Translate

Tuesday, April 21, 2015

How social can work with math

Mathematics is one of those rare areas of human endeavor where a result is absolutely right or wrong.

Absolutes are a special kind. And nothing any human being can do can shift that absolute reality one way or the other. Either the math works or it does not.

But people can SAY anything, which is where social effects can be fascinating.

If you argue against an absolute truth, you can not only destroy the trust of others in your opinion, but also your trust in your own opinion, within yourself.

Regardless, the math doesn't care.

Social helps though, I think. Talking out ideas can be extremely useful. But you have to keep it in perspective.

Or in other words, you can waste your time arguing with people about math or you can enjoy your time finding it.

And if the mathematics works, trust human curiosity and need, but don't worry about it.

Want to see an example of an absolute mathematical truth?

Here's one:

If x2 + y2 = z2 then (v2 - 1)z2 - 2xy = 0(mod x+y+vz), where v can be any value.

And I have up a post stepping through in enough detail to make certain, well, of absolute certainty: Example showing truth logic and absolute proof

There is a plodding quality to valid mathematics where you can just check step-by-step, and every piece must be perfect. I find that comforting. But to some the effort may seem difficult, but I think it is refreshing. Passion for what you do, gives you the energy to do what it takes.

Maybe social can help reinforce views that give permission to escape the work, if people tell each other what they wish to hear.

But the math doesn't care how hard you think it is, or not.

The proper point of view with mathematics is: absolute proof.

Accept nothing less.


James Harris

Thursday, April 16, 2015

BQD Iterator and split points

How can you get multiple solutions for a sum of squares? Turns out an easy way is with my binary quadratic Diophantine iterator or BQD Iterator for short:

If: u2 + Dv2 = F

then it must be true that

(u-Dv)2 + D(u+v)2 = F(D+1)

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

The BQD Iterator lets you start with one known solution and get new ones by iterating. And at each iteration you have split points, where you may get two new solutions.

For example let u = 1, and v = 2, and D = 4, where I'm picking to get a sum of squares.

Then 1 + 4*4 = 17, so F = 17. Now you can get new values.

The first iteration then with all positives is:

(-7)2 + 4(3)2 = 17*5

But I have a split point around negatives, as u = -1 and v = 2, will also give F = 17, since you're squaring, so using those:

(-9)2 + 4(-1)2 = 17*5

So each iteration actually gives me two potential solutions, and you can find multiple solutions that sum to give the same thing.

To demonstrate I'll give one more iteration with each.

For (-7)2 + 4(3)2 = 17*5, next iteration:

(-19)2 + 4(-4)2 = 17*5and (-5)2 + 4(10)2 = 17*52


For (-9)2 + 4(-1)2 = 17*5, next iteration:

(-5)2 + 4(-10)2 = 17*5and (13)2 + 4(8)2 = 17*52


Which is interesting as ignoring signs, one solution shows up twice. So you don't always get unique solutions or you'd have the total number doubling at each split point. Also if u = v, or -v, you will not have a split point either.

And just using positives so I can make it look pretty, and pulling the 4 into the square, I have:

192 + 82 = 52 + 202 = 132 + 162 = 17*52

Which demonstrates how to use split points to generate multiple sums of squares for the same value.

The power of D+1 matches iterations, and maximum number of solutions for n iterations is 2n but as we see above, you can get the same solution more than once, or have u = v, or -v. So though that maximum value is 4, we only had 3 uniques by sign with our example.

You can do more than a sum of two squares though, and I have a post showing how to sum a specific number of squares, so you can go looking for multiple solutions for exactly the number of squares you decide.

That could be useful for many fun things.


James Harris

Wednesday, April 15, 2015

Celebrating twenty years of searching

Twenty years ago I was walking around thinking to myself about wanting to engage in the mental exercise of considering old math problems considered hard. Didn't write anything down about those thoughts, but know I wanted to keep up with how much time had passed in case I needed to convince myself to quit, and remember April 15, 1995, as the day.

So I celebrate today as the start. Though I admit to myself I'm not completely sure. I'm sure it's close enough, and could be the correct day too. Good enough for me.

This time I'm twenty years distant and talking about it publicly, as that seems like a milestone worth putting out there.

So here's a post. That was fun.


James Harris

Tuesday, April 14, 2015

Summarizing non-polynomial factorization conclusions

Remarkably pushing a simple algebraic factorization beyond polynomials leads to far reaching conclusions, including the find of numbers which escaped prior non-field ring classification schemes. Here I'll summarize by pulling key points from my post which steps through the full argument.

Deliberately I use a very simple start, as simple as I could get with:

P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

Introduce k, where k is a nonzero integer, and not 1 or -1, and new functions f1(x), and f2(x), where:

g2(x) = f2(x) + k-2 and g1(x) = f1(x)/k,

which gives me the now symmetrical form:

k*P(x) =  (f1(x) + k)(f2(x) + k)

And introduce H(x), where I like using the capital letter here for visual reasons, but it is not to signify H(x) must be a polynomial, where:

f1(x) + f2(x) = H(x)

So I can find:

f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0

And you can solve for f1(x) using the quadratic formula:

f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2

This result tells us that the f's are in the ring of algebraic integers if H(x) is.

It also shows that the f's are not linear equations unless the g's are, when k does not equal 1 or -1. So they have this square root which will only resolve in general for all integer x, if the g's are linear with integer coefficients and our original factorization is a polynomial one.

But of course polynomial factorizations are well understood and boring, so it is more interesting to move to non-polynomial ones, where now we can see that the f's will still be algebraic integers, as long as H(x) is, but they will tend to be non-rational for integer x, because the square root will force it.

And H(x) has constraints as it must fit with the given equations, for instance H(0) = -k + 2, is a requirement from them.

Let's look at a non-polynomial factorization example.

P(x) = 175x2 - 15x + 2,

and let k = 7.

And further I have:

7*P(x) = 7(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x) + 7)

So f1(x) = 5a1(x), and f2(x) = 5a2(x), where the a's are roots of:

a2 - (7x-1)a + (49x2 - 14x) = 0

Which means H(x) = 5(7x - 1), as you can see the a's sum to (7x-1), and just multiply that by 5, to get the sum of the f's, as H(x) =  f1(x) + f2(x).

However, I didn't just make up an H(x), but worked it out from a non-polynomial factorization I had found already from a much more difficult argument, by which I found the a's. Turns out I had that first, which seems to be a functional approach to solving these type of equations, which requires use of a special construction.

That's because things have to be highly particular if you will have g's which are still like integers, even though non-rational. And I've explained that in detail in an earlier post.

So now what do we know? Well with an example I have values for the f's, which are indeed algebraic integers for algebraic integer x. I know the underlying factorization is:

P(x) = (g1(x) + 1)(g2(x) + 2)

And I can pick a value, say x = 1, for which the a's are 3+sqrt(-26) and 3-sqrt(-26), which tells me the f's are:

5(3+sqrt(-26)) and 5(3-sqrt(-26))

Notice at this value, the polynomial equals P(1) = 175 - 15 + 2 = 162, so we've just factored 162, but we can see a factorization by multiplying by 7:

7*162 = (5(3+sqrt(-26)) + 7)(5(3-sqrt(-26)) + 7)

And we know these connect back to the g's, so here is an existence which means the g's must exist, but my choice of 7 was arbitrary. The underlying factorization doesn't know or care what I multiply times it. And it was just a tool to get to something visible at x=1 by forcing symmetry, and in fact we can use k, where k is a nonzero integer not 1 or -1, with the same set of g's, where for each one, the end result will be algebraic integer f's. So since an infinite number of integers would work we have proof that the g's must be integer like themselves.

So let's label a set of these numbers whose existence and integer like properties we now have proven, o1 and o2, where these are solutions for the g's, to put them in a more mathematical statement.

If o1 is the non algebraic integer, then 7o1 and in general, ko1 is an algebraic integer for k, nonzero and not 1 or -1.

So there is NO WAY the o's can be anything like a fraction, as of course if say, there were a 7 in a denominator then multiplying by some other number wouldn't remove it. And no number could have every possible k in its denominator.

So with absolute mathematical certainty we have new numbers, which are like integers, despite being non-rational, like gaussian integers, but which, unlike gaussian integers, are NOT algebraic integers!

We have new ones to add to the realm of integer-like numbers.

But what about fields? Does this impact them?

And it is easy to show why fields are not impacted, as consider o2/o1, which might not seem to be in the field of algebraic numbers, but you just have to multiply top and bottom by some nonzero integer not 1 or -1, like:

2o2/2o1

and you have a ratio of algebraic integers, showing that it is indeed an algebraic number.

With no impact on fields, these numbers take you stepping outside of the boundaries of polynomial factorizations to find them, which we did by pushing an asymmetrical quadratic factorization into non-polynomial factorizations.

So this approach allows studying the underlying asymmetrical form by moving to a symmetrical form using basic algebra.

And after pondering them for some time I came up with a more robust classification scheme with what I decided to call the object ring.

So why was this not shown by someone else before me?

Well, I did it by coming from a different direction, and wonder what might have happened if instead I'd started with:

P(x) = (g1(x) + 1)(g2(x) + 2)

Turns out that the function I call H(x) is rather tricky, and in fact if you just pick values for it, you can end up outside of even the ring of objects. I've never picked values for it, but found what it had to be from a non-polynomial factorization I had from another way.

So then with one in hand, I eventually worked out the other details.


James Harris

Monday, April 13, 2015

Generalizing a non-polynomial factorization

Going to step through a generalization of the factorization of a polynomial.

P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

(Updating October 26, 2017, to correct wrong beliefs about existence in ring of algebraic integers. Was able to simply remove from this earlier section so was easy.)

I pushed the envelope by imagining the quadratic is what's called primitive and irreducible over Q, which means that it does NOT factor into polynomials.

And I showed stepping through some simple algebraic manipulations where I multiplied by 7, but this time I will multiply by k, where k is a nonzero integer, and not 1 or -1:

Introduce new functions f1(x), and f2(x), where I'll use the second one first, as let

g2(x) = f2(x) + k-2.

Now make the substitution:

P(x) = (g1(x) + 1)(f2(x) + k - 2 +  2) = (g1(x) + 1)(f2(x) + k)

Multiply both sides by k:

k*P(x) =  (kg1(x) + k)(f2(x) + k)

Now, let g1(x) = f1(x)/k, and make that substitution to get:

k*P(x) =  (f1(x) + k)(f2(x) + k)

Now multiply it out:

k*P(x) = f1(x)*f2(x) + k(f1(x) + f2(x)) + k2

And introduce H(x), where I like using the capital letter here for visual reasons, but it is not to signify H(x) must be a polynomial, where:

 f1(x) + f2(x) = H(x), so: f2(x) =  -f1(x) + H(x),

and make that substitution, to get:

k*P(x) = f1(x)*(-f1(x) + H(x)) + kH(x) + k2

So you have:

k*P(x) = -f12(x) + H(x)f1(x) + kH(x) + k2

Which means:

f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0

And you can solve for f1(x) using the quadratic formula:

f1(x) = (H(x) +/- sqrt(H2(x) + 4kH(x) + 4k2 - 4k*P(x)))/2

which is:

f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2

And we can see when that is rational.

Looking at our trivial example again, H(x) = kx + x - k + 2 = (k+1)x - k + 2

And inside that square root then is:

((k+1)x + k + 2)2 - 4k(x2 + 3x + 2)

which is:

(k+1)2x2 + 2(k+1)(k+2)x + (k+2)2 - 4kx2 - 12kx - 8k

which is:

(k-1)2x2 + 2(k-1)(k-2)x + (k-2)2  = ((k-1)x + k-2)2

And making that substitution:

f1(x) = ((k+1)x - k + 2 +/- ((k-1)x + k-2))/2 = kx or x - k + 2 as required.

So only polynomial factorizations can remove that square root for all integer x.

Here is a non-polynomial factorization example though.

Where I use:

P(x) = 175x2 - 15x + 2, and k = 7

And further I have:

7*P(x) = 7(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x) + 7)

So f1(x) = 5a1(x), and f2(x) = 5a2(x), where the a's are roots of:

a2 - (7x-1)a + (49x2 - 14x) = 0

Where notice H(x) = 5(7x - 1), as you can see the a's summed in there, and just multiply that by 5, to get the sum of the f's.

That non-polynomial factorization though from which I worked out what the value for H(x) must be comes from my use of a special construction.


James Harris

Sunday, April 12, 2015

Classifying some numbers

Careful study of a seemingly simple algebraic expression can reveal some numbers with interesting properties outside of currently established non-field ring classification schemes.

With a quadratic polynomial, we can consider its factorization in general with:

P(x) = (g1(x) + 1)(g2(x) + 2)

where, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

Values for the g's are trivial for a polynomial factorization.

For example, if g1(x) = x, and g2(x) = x, then, of course P(x) = x2 + 3x + 2.

P(x) = (x+1)(x+2)

But if P(x) does not factor into polynomial factors, then the g's cannot be algebraic integers, which I've shown by multiplying by a constant integer where the result must be an algebraic integer.

For example,

11*P(x) =  (f1(x) + 11)(f2(x) + 11), where g1(x) = f1(x)/11 and g2(x) = f2(x) + 9.

For algebraic integer x, the f's are in general within the ring of algebraic integers, which I've proven in detail in a post:

somemath.blogspot.com/2015/04/why-in-ring-of-algebraic-integers

But while the f's are algebraic integers for algebraic x, the g's can't both be, which is from the forced asymmetry. That's why I used 1 and 2, as I wanted something simple, and that's about as simple as I could get:

P(x) = (g1(x) + 1)(g2(x) + 2)

So consider some nonrational solutions for the g's, which I'll call o1 and o2, which would be for some particular nonzero, algebraic integer x.

I have that one is an algebraic integer, while the other is not, and does not fit within any currently established ring.

But if I multiply it, for instance, by 11, then it is an algebraic integer. But if I multiply the same number by 7, then it is also an algebraic integer. And in general I can multiply the same number by any nonzero integer not 1 or -1, and it will be an algebraic integer.

That is to say, if o1 is the non algebraic integer, then 7o1 and 11o1 are, and in general, ko1 is an algebraic integer for k, nonzero and not 1 or -1, while o1 is not:

k*P(x) =  (f1(x) + k)(f2(x) + k)

where g1(x) = f1(x)/k and g2(x) = f2(x) + k-2

And then it is clear why fields are not impacted, as consider o2/o1, which might not seem to be in the field of algebraic numbers, but you just have to multiply top and bottom by some nonzero integer not 1 or -1, like:

2o2/2o1

and you have a ratio of algebraic integers, showing that it is indeed an algebraic number.

So this approach allows studying the underlying asymmetrical form by moving to a symmetrical form using basic algebra.

And it's nice to mark over 10 years since I came up with a more robust classification scheme with the object ring.

That is the first post on this blog made over 10 years ago. Yeah, it's that important.

Probably is the reason this blog exists, because it was long enough ago, I'm not sure all of what I was thinking at the time. But since I made it the very first post, got a feeling it had much to do with my motivations.

So I'd also like to celebrate that special event.

And will note coming up on a 20 year celebration this month.

That there are numbers which behave like integers, which are themselves not rational, which are NOT algebraic integers is fascinating.

And since they are not algebraic integers it's hard to see them directly. What you can do is get a solution for the f's and divide off the constant multiplier from both solutions knowing it works for one, and for the other you have something that is fractional, but that doesn't tell you which is which.

I've been thinking about these numbers for a while now, over a decade.

So how were they missed before? If you step through the tests done on the ring of algebraic integers, it's easy to see how they don't catch these numbers.

So I came up with a more inclusive ring classification and call these numbers along with integers--objects.

Objects replace algebraic integers at the foundations of numbers. The object ring includes the ring of algebraic integers plus the additional numbers previously missed.


James Harris

Saturday, April 11, 2015

Why in ring of algebraic integers?

One of my favorite demonstrations uses a seemingly simple construction:

P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

Turns out that expression can be pushed out of the ring of algebraic integers.

For example, like have g1(x) = x, and g2(x) = x, then, of course P(x) = x2 + 3x + 2.

But I pushed the envelope by imagining the quadratic is what's called primitive and irreducible over Q, which means that it does NOT factor into polynomials, where an additional factor can force outside the ring of algebraic integers.

And have shown where I multiplied by 7, but this time I will multiply by 11,which is the additional factor:

Introduce new functions f1(x), and f2(x), where I'll use the second one first, as let

g2(x) = f2(x) + 9.

Now make the substitution:

P(x) = (g1(x) + 1)(f2(x) + 9 +  2) = (g1(x) + 1)(f2(x) + 11)

Multiply both sides by 11:

11*P(x) =  (11g1(x) + 11)(f2(x) + 11)

Now, let g1(x) = f1(x)/11, and make that substitution to get:

11*P(x) =  (f1(x) + 11)(f2(x) + 11)

Now multiply it out:

11*P(x) = f1(x)*f2(x) + 11(f1(x) + f2(x)) + 121

And introduce H(x), where I like using the capital letter here for visual reasons, but it is not to signify H(x) must be a polynomial, where:

 f1(x) + f2(x) = H(x), so: f2(x) =  -f1(x) + H(x),

and make that substitution, to get:

11*P(x) = f1(x)*(-f1(x) + H(x)) + 11H(x) + 121

So you have:

11*P(x) = -f12(x) + H(x)f1(x) + 11H(x) + 121

Which means:

f12(x) - H(x)f1(x) - 11H(x) - 121 + 11*P(x) = 0

And you can solve for f1(x) using the quadratic formula.

Notice you can see the ambiguity between the f's right there, which prevents you from knowing which one has 11 as a factor, as of course looks the same with f2.

And if H(x) is in the ring of algebraic integers, since the coefficients of P(x) are integers, f1(x) must be in the ring of algebraic integers.

Notice you can start with:

P(x) = (g1(x) + 1)(g2(x) + 2)

and multiply by 13 if you wish, as of course it is bare. What you multiply with is human choice. I've chosen 7 and 11, for my own reasons. And you may now wonder, so how do you pick H(x)?

Does it matter?

One way of looking at it, going in this direction, you can choose H(x) and in that way force the g's, which means it is a handle on the original factorization.

Your choice of H(x) tells the math how the original factorization must go.

Are there any circumstances under which the f's that result from your choice could NOT be in the ring of algebraic integers?

Well the only way is if you deliberately make H(x) not be in the ring of algebraic integers by picking coefficients outside of it, but even if you try I think it's kind of hard to do and have all the equations work. For instance since f1(0) = 0, and f2(0) = -9, you know it must be that H(0) = -9.

Actually I don't know if you can force H(x) to not be an algebraic integer and have all the equations work. But even if you can, you'd have to be an ornery (and clever) person who just made that happen.

With my own research H(x) has been linear with integer coefficients.

Those original g's are weird then, eh? Solutions for algebraic integer x cannot all be seen, as they are not both expressible as algebraic integers, so can't be seen directly, but you can multiply the one that is not by an integer and see the result!

And they're not too picky about the integer either. It just needs to be nonzero and not 1 or -1, and they will appear, as the result is an algebraic integer.

I came up with the object ring to categorize these peculiar numbers, and call them and integers, objects. So you can say mathematically that given an object o1, which is a solution to one of the g's from my original factorization which is not an algebraic integer, it is visible with 7o1 or 11o1 with the examples I've used. It doesn't care.

Oh yeah, so what about o2? Well it is an algebraic integer, but which is which? No way to tell. But, yup, o1o2 is an algebraic integer.

Non algebraic integer objects are fascinating critters!

Their refusal to be seen directly is NOT intuitive.

These non-rational objects which are not algebraic integers are an undiscovered country of number theory. If you don't believe they exist and are being befuddled by the idea of fractions then how can I multiply the same number by 7 or 11 to get an algebraic integer? Would it help if I did it again with 9? Or 16?


James Harris

Sunday, April 05, 2015

My functional math perspective

My focus is on math with which you can DO something, so for instance, what if you like summing distinct squares to get a square?

For example:

42 + 6+ 10+ 14862 = 882

and

862 + 129+ 215+ 301+ 8812 = 9682

Why? Why ask why? Maybe you just like to do it for fun! That's a good enough reason for me. But how do you find them?

I can tell you a way I found with my research:

somemath.blogspot.com/2015/01/numbering-sum-of-squares.html

Using the techniques at that post you could make a sum of 100 distinct squares to give a square. Why do such a thing? I wouldn't know. I haven't done it, and I discovered the mathematical machinery.

That's a fun thing to be able to say too. Discovered the mathematical machinery. Being able to say that and it be true is living the dream.

So yeah, I discovered the mathematical machinery.

That's a powerful thing to be able to put up publicly. Think about it.

I like giving people mathematical tools. It's a quirk of mine.

Then ask yourself, do you really think I'm struggling to get known?

Nope. I can probably strut into any math department in the world and demonstrate half a dozen "impossible things" but there's no motivation.

Why would I bother? What does any math department in the world have that I'd want?

What can they give me?

Answer, nothing at all.

And I guess that can sound harsh, but I've tried to explain. Short of it is, I'm not a mathematician. I don't have a career in that area. Mathematicians are at best just competitors to me. I have no need nor interest in hanging out with rivals. And there really is nothing they can give me in return for spending time with them, anyway. Where thanks to the web, I don't need their recognition either.

More importantly, what can you give yourself? Love of discovery, passion for mathematics, and the thrill of playing with numbers.

Giving people the tools to pursue their interests? Now THAT is something.

The functional perspective is a great one, and can mean that years from now if I do feel like doing something with some particular math discovery of mine, I know the option is there. What might I do? Who knows. It's open ended. But without these tools, how?

For me it was often frustrating in the past to just want to play with integers and read through massively complex mathematical tomes that often wouldn't even give me a way to do that, so I couldn't get that pleasure of watching the numbers DO something interesting to me. So you do all that work to understand something you can't even really use! Why bother?

But with my own research NONE of it is that way. And I wouldn't have it any other way.

I love being able to DO things with integers. For me it is a passion.

Which gives the best benefit of a functional perspective: the math has to actually work.


James Harris

Thursday, April 02, 2015

Want a PDF?

There are posts which I've turned into PDF's, where you can download from my math group:

groups.google.com/d/forum/mymathgroup

And I try to say there when a post has a PDF attached.

You can also comment there though it has been a very quiet spot. Not a big deal. Have had it for some years, where its primary purpose is for later, so for instance, if someone thinks he's found something mathematically important which he wants to show me, I'd refer him to post on my math group.

I have no intention of ever considering math ideas presented to me by others as, well, I'm not a mathematician. However, I can provide a forum.


James Harris