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TruthlessHero
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 Re: Stuck: How can you guess that?   « Reply #25 on: Oct 5th, 2006, 12:50pm » Quote Modify

Do you mean 100 digits or 100 thousand digits? Becaue I know people have memorized more than 100 digits of pi. Also, I believe it was my idea for the sqrt not honkeyboy's. I still stick to that answer even though Icarus doesn't think it's right.
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jollytall
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 Re: Stuck: How can you guess that?   « Reply #26 on: Oct 6th, 2006, 12:42am » Quote Modify

Sorry TruthlessHero, I missed it up. Of course credit should go, where it is due.

I meant 100 thousand digits obviously (decimal and thousand separators are different in different languages/local practices and by mistake I used the Hungarian thousand separator rather than the English). Still I hope you did not read 100.000 digits but 100,000 digits, since the number of digits is always integer, hence there would be no meaning of decimals.

Furthermore 100 digits even I could remember if invested some time, but the 100,000 number I found astonishing.
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TruthlessHero
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 Re: Stuck: How can you guess that?   « Reply #27 on: Oct 6th, 2006, 3:44am » Quote Modify

True, 100k is most impressive.
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grungy
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 Re: Stuck: How can you guess that?   « Reply #28 on: Dec 17th, 2006, 2:38am » Quote Modify

Yeah, but doesn't Pi start to repeat after a few hundred digits?  That, admittedly, would make it easier.  Although, 100K still is freaky... wouldn't he at some point screw up?
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towr
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 Re: Stuck: How can you guess that?   « Reply #29 on: Dec 17th, 2006, 3:06am » Quote Modify

on Dec 17th, 2006, 2:38am, grungy wrote:
 Yeah, but doesn't Pi start to repeat after a few hundred digits?
That would mean it was rational. It actually never repeats.
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Icarus
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 Re: Stuck: How can you guess that?   « Reply #30 on: Dec 17th, 2006, 8:25pm » Quote Modify

The several billion digits of pi that have been calculated have been studied several times for any hint of a pattern - even a subtle one. None has ever been found. And as towr indicates, we can prove that it never repeats.
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Gollelio
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 Re: Stuck: How can you guess that?   « Reply #31 on: Dec 27th, 2006, 7:03pm » Quote Modify

Hello and Merry Xmas!
I saw this forum while searching for some riddles to solve and i can say that i like the whole concept.

Now to the riddle. I think a question could be "If n is the number you are thinking of, is 1/(n-2) positive?"

So

If n=1, 1/(1-2) negative, so No.
If n=2, 1/(2-2) indefinite, so Don't know if it's positive.
If n=3, 1/(3-2) positive, so Yes.

The hint in the original post was very helpfull though, so i was searching for a form of indefinability.

As i quickly scanned through the answers later i saw a correct answer, the one with the complex numbers and another that tried to use the 1/0 but since the question had = and not > the 1/0 would count as no.

Edit: I saw the answer with the kidneys too. Very good answer!

 « Last Edit: Dec 27th, 2006, 7:08pm by Gollelio » IP Logged
Icarus
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 Re: Stuck: How can you guess that?   « Reply #32 on: Dec 27th, 2006, 10:57pm » Quote Modify

Sorry - that doesn't work. Because 1/(2-2) is undefined, it is not positive (it is not negative either, or any thing else that requires it to exist), so your answers are
n=1: "No"
n=2: "No"
n=3: "Yes"

The answer with complex numbers suffers the same problem. i is not > 0, so the answer for 2 there is also "no", not "I don't know".

One question would be: Do any non-real zeros of the Reimann zeta function have real part less than (n-1)/2?

Until someone proves or disproves the Reimann hypothesis, the answers have to be:
1: no
2: I don't know
3: yes

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BNC
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 Re: Stuck: How can you guess that?   « Reply #33 on: Jan 1st, 2007, 3:54am » Quote Modify

on Dec 27th, 2006, 10:57pm, Icarus wrote:
 One question would be: Do any non-real zeros of the Reimann zeta function have real part less than (n-1)/2?   Until someone proves or disproves the Reimann hypothesis, the answers have to be: 1: no 2: I don't know 3: yes

Being honest and truthful, I would answer:
1: I don't know
2: I don't know
3: I don't know

...
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Icarus
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 Re: Stuck: How can you guess that?   « Reply #34 on: Jan 1st, 2007, 8:36am » Quote Modify

You could include a short description of the Riemann hypothesis (my apologies to Riemann for mispelling his name in my earlier post) in the question, but then it boils down to the same thing as Paul Hammond's suggestion.

Of all the approaches so far, I like Paul Hammond's best: "I'm thinking of either 1.5 or 2.5 - is your number greater than mine?"

It is a variant of the Unsolved problem method, but in this case, you can be sure that the other person doesn't secretly have a solution to the problem, and is aware of what the problem is.

And I can't agree with Barukh about it suffering from the "is that really one question" shortcoming.

-----------------------------------------------------------------

Just for the record: The Riemann zeta function is defined for real s > 1 (using s as the variable is traditional) by
(s) = 1-s + 2-s + 3-s + ...
It can be extended analytically to the entire complex plane except s=1. It is fairly easy to show that (s) = 0 if s is a negative integer, and (s) is non-zero for all other real s. It has also been shown that the non-real zeros of (s) must have 0 < Re(s) < 1, where Re(s) is the real part of s.

The Riemann Hypothesis says that for all non-real zeros of (s), Re(s) = 1/2.

This is important because of a relationship that Euler discovered:

1/(s) = (1-2-s) (1-3-s) (1-5-s) (1-7-s) (1-11-s)...,

where the product is for all primes in the base. Riemann used this relation to prove an early version of the Prime number theorem (whose full version he also conjectured), which gives the overall density of primes. If we could be sure that all non-real zeros of lie on the line Re(s) = 1/2, a large number of stronger results will follow. It is quite common in Analytic Number Theory to see a theorem begin with "If the Riemann Hypothesis is true, then ...".
 « Last Edit: Jan 1st, 2007, 8:43am by Icarus » IP Logged

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 Re: Stuck: How can you guess that?   « Reply #35 on: Jan 2nd, 2007, 1:30am » Quote Modify

Thanks for the explaination, Icarus
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Locke64
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 Re: Stuck: How can you guess that?   « Reply #36 on: Jan 5th, 2007, 5:25pm » Quote Modify

I don't see how there can possibly be any better question than
"Is your number greater than a number that is randomly selected between 1.5 and 2.5?".  There may be more that accomplish the same thing, but I don't know how they could possibly be better, as this gets the job done.  I think I'm being a bit redundant... :(
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Three Hands
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 Re: Stuck: How can you guess that?   « Reply #37 on: Jan 9th, 2007, 1:15am » Quote Modify

Well, if he randomly selects a number between 1.5 and 2.5, then he can provide a "yes" or "no" answer for if he is thinking of "2", thus creating some problems...

I think you could do better if you say "I have randomly selected a number between 1.5 and 2.5. Is the number you are thinking of greater than the number I have selected?", although I suspect someone will come up with a suitably devious way around this...
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JiNbOtAk
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 Re: Stuck: How can you guess that?   « Reply #38 on: Jan 30th, 2007, 5:52pm » Quote Modify

If 3 is not the number that you have in mind, is the number odd ?

Yes : 1
No  : 2

Now, what is wrong with that approach ?
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 Re: Stuck: How can you guess that?   « Reply #39 on: Jan 30th, 2007, 6:42pm » Quote Modify

on Jan 30th, 2007, 5:52pm, JiNbOtAk wrote:
 If 3 is not the number that you have in mind, is the number odd ?

Well if they have 3 in mind than they would say "No".  Or am I misreading this?
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Icarus
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 Re: Stuck: How can you guess that?   « Reply #40 on: Jan 30th, 2007, 7:42pm » Quote Modify

Actually, my interpretation is that if 3 is the number, they are free to answer in any way they like.

An if-then statement is true when the if clause is false, regardless of the truth value of the predicate.

Answering your question "Yes" is equivalent to saying "If my number is not 3, then it is odd" - true for 1 or 3.
Answering "no" is equivalent to saying "If my number is not 3, then it is not odd" - true for 2 or 3.
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CowsRUs
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 Re: Stuck: How can you guess that?   « Reply #41 on: Feb 8th, 2007, 5:19pm » Quote Modify

Is it Three?
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maxmin
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 Re: Stuck: How can you guess that?   « Reply #42 on: Feb 9th, 2007, 4:25am » Quote Modify

Is F(n) possitive if
F(x) < 0 if x < 1.5
F(x) > 0 if x > 2.5 ?

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Icarus
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 Re: Stuck: How can you guess that?   « Reply #43 on: Feb 9th, 2007, 8:04pm » Quote Modify

on Feb 9th, 2007, 4:25am, maxmin wrote:
 Is F(n) possitive if   F(x) < 0 if x < 1.5 F(x) > 0 if x > 2.5 ?

n=1: "No"
n=2: "No"
n=3: "Yes"

To be positive F has to be defined, since F is not defined for 2, it is not positive there.
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CowsRUs
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 Re: Stuck: How can you guess that?   « Reply #44 on: Feb 11th, 2007, 1:28pm » Quote Modify

Of course you can always ask if:
Is the number prime?
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TruthlessHero
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 Re: Stuck: How can you guess that?   « Reply #45 on: Feb 11th, 2007, 1:35pm » Quote Modify

on Feb 11th, 2007, 1:28pm, CowsRUs wrote:
 Of course you can always ask if: Is the number prime?

Not really:

1:
2: Yes
3: Yes

I'm not sure what 1 is technically considered, but since 2 and 3 are both prime, it doesn't matter.
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CowsRUs
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 Re: Stuck: How can you guess that?   « Reply #46 on: Feb 11th, 2007, 2:14pm » Quote Modify

forgot prime And odd or a lot of And combinations
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Icarus
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 Re: Stuck: How can you guess that?   « Reply #47 on: Feb 11th, 2007, 8:27pm » Quote Modify

on Feb 11th, 2007, 1:35pm, TruthlessHero wrote:
 I'm not sure what 1 is technically considered, but since 2 and 3 are both prime, it doesn't matter.

1 is considered a "unit", not a prime. For the natural numbers, this is a trivial distinction, but the concept of primes and factorization can arise in more general settings, called "Unique Factorization Domains", or UFDs (the set of natural numbers is not a UFD itself, but the full set of integers is). In UFDs the various "numbers" fall into three types: units, primes, composites. Units are "numbers" that have an inverse. You can always "factor" them out of any number by multiplying by the inverse. Factorization is only "unique" up to multiplication of the factors by units. Primes are numbers x such that if x = y*z, then either y or z is a unit. Composites are everything else.

After the integers, the next example of a UFD is the Gaussian integers: complex numbers of the form a+b, with a and b integers. The units of the Gaussian integers are 1, -1, , -. The primes turn out to be the natural prime numbers of the form 4k+3 (and their opposites and multiples by and -), and all numbers of the form a+b, where a2 + b2 is a natural prime of the form 4k+1.

Other well known examples:
The set of all polynomials in one variable with complex coefficients. The units consist of the complex numbers (which are polynomials of degree 0). The primes are all the linear polynomials (those of the form ax + b).

The set of all real polynomials in one variable. Units are all real numbers. Primes are linear polynomials, and quadratic polynomials with non-real roots.

The set of all polynomials in one variable with integer coefficients. The units are 1 and -1. And just like the natural numbers, identifying exactly which are prime is not an easy task.
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TruthlessHero
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 Re: Stuck: How can you guess that?   « Reply #48 on: Feb 11th, 2007, 8:36pm » Quote Modify

on Feb 11th, 2007, 8:27pm, Icarus wrote:
 1 is considered a "unit", not a prime. For the natural numbers, this is a trivial distinction, but the concept of primes and factorization can arise in more general settings, called "Unique Factorization Domains", or UFDs (the set of natural numbers is not a UFD itself, but the full set of integers is). In UFDs the various "numbers" fall into three types: units, primes, composites. Units are "numbers" that have an inverse. You can always "factor" them out of any number by multiplying by the inverse. Factorization is only "unique" up to multiplication of the factors by units. Primes are numbers x such that if x = y*z, then either y or z is a unit. Composites are everything else.   After the integers, the next example of a UFD is the Gaussian integers: complex numbers of the form a+b, with a and b integers. The units of the Gaussian integers are 1, -1, , -. The primes turn out to be the natural prime numbers of the form 4k+3 (and their opposites and multiples by and -), and all numbers of the form a+b, where a2 + b2 is a natural prime of the form 4k+1.   Other well known examples: The set of all polynomials in one variable with complex coefficients. The units consist of the complex numbers (which are polynomials of degree 0). The primes are all the linear polynomials (those of the form ax + b).   The set of all real polynomials in one variable. Units are all real numbers. Primes are linear polynomials, and quadratic polynomials with non-real roots.   The set of all polynomials in one variable with integer coefficients. The units are 1 and -1. And just like the natural numbers, identifying exactly which are prime is not an easy task.

Interesting. Might I inquire as to what math degrees you have, if any?

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Whiskey Tango Foxtrot
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 Re: Stuck: How can you guess that?   « Reply #49 on: Feb 11th, 2007, 9:52pm » Quote Modify

Be careful, Hero.  I don't think you meant to sound pretentious, but it came out that way.  There's been a rash of Icarus-doubters recently and I know I would be past the point of mere annoyance.
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