1 as long as .111...=1/9, .333...=1/3, etc.
It works for me. Some people have trouble with infinity, as in the infinite 9s after the decimal point. I can't picture infinity in my mind, but I can still work with it as a concept or tool.

Fred
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  #45 (permalink)  
Old 16-March-2005, 04:24 PM
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Quote:
Originally Posted by Nowhere Man
Quote:
Originally Posted by W.F. Tomba
But that in itself implies that .999...=1.
Which is the question that started the whole thing.
Right, and it's exactly what the demonstration was trying to show. I'm saying that what you identified is not a problem, but another way of showing that .999...=1.
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  #46 (permalink)  
Old 16-March-2005, 04:27 PM
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Quote:
Originally Posted by Nowhere Man
ToSeek: I beat you!

In the summation part, one must be careful how one simplifies the series:
  • 9 * ((1/10) / (1 - 1/10))
  • 9 * ((1/10) / (9/10))
  • 9 * 1/9
At this point you can either say [...]
  • 9 * 0.111111111... = 0.999999999...
Which leads to the trivial conclusion 0.999... = 0.999...

Quote:
Originally Posted by Nowhere Man
[or] 9/9 = 1
Which proves what we wished to prove.

No problem there. 8)
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  #47 (permalink)  
Old 16-March-2005, 06:04 PM
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Yeah, the first one is definitely a non-rigorous way, but I always thought it was quite elegant and simple. The second one is rooted in more complex mathematics, but is certainly rigorous provided that you accept what calculus has shown about infinite series and things like that.
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  #48 (permalink)  
Old 16-March-2005, 06:32 PM
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I don't know if that IS calculus!!!




If you want another problem, try proving whether or not prime numbers ever end, or do they just keep finding higher and higher prime numbers, the more powerful computers get...?




8)



Quote:
Infinite Series
"In his first paper on the Calculus (1669), Newton proudly introduced the use of infinite series to expedite the processes of the calculus... As Newton, leibnitz, the several Bernoullis, Euler, d'Alembert, Lagrange, and other 18th-century men stuggled with the strange problem of infinite series and employed them in analysis, they perpetuated all sorts of blunders, made false proofs, and drew incorrect conclusions; they even gave arguments that now with hindsight we are obliged to call ludicrous."
>From "MATHEMATICS: The Loss of Certainty" by Morris Kline.
Ok, I'm wrong again....!
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  #49 (permalink)  
Old 17-March-2005, 06:41 PM
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Quote:
Originally Posted by Frog march
If you want another problem, try proving whether or not prime numbers ever end, or do they just keep finding higher and higher prime numbers, the more powerful computers get...?
The infinitude of primes has been known for thousands of years. And its an easy proof. In general, for any positive integer n (prime or otherwise), the prime factors of n!+1 are all > n.

Examples:

n=5, n!+1 = 121, Factors are 11,11
n = 6, n!+1 = 721, factors are 7,103
n=10, n!+1 = 39916801, prime
n = 12 , n!+1 = 479001601 , factors are 13,13,2834329


If the primes are finite, then there is a largest one. Call it p. Then p!+1 is not divisible by any number <= p. Therefore the prime factors of p!+1 are all > p, which contradicts the notion that p is the largest prime.
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  #50 (permalink)  
Old 17-March-2005, 10:30 PM
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Try to prove or disprove that R and R^2 and C are the same size, were R is the real number line, R^2 is the Plane of all real numbers, and C is the plane of complex numbers. Prove the set R, R^2, and C are all the same size.

Then prove, That for all n,x,y,m size of C^n = size C^x = size R^y = size R^m where n,x,y,m can be any number.
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  #51 (permalink)  
Old 17-March-2005, 10:38 PM
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Quote:
Originally Posted by VTBoy
Try to prove or disprove that R and R^2 and C are the same size, were R is the real number line, R^2 is the Plane of all real numbers, and C is the plane of complex numbers. Prove the set R, R^2, and C are all the same size.

Then prove, That for all n,x,y,m size of C^n = size C^x = size R^y = size R^m where n,x,y,m can be any number.
Why?