Maths Query

[Requiemdk]

The problem he posted is known as the Collatz Conjecture (naughty naughty fella -.-) and is so far unproven though experimental evidence supports it, much like Goldbach's and the Twin Prime Conjectures (students are introduced to these 2 very simple conjectures very early in their school life but usually forget them once they hit sec school). Actually, just asking someone to prove something as seemingly trivial an observation as the Goldbach Conjecture would be enough. In comparison, the Collatz Conjecture looks a lot more confusing to laymen.

 

There're frequently smaller breakthroughs in Maths, so it's not true that Maths isn't exciting just because the Riemann-Zeta Hypothesis doesn't look like it'll be proven anytime soon. It just happens to be a famous problem with a million dollar bounty but there're other equally exciting problems to work on. There're enough young mathematicians who make the mistake of gunning for famous problems right from the start before contributing much to the field.

Edited January 25, 2008 by Requiemdk

[Redwood]

Thanks for directly to the site.

 

The question posed in the forum by unclewolve should be the modified form of the "Collatz Conjecture" which accordingly to the article, have a proof.

 

Seems that you're into Mathematics also! Are you a Math major, Still studying?

[Requiemdk]

Teaching currently, but studying on the side too to prepare for further studies. Are you sure the Collatz Conjecture has been proven? I've read a few articles before a couple of years ago that claimed to have proven it, but I never saw any that were refereed and defended. Any link for reading?

 

Also, about the question in the original post, I don't get why that triangle is isosceles. Drawing any circle with a diameter of 18 will give all the triangles with that configuration. I can't see why we're assuming that it's isosceles unless there's some information that I'm missing out on...

[Redwood]

You teaching in the university? Doing Master/PHD in Maths? If you're in uni. there are tons of research material/papers accessible in the school database.

 

The function given in the forum was f(x)=(3x+1)/2, thus this is the modified version of the problem in "Collatz Conjecture" where the original is f(x)=3x+1. The article mentioned that the proof is simple and can be verified by hand. Normally for problems in number theory solution are not easily found in the internet, even for the simple one. Can you provide your own version of proof to enlighten me, you should be much experience since you teach.

 

I did not tried the triangle problem.

[UncleWolve]

Surprised that it took these while for someone to shout Collatz! ... 3X+1~~~!!!

[Redwood]

Seems like the modified "Collatz Conjecture" is being proven.

"Terras Theorem"

 

http://www.ericr.nl/wondrous/terras.html

[Redwood]

Could this be the solution,

http://www.ericr.nl/wondrous/terras.html

[Requiemdk]

It seems to me that the Terras Theorem is a way of stating that if there exists numbers for which the iteration fails, it'll only be a finite number of them.

 

Anyway, I have to apologize for not reading properly. I saw (3x+1) and immediately assumed that it was the Collatz Conjecture.

 

The iteration presented by Terras is another type of Collatz function though and it seems like proving that it terminates at 1 for all positive integers conclusively would be another intractably hard problem.

 

Edit: Erm and I'm not teaching in a uni. haha. If I were, I'd probably have better answers to these things.

Edited January 25, 2008 by Requiemdk

[T_k_w]

i read the link you posted, quite cool, the way they use parity vectors.

 

The terras Theorem states:

Let M be a positive integer.

Let D(M) be the fraction of numbers < M that do not have finite stopping time.

Then the limit of D(M) for (M tends to 0) is equal to 0.

 

That means if we select an arbitary subset of positive integers, we don't know how many within the set will diverge, but if we look at the entire set of positive integers, (the number of integers that will not diverge) drops to (0+)? (wtf?)

 

So is this terras theorem a recognised proof?

 

when i bohliao, i like to go to this site to look look see see.

http://mathworld.wolfram.com/CollatzProblem.html

 

Quoting mathsworld,

"Conway (1972) also proved that Collatz-type problems can be formally undecidable."

 

"...Because of the difficulty in solving this problem, Erdos commented that "mathematics is not yet ready for such problems" (Lagarias 1985)."

 

Haha...

Edited January 25, 2008 by T_k_w

[Redwood]

Maybe try a few simple problems since this is a maths tread,

 

1. Proof that the increment product of n consecutive number divided by n factorial is an integer.

Example,

suppose n=3, start number is 6 thus 6*7*8/(3!) = 56 which is an integer.

suppose n=5, start number is 9 thus 9*10*11*12*13/5! which is an integer also.

 

2. Given a number, if sum of all digits is divisible by 3 then the number is divisible by 3 also.

Example, number = 123, digit = 1+2+3=6, since 6 is divisible by 3 implies 123 is also divisible by 3.

[T_k_w]

1. Proof that the increment product of n consecutive number divided by n factorial is an integer.

Example,

suppose n=3, start number is 6 thus 6*7*8/(3!) = 56 which is an integer.

suppose n=5, start number is 9 thus 9*10*11*12*13/5! which is an integer also.

 

 

how you determine 6 and 9? arbitary?

[T_k_w]

2. Given a number, if sum of all digits is divisible by 3 then the number is divisible by 3 also.

Example, number = 123, digit = 1+2+3=6, since 6 is divisible by 3 implies 123 is also divisible by 3.

 

I think i proofed this before(cannot remember, should be when i start giving tuition because got student ask before), I teach this to my students the moment they enter sec 1. Very useful for emaths.

Edited January 25, 2008 by T_k_w

[Redwood]

you're right arbitrary. can be any number.

Any maths problem you would like to pose?

 

"Requiemdk" mentioned that the Terras is also intractable.

[T_k_w]

I like to(try and) solve questions(for fun), but I don't think i am zhai enuff to think of conjectures for others to prove...

 

Because I work as computer programmer, i find that doing this keeps brain from rusting, just like solving sudoku.

[Requiemdk]

1. Any general number like that is just k(k-1)...(k-n+1)/n! which is equivalent to kCn, i.e. the number of ways of choosing n objects out of k distinct objects and is hence an integer.

 

2. Should be a consequence of the fact that 10=1(mod3) (sorry, no congruence sign... haha). I suppose a less technical proof would be to observe that 10=3(3)+1, 100=3(33)+1, 1000=3(333)+1, etc.

 

So if we let the digits of n dcba for example (i.e. 1234: d=1, c=2, b=3, a=4), then n = d(1000)+c(100)+b(10)+a = d(3(333)+1)+c(3(33)+1)+b(3(3)+1)+a = 999d + 99c + 9b + (a+b+c+d) and from there, if a+b+c+d is divisible by 3, then the whole number is too.

 

Extension of 1 - prove that (nk)!/((k!)^n) is an integer.

[Redwood]

Wah you programmer, must be very good in those programming language. Have you tried writing program to solve sudoku?

[Redwood]

[T_k_w]

for k numbers and abitary n:

there exist an integer x such that x= n*(n-1)*..*(n-(k-1))/k!

 

since n*(n-1)*..*(n-(k-1)) = n!/(n-k)!

 

therefore n(n-1)*..*(n-(k-1))/k! = n!/[k!*(n-k)!]

 

so we can simplify the question as

for k numbers and abitary n:

there exist an integer x such that x= n!/[k!*(n-k)!]

 

can use induction i think... let me sleep on it.