The recurring decimal

The recurring decimal

Posers and Puzzles

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15 Dec 11

Originally posted by mikelom
I think you're right, but didn't explain it in a more simplistic way.

A picturing point of view is for repetitive halfs which could be endless, and yet we know there is an end.

Consider the analogy of throwing a stone into the air. We know to come down, it must come half way as high as it went. It must come half way again, and then half of that distanc ...[text shortened]... fore, this infinite half way argument is flawed, same as 0.9 recurring doesn't = 1.

-m. 😉
Nice one, we can even assume that the stone comes down 8/9ths of the remaining way each time, and it still definitely hits the ground!

F

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16 Dec 11

This 'paradox' is known from the ancient greek times. At that time, without knowledge of infinitesimality, it was really a paradox. Nowadays it isn't.

Those who don't know about infinity cannot understand why 0.99999... is exactly equal to 1.00000...

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17 Dec 11

Originally posted by FabianFnas
This 'paradox' is known from the ancient greek times. At that time, without knowledge of infinitesimality, it was really a paradox. Nowadays it isn't.

Those who don't know about infinity cannot understand why 0.99999... is exactly equal to 1.00000...
Yeah, it's Zeno all over again, and a tenth again, and a hundredth again, and a thou...

By the way, welcome back.

Richard

Art, not a Toil

60.13N / 25.01E

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17 Dec 11

You're, of course, right, Joe. Sorry about that. n+1, not n.

A modern variant of the dilemma..

Ask a taxi driver how far is the distance to to four blocks east and three north is, and he'll say seven. What if he's allowed a route closer to how-the-bird flies? even if you zigzag 400 times east and 300 times north, a hundredth of a block at a time, he'll still say seven. Four million tiny increments east and three million north, and the route is still seven. Arranging those terms so that the taxi will stay as close to the direct possible route as possible has the route approach the hypotenuse of a triangle with other sides three and four.

And thus.. with n the number of turns.. as n approaches infinity, lim 7 -> 5. ..?

Obviously the reason is not the same, but it goes well with fixing the never-ending problem of having high-school students suggest that sqrt(a+b) = sqrt(a) + sqrt(b), just like sqrt(ab) = sqrt(a) sqrt(b).

Some have learned to avoid that from having them find the mistake in the "proof" that 7 = 5;

7 = 4 + 3 = sqrt (16) + sqrt (9) = sqrt (16 + 9) = sqrt(25) = 5.

R
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17 Dec 11

Originally posted by talzamir
You're, of course, right, Joe. Sorry about that. n+1, not n.

A modern variant of the dilemma..

Ask a taxi driver how far is the distance to to four blocks east and three north is, and he'll say seven. What if he's allowed a route closer to how-the-bird flies? even if you zigzag 400 times east and 300 times north, a hundredth of a block at a time, he'l ...[text shortened]... he "proof" that 7 = 5;

7 = 4 + 3 = sqrt (16) + sqrt (9) = sqrt (16 + 9) = sqrt(25) = 5.
So which is the "true" distance. 7 or 5?

R
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19 Dec 11

Originally posted by talzamir
You're, of course, right, Joe. Sorry about that. n+1, not n.

A modern variant of the dilemma..

Ask a taxi driver how far is the distance to to four blocks east and three north is, and he'll say seven. What if he's allowed a route closer to how-the-bird flies? even if you zigzag 400 times east and 300 times north, a hundredth of a block at a time, he'l ...[text shortened]... he "proof" that 7 = 5;

7 = 4 + 3 = sqrt (16) + sqrt (9) = sqrt (16 + 9) = sqrt(25) = 5.
Does the limit really apply...I mean its only a "limit" in the visual sense( not in a calculus sense), regardless of "n" in this scenario the distance remains constant, at 7.

F

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20 Dec 11

Originally posted by joe shmo
Does the limit really apply...I mean its only a "limit" in the visual sense( not in a calculus sense), regardless of "n" in this scenario the distance remains constant, at 7.
A "limit" that differs the visual sense from the calculus sense is always wrong. When it does it is an origin of a paradox.

The Zeno paradox is an example of this. Visually it seems right, but calculus shows what is really happening.

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20 Dec 11
2 edits

Originally posted by FabianFnas
A "limit" that differs the visual sense from the calculus sense is always wrong. When it does it is an origin of a paradox.

The Zeno paradox is an example of this. Visually it seems right, but calculus shows what is really happening.
I think it is extremely hard to prove pythagoras' theorem via calculus, which is the same as trying to prove that the lower limit of the taxicab's journey is 5 units (without invoking pythagoras).

The vastly easiest approach is to prove pythagoras using some other technique (e.g a dissection) and then use pythagoras to derive the limit.

Here is one attempt to prove pythagoras with calculus, good luck following it!

http://www.scribd.com/doc/30552/A-Calculus-Proof-of-the-Pythagorean-Theorem

F

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21 Dec 11

Originally posted by iamatiger
I think it is extremely hard to prove pythagoras' theorem via calculus
I think not. As I recall I was presented the proof of Theorem of Pythagoras in an understandable way in the school before the age of 16.

To understand that by only going west and north n times it is not possible that the distance between two points is different regardless of n. When you know the concept of infinity it is obvious. Else it is impossible to understand.

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23 Dec 11

Originally posted by FabianFnas
I think not. As I recall I was presented the proof of Theorem of Pythagoras in an understandable way in the school before the age of 16.
But was that a proof using calculus? That is the salient point. There are many relatively simple ways of proving Pythagoras. Using calculus is probably not one of them.

Richard

F

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26 Dec 11

Originally posted by Shallow Blue
But was that a proof using calculus? That is the salient point. There are many relatively simple ways of proving Pythagoras. Using calculus is probably not one of them.

Richard
Of course it was. It have been repeated many times thereafter, the last time in University calculus. Along with numerous other calculus proofs.

Why? Don't you believe in the pythagoras theorem?

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29 Dec 11

Originally posted by FabianFnas
Of course it was. It have been repeated many times thereafter, the last time in University calculus. Along with numerous other calculus proofs.

Why? Don't you believe in the pythagoras theorem?
Interesting
Could you show us this proof or give a link to it?

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Shanghai

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29 Dec 11

Off the top of my head I can only think of using the arc length formula to prove pythagoras using calculus. This is one extremely bad flaw in that my proof for the arc length formula assumes pythagoras so it is a circular argument.

I am wondering though what is elegant and satisfying about proving pythagoras using a more complicated way? Mathematical proofs are at their best when they are concise and easy to understand.

Quiz Master

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29 Dec 11

http://www.cut-the-knot.org/pythagoras/index.shtml

F

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29 Dec 11

Originally posted by iamatiger
Interesting
Could you show us this proof or give a link to it?
Google yourself. Ask your math professor. Look it up in the math book.

So you don't believe in the pythagoras theorem...