I thought I should start a thread devoted solely to discussing mathematics and science.

Here's my current idea:

Imagine a building with various cubicles in columnn and rows. Let the amount of total desks (the amount of rows multiplied by the number of columns) be one number and the perimeter of the cubicles be another. Is there another number of cubicles whose numbers match up with the originals?


Mathematically:


A = {a_0, a_1, a_2, ..., a_n | a_i > 1 }
B = {b_0, b_1, b_3, ..., b_n | b_i > 1 }


?
Σ a_i = Σ b_i
Π a_i = Π b_i



Pigeon Holing:

n is the amount of objects
m is the amount of holes

If n>m, then at least one object will occupy the same hole.

Reapplied, we have:
n be the total amount of numbers
m be the total amount of outcomes

n = inf.
m tends toward infinity

Is n>m?
If so, then there must be two sets whose multiplicational and additional values equal each other.



If the two numbers are unique to each set, then a large array of numbers can be reduced to only two terms.

(This also has some relation to P vs. NP, but I'd rather not get into that)

---

My second idea:

Scatter Plot Accurate Lining

Both of these ideas I developed when I was in the car on a long trip. I didn't have any mathematical tools handy, so, I don't know if this was already done or not.

--Shorthand Method
The shorthand method involves finding the slope between every two consecutive points.


[m 1 b]


[-Σ m_i/n 1 Σ b_i/n]

Where:




So:



--Full Method
The Full Method involves finding the slope between EVERY two point combinations.






y = mx + b




Now, I'm not sure if my sigma notation is right since I've never had formal training in sigma and pi notation. Also, I'm not sure if this has been done before, nor if it can be reduced.

Post has been edited 2 time(s), last time on Apr 29 2008, 5:00 am by A_of-s_t.

I don't quite get what you mean with the first thing. Do you mean having two different number sets, that when you sum up all their numbers the result is the same and also if you multiply all of them?

If so, the sets are not unique:


Example:


So yeah they're not unique in R, not sure if you're talking about natural or integer numbers, there it can get a lot more tricky.

And about the pidgeon holing principle, you cannot make n REACH infinity, because once it reaches the infinite there's nothing that ensures you that the principle still holds...

I meant:

Imagine that you have two set of numbers. We know that we can have two sets have the same value when multiplied together. We also know that we can have the same values when added together.

Ex.

2x2 = 1x4

2+2 = 1+3

But are there two sets of numbers that can have BOTH of these two values equal?


And... we know that the the whole numbers greater than one ( let the sets be comprised of whole numbers greater than one ) is infinite. And we know that the sets of numbers produced is infinite, but is less than the infinity produced by the numbers of whole integers...

I've been thinking about this. If the numbers can be any real number, the equation I gave you in my first post gives you all the solutions. But if the numbers have to be integer or integer and possitive then the question becomes a whole lot harder to answer.

A = { 6 }
B = { 1, 2, 3 }

Those two set have the same multiplication and addition values. What I don't know is if the sets that have that property are finite or infinite. Or if for any set you can find another one with that property...

P vs. NP is not worth your time... Nobel prize stuff isn't going to be solved over sen sorry to crush your dreams lol. But the pigeon hole theory is a little flawed...

That doesn't apply because Inf will always ruin your theory. You cannot have n=Inf, because if m=Inf, n cannot be >Inf. Also, nothing eliminates the possibilities of n=A , m= >A. It says "If n>m", but it doesn't cover an if not...

Any outstanding rules I missed?

Post has been edited 1 time(s), last time on Apr 30 2008, 5:40 pm by Jello-Jigglers.

Actually there are different infinity levels, but it is not as simple as you'd think.

For example:
A = { 1, 2, 3, 4, 5, 6, .... }
B = { 1², 2², 3², 4², 5², ... }

It's surprising that both A and B have exactly the same number of elements. It seems illogical, but actually it is not so, because you can "compare" them relating their members with a one-by-one relation:

1 - 1²
2 - 2²
3 - 3²
...

For every member in A you can find a partner in B and every member in B also has a partner in A. So A and B are equally sized.

The pidgeon hole theory is not flawed, you just have to be careful when dealing with infinity, because there might be the same number of holes and pidgeons when seems otherwise.

I can explain more stuff about the cardinal theory (that deals with the size of sets) if you want...

en.wikipedia.org/wiki/Perfect_number
http://www.research.att.com/~njas/sequences/A000396

Yeah, I already knew about perfect numbers, but not all perfect numbers would work, since if they have more than two factors (like 28 = 2*2*7) when you multiply all their divisors you won't get the number again:

1 + 2 + 2*2 + 7 + 2*7 = 28
1 * 2 * 2 * 2 * 7 * 2 * 7 = 784

That's why I didn't say anything about perfect numbers...

You said perimeter. Perimeter of 2x2 is 8 (2 + 2 + 2 + 2) or 2x(2 + 2)

Either way, it seems an interesting question. I tried my hand, but the proof, one way or the other, is likely not straightforward. Solving 2 equations in 4 unknowns doesn't really work, so another method would be required. I tend to think that such numbers do exist, since I didn't run into any apparent contradictions when running through it.


Edit: Clokr_, You start with the divisors (or factors), list them, then sum them. If they sum to the number, that number is a Perfect Number. You don't start with sums and then multiply them. (Unless you're trying to communicate something entirely different, then I have no idea what you're talking about)

Post has been edited 1 time(s), last time on May 1 2008, 12:33 am by frazz.

@Clokr_ -- Ah, I see. I was making the assumtion that the number of outputs of the function are less than the inputs of the function...
So, if it were to work, there would need to be a proof that (x, f(x)) is not a one to one function. Which, I think can be properly proved since:

9 * 4 = 36 == 3 * 3 * 2 * 2
6 * 6 = 36 == 3 * 3 * 2 * 2

So, if the function is not one to one, then the out put is less than the input. I'm not sure, since I've never had formal teaching in pigeon holing, but that shows us that n>m.


@frazz -- it is a tricky problem since four variables are involved and I have yet to find a graphing calculator that can use sigma or pi notation -- let alone be able to graph four pieces of information.

Here's a proof I started:

A = {a1, a2, a3, ..., an}
B = {b1, b2, b3, ..., bn}

Assuming that this is true:
Σ a_i = Σ b_i
Π a_i = Π b_i


Σ a_i - Σ b_i = 0

Π a_i - Π b_i = 0

Σ a_i - Σ b_i = Π a_i - Π b_i

Σ (a-b)i = Π (a-b)i


I'm trying to prove that a_n must equal b_n to prove the theory that each set of numbers have a unique set of two numbers.

Wait wait. What are you trying to prove or disprove? What is the hypothesis? You seem to have strayed from the area-perimeter concept.

If so, proving that the set A equals the set B will prove that there are no such pairs of sets.
If you want merely to prove the existence of some pairs, that might be a bit tougher. It would be best to suppose that no such pairs exist, and follow that to a contradiction (I think).

A quick check could be done with computers, which would likely yield an example of such sets (boring...).

A_of-s_t was asking if we could find two number sets A = { a1, ..., an }, B = { b1, ..., bm } so that:
a1 + ... + an = b1 + ... + bm
a1 * ... * an = b1 * ... * bm

So I told him that the two sets { 6 } and { 1, 2, 3 } make those two equations true. However, I was saying that not every perfect number would work, because the second equation wouldn't be true with most of them.


Two sets A and B are equally sized if you can find a bijective function from A to B (f: A -> B). A bijective function is a function that relates their elements one-to-one.

However, the inverse of the theorem would look this way:
Two sets A and B are not equally sized if you can't find a bijective function from A to B.

Notice that proving that there exists a function from A to B that is not bijective doesn't mean that they're not equally sized. You have to prove that there isn't any bijective function at all.


You could easily build a C program that prints a value list to a file, then use a graphing program to draw the results. But anyway, if you use a graphic calculator, wouldn't you be accepting any real number (decimal numbers included)?
If so, I've already given you a proof in post #1 that there exists infinite sets that make true those two equations. For example:

And if you only want integer numbers I don't know how a graphic calculator will be useful...


Sigma and Pi notations are not so good... Whenever I have to operate with them I usually use the ... notation since it is a lot more intuitive.


Σ a_i - Σ b_i = Π a_i - Π b_i

Σ (a-b)i = Π (a-b)i


Σ a_i - Σ b_i = a1 + a2 + ... + an - b1 - b2 - ... - bn = (a1 - b1) + (a2 - b2) + ... + (an - bn) = Σ (a_i-b_i)
Π a_i - Π b_i = a1 * a2 * ... * an - b1 * b2 * ... * bn but this does not equal (a1 - b1)*(a2 - b2)*...*(an - bn) = Π (a_i-b_i)

So your proof is wrong.

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Post has been edited 1 time(s), last time on May 2 2008, 10:09 pm by Doodle77.

I ran a program, doing two pairs will not yield results early on. I may try to refine my algorithm and see if I can go above the first few thousand numbers.
Either way, it looks unlikely (for pairs). It also seems intuitively unlikely to me. Think about it. If you have two pairs that multiply to the same number, they're all factors. If they're different though, they're either closer together or farther apart. Like 6x1 and 2x3. When closer together, they seem to sum to a lesser number in general. They should never sum to the same number.

Meh, SEN's not the best place to write math stuff... ASCII text makes reading hard. Anyway, if you're focusing on pairs doing what I did on post #1:

If we have a pair A = { x', y' } with a = x' + y' and b = x'y'
We want to find another pair B = { x, y } so that:
x + y = a
xy = b

x = a - y
(a - y)y = b
ay - y² = b
y² - ay + b = 0
y = (a ± sqrt(a² - 4b))/2 = (x' + y' ± sqrt(x'² + 2x'y' + y'² - 4x'y'))/2 = (x' + y' ± sqrt(x'² - 2x'y' + y'²))/2 = (x' + y' ± sqrt((x'-y')²))/2 = (x' + y' ± (x' - y'))/2

Case 1:
y = (x' + y' + x' - y')/2 = 2x'/2 = x'
x = a - y = x' + y' - x' = y'

Case 2:
y = (x' + y' - x' + y')/2 = 2y'/2 = y'
x = a - y = x' + y' - y' = x'

So you were right, given a pair there's no other pair with that property. However when talking about triplets there are infinite ones for whose that property holds. What I don't know if there's any integer one...

Hey, it's our old buddy quadratic formula. Fun.
Any integer one? You mean like for any set of two n-tuples? Or, even more generically, m sets of n-tuples?

For any given n-tuple with n > 2 you can find infinite n-tuples with that property, since you only have two equations to work with, so you can only calculate the value of 2 numbers, and the rest can be whatever you want them to be.
What I don't know is if there are pairs of n-tuples of integer numbers with that property.

If there are, it doesn't seem apparent.

If the n-tuple contains the zero then finding another one with integer numbers is not hard:

A = { 0, 2, 3 }
B = { 0, 1, 4 }

If they do not I have no idea how you can either prove that it's possible or not to find another n-tuple.