Pondering my navel regarding Number Theory

[maddog]

Hi All,

 

I'm back!

 

I've been teaching my self on Lie Groups which has been having me reviewing some

concepts in Abstract Algebra. So recently I was reading through book (part of the

series Algebra through Practice, T.S. Blyth, E.F. Robertson), Vol 6, Rings, Fields &

Modules. I was learning that the definition of a Module is like a Vector Space for

Rings (instead of Fields). -- This got me to thinking (I don't know why ???)

 

- - - - - - -

 

If an Odd * Odd is Odd, Even * Even is Even, Even * Odd is Even

then why aren't there more Evens than Odds ?

 

I understand that using a correspondence mapping, you can prove that there are

as many Evens as there are Odds. I was just wondering.... ? ;)

 

Maddog

[CraigD]

Hi Maddog, and welcome back (it’s been a year and 6 days!).

If an Odd * Odd is Odd, Even * Even is Even, Even * Odd is Even

then why aren't there more Evens than Odds ?

This is some serious numeric naval gazing indeed, but one for which I think the rabbit hole/naval goes not too deep - at least at first.

 

Assuming we mean even and odds as subsets of the positive integers, I think the answer can best be found by considering how the integers are “generated”. Taking a couple of the Peano postulates – the number 0 exists, and every number has a successor (1’s is 2) - as the answer to that “how” question, we can see that they’re generated by the unary operation of incrementation. Clearly, at no point in generating the integers is the count of the evens more than 1 greater than the count of odds.

 

Now, if instead of the usual Peano postulates and integers, we generated a set of numbers (which could be a group under various operations) with the postulates that 2 and 3 exist, and other numbers are generated the the binary operation of multiplication, then as your odd*odd/even*even/even*odd observation suggests, there would be more evens than odds – infinitely many, I’ll hazard an unproven guess.

 

For example, after 0 “generations”, the partial set is {2, 3}. After 1 generation, it’s {2,3,4,6,9}. After 5 “generations”, it’s {2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, ... 823564528378596, 1235346792567894, 1853020188851841}, 528 of which are even, 32 odd.

 

A question that tantalizes me is: if the generator function used is not n+1 or n*m, but n*m+1, are there more evens than odds, or vice versa? I’d bet on there being more odds, but with the calculator I have on hand now, I can tell only that the even vs. odd ratio for successive generations is 0 v 1, 1 v 1, 1 v 3, 6 v 5, 15 v 45, and 948 v 641.

[Turtle]

Hi All,

 

...

If an Odd * Odd is Odd, Even * Even is Even, Even * Odd is Even

then why aren't there more Evens than Odds ?

 

I understand that using a correspondence mapping, you can prove that there are

as many Evens as there are Odds. I was just wondering.... ? :confused:

 

Maddog

 

i ran across the even/odd density question a month or so back. in the interest of brevity, i'll just quote what i posted on it elsewhere.

 

 

see if this helps. :( >>

Natural density - Wikipedia, the free encyclopedia

If we pick randomly an integer from the set [1,n], then the probability that it belongs to A is the ratio of the number of elements of A in [1,n] to the total number of elements in [1,n]. If this probability tends to some limit as n tends to infinity, then we call this limit the asymptotic density of A. We see that this notion can be understood as a kind of probability of choosing a number from the set A. Indeed, the asymptotic density (as well as some other types of densities) is studied in probabilistic number theory. ...

 

oddly enough, and since you mentioned them, using a different density calculation gives unequal densities for the evens & odds. :eek: Schnirelmann density - Wikipedia, the free encyclopedia

Consequently, the Schnirelmann densities of the even numbers and the odd numbers, which one might expect to agree, are 0 and 1/2 respectively. Schnirelmann and Yuri Linnik exploited this sensitivity as we shall see. ...

 

that's it. . . . .

[Qfwfq]

If an Odd * Odd is Odd, Even * Even is Even, Even * Odd is Even

then why aren't there more Evens than Odds ?

Actually, you include Even * Odd is Even but omit Odd * Even is Even. This would make for one out of four numbers being odd, instead of the correct one out of two.

 

The even odder thing about it is that the above conclusion rests on the tacit assumption that there are as many evens as odds. IOW you get the one out of four as a thesis by assuming the one out of two as hypothesis. :naughty:

 

The short story is that [imath]\mathbb{N}_0^2[/imath] isn't [imath]\mathbb{N}_0[/imath]. Indeed most numbers appear more than once, as a product, in the cartesian square. No doubt even ones must be repeated even more, compared to odd ones.

[Ben]

The short story is that [imath]\mathbb{N}_0^2[/imath] isn't [imath]\mathbb{N}_0[/imath]. Indeed most numbers appear more than once, as a product, in the cartesian square. No doubt even ones must be repeated even more, compared to odd ones.

I don't understand what you are saying here! What is [math]\mathbb{N}_0[/math]?

 

Where does the Cartesian product enter the picture? Is it not the case that the elements in [math]\mathbb{N} \times \mathbb{N}[/math] are the ordered pairs [math](a,b) \ne (b,a)[/math]? But how is it relevant?

 

As for the OP, Maddog says he is aware of the proof that every countably infinite set can be placed in one-to-one correspondence with a proper subset of itself. Since the odds and even are proper subsets of [math]\mathbb{Z}[/math] and since the intersection of odds and evens is empty, it follows that they have the same cardinality as sets

[Qfwfq]

I don't understand what you are saying here! What is [math]\mathbb{N}_0[/math]?

Fairly standard notation for the set of natural numbers excluding zero.

 

Where does the Cartesian product enter the picture? Is it not the case that the elements in [math]\mathbb{N} \times \mathbb{N}[/math] are the ordered pairs [math](a,b) \ne (b,a)[/math]? But how is it relevant?

He is counting two-factor products rather than the numbers themselves. This is what makes for a density of one in four instead of one in two. Of course, if one doesn't consider ordered pairs but just pairs, it makes for one in three and of course it isn't really the cartesian square. In either case it isn't the actual density of even numbers.

 

As for the OP, Maddog says he is aware of the proof that every countably infinite set can be placed in one-to-one correspondence with a proper subset of itself.

Well, not with finite proper subsets, however...

Since the odds and even are proper subsets of [math]\mathbb{Z}[/math] and since the intersection of odds and evens is empty, it follows that they have the same cardinality as sets

Yes this is trivially so but the infinite cardinalities being equal doesn't clear up the matter. One might as well say there are just as many multiples of googolplex as there are rationals, but this doesn't clean up his conundrum about densities.

[Ben]

{[math]\mathbb{N}_0[/math] is}Fairly standard notation for the set of natural numbers excluding zero.

Not where I live, but notation is arbitrary, so let's use yours

 

He is counting two-factor products rather than the numbers themselves.

Does this make any difference? [math]\mathbb{N}_0[/math] (in your notation) is a ring, that is, closed under addition and multiplication. That is, for any [math]n \in \mathbb{N}_0,\,\, n_1*n_2 \in \mathbb{N}_0[/math]

 

Suppose that [math]n \in \mathbb{N}_0[/math] is exactly divisible by 2 (even), then obviously [math]n+1 \in \mathbb{N}_0[/math] is not (odd).

 

Now since [math]\mathbb{N}_0[/math] is countable and infinite, then the "density" of all [math]n \in \mathbb{N}_0[/math] is precisely the density of all [math]n+1 \in \mathbb{N}_0[/math].

 

By the closure axiom, the multiplication operation cannot possibly change the density of the evens and odds, since the operation is for sure an element in [math]\mathbb{N}_0[/math], whether it be even or odd

 

In short, I believe you are chasing red herrings.....er, wild geese? Whatever

[Qfwfq]

In short, I believe you are chasing red herrings.....er, wild geese? Whatever

'twas not I, 'twas Maddog!

 

Strictly, the set of naturals without the additive neutral is not a ring, but I don't see what difference this makes. Surely we all agree that the density of odds and evens is the same... in the set of naturals. Maddog's conundrum was due to the fact that they don't have the same density in the cartesian square. Dilemma solvit, pro quid disputare?

[Ben]

Because I have no idea WTF you are talking about! It must be to do with the fact that I don't know as much math as I pretend.

 

First, by "Cartesian square" I assume you mean Cartesian product. Yes or no?

 

Second, if yes, then (using your horrid notation), this product is given by [math]\mathbb{N}_0 \times \mathbb{N}_0 \ni (n_i,n_j)[/math] for all [math]i,\,j [/math]. Whereas "oddness" and "evenness" makes sense in [math]\mathbb{N}_0[/math], I doubt it makes sense in the Cartesian product. Does it?

 

So what is meant by the claim that

they {odds and evens} don't have the same density in the cartesian square.

I can make neither head, tail nor anything else of this. Please explain.

 

Either way, define a closed, binary operation called "multiplication" such that [math]*: \mathbb{N}_0 \times \mathbb{N}_0 \to \mathbb{N}_0,\,\, *(n_i,n_j) = n_i *n_j = n_k \in \mathbb{N}_0[/math] where we may NOT have that [math]n_k =1[/math] i.e. multiplicative inverses don't exist (since all elements are strictly positive integers).

 

If the density of odds and evens in [math]\mathbb{N}_0[/math] is equal before entering into the product, how can they be different after under the scenario above?

[CraigD]

By the closure axiom, the multiplication operation cannot possibly change the density of the evens and odds, since the operation is for sure an element in [math]\mathbb{N}_0[/math], whether it be even or odd

I agree.

 

However, what I got from maddog’s original post, and put into post #2, is that reasonable interpretation of a set defined somehow by the math-y aphorism “Odd * Odd is Odd, Even * Even is Even, Even * Odd is Even” is not the positive integers ([math]\mathbb{N}_0[/math]), but a peculiar set defined by at least 2 initial members greater than 1, including at least one odd and one even (in the examples I gave, I used 2 and 3) and a “generator” [imath]a_k = a_i \cdot a_j[/imath], where i and j may be equal. Such a set [imath]A[/imath] (which is a group) is “full of gaps”, and

[math]\frac{|O_n|}{|E_n|} \to 0[/math] as [math]n \to \infty[/math]

where [math]O_n = \{a : a \in A, a \le n, a \,\mbox{is odd}\}[/math] and [math]E_n = \{a : a \in A, a \le n, a \,\mbox{is even}\}[/math].

 

In short, I’m saying that [math]\mathbb{N}_0[/math] has about equal counts of even and odds because its generator is [imath]n_{i+1} = n_i +1[/imath], while [imath]A[/imath] doesn’t, because its generator is inspired by the original post’s aphorism.

[Qfwfq]

First, by "Cartesian square" I assume you mean Cartesian product. Yes or no?

A square is a special case of a product, isn't it?

 

I doubt it makes sense in the Cartesian product. Does it?

It's simple, despite the confusing fact that the word product can also be used to indicate the result of the same product. But then again what else could the OP have meant by the aphorism?

 

2*9=18 is a product. Is it odd or is it even?

 

7*3=21 is a product. Is it odd or is it even?

 

If the density of odds and evens in [math]\mathbb{N}_0[/math] is equal before entering into the product, how can they be different after under the scenario above?

Math is timeless, but I suppose you mean domain and image. The [math]*[/math] that you define is not injective and I don't see why it should conserve density.

 

Even restricting products to 2 factors, how many of them give the result 7159 and how many give 7168?

[Ben]

A square is a special case of a product, isn't it?

 

Well yes and no. If you mean a square is the arithmetic product (i.e. multiplication) of its sides, then yes. Otherwise no.

 

Look I am sure you don't need this patronage, but let me emphasize this for general consumption.

 

 

Suppose that [math]S ni a,,,b[/math] is a set. Then the Cartesian product is defined to be the NEW set [math]S times S[/math] whose elements are the ordered pairs [math](a,B) ne (b,a)[/math].

 

 

Notice that NO operation is implied. But once we have a closed binary operation defined on [math]S times S[/math], call it [math]circ[/math], then we may very well have that [math]circ: S times S to S,,, circ(a,B) = a circ b = circ(b,a) = b circ a[/math]. This is called a commutative operation.

 

 

the word product can also be used to indicate the result of the same product.

What's meant by the "result of the same product"? Lost me there!!

 

 

Math is timeless, but I suppose you mean domain and image.

No, I do not. The domain of * is the Cartesian product, where I believe "oddness" and "evenness" make no sense. The codomain is precisely the exact same set that enters into the Cartesian product. I repeat, if the density of odds and evens is equal in those sets that enter into the Cartesian product, the domain, how can they be different in the codomain?

The [math]*[/math] that you define is not injective

Of course it's not dear boy, in fact I doubt whether any closed binary operation on any Cartesian product ever is.

 

 

Anyhoo....following Graig's excellent post, let me muddy the waters still further.

 

 

By the Prime Factorization Theorem, we know that any integer whatever can be written as the arithmetic product of positive primes together with the units [math]pm1[/math]. This suggest that the set [math]mathbb{Z}[/math] endowed with multiplication as an operation is a monoid (identity but no inverse) and can be generated, under multiplication, by the subset [math]{-1,1,P^+}[/math] where [math]P^+[/math] denotes the set of all positive primes.

 

 

Notice that [math]P^+[/math] is infinite and yet the only non-odd element is [math]2[/math]. Likewise, the set [math]mathbb{Q}[/math] of all rational numbers is a group under multiplication (inverses exist), and is generated by the same subset.

 

 

How weird is that?

[maddog]

I see my simple little post has created quite a controversy. B)

 

From Qfwfq and Ben's posts, I see I have made a tacit assumption on * (my binary

operation) that my group in question was abelian (Z being the Integers).

 

Were I to be more general then yes, "density" could have other implications.

 

i.e. when a * b is not b * a.

 

By Closure if a, b in Z then a * b, b * a in Z (even if not equal)

 

I was only musing that because of subdividing Z into two groups (even or odd)

yield three ways to determine the result.

 

This would imply somehow the that the balance between them are not equal.

 

My countability argument assumed an abelian ring thereby allowing communativity

and thereby the canceling of cross terms.

 

maddog

 

ps: did ya' ever that there are two groups of people ?

People who divide people into two groups and those that don't.... B)

[Qfwfq]

I divide people in this thread into those who keep missing the point, and myself...

I see my simple little post has created quite a controversy. ;)

Yes, indeed! :D

 

I have made a tacit assumption on * (my binary

operation) that my group in question was abelian (Z being the Integers).

This isn't the actual crux of the matter, commutativity only implies that one may reduce the cartesian...:doh: oooops, product, ehem, to all the diagonal and half the off-diagonal elements (pairs rather than ordered pairs). Now, abelian or not, it's obvious that for any given choice of this binary operation, it magically makes sense to divide the domain into "odd" and "even" according to the image of each pair. After all, that's what the aphorism means...:shrug:

 

...and is generated by the same subset.

 

How weird is that?

It's only weird if you don't count in the inverses of the primes, but even so these aren't defined until we have [imath]\mathbb{Q}[/imath] already, or if you define them by fiat, so it isn't the best construction. I much prefer equivalence classes over [imath]\mathbb{Z}\times\mathbb{Z}_0^+[/imath]

 

Ben, is 25 a square?

[Ben]

maddog I wouldn't call in controversy as such, rather more a discussion. I hope all agree it has been good-natured?

 

makes sense to divide the domain into "odd" and "even" according to the image of each pair. After all, that's what the aphorism means

I don't believe it does. Let's stick with the integers [math]\mathbb{Z}[/math] for convenience. Let's also forget all about Cartesian products and the like, and take the simple-minded view.

 

Now choose completely arbitrary elements [math] x,\,\,y \in \mathbb{Z}[/math]. If [math]x^* y[/math] is even, how do we know whether these are both even or one is even and one odd? This is the crux of the aphorisms.

 

I contend this fact shows that the image set cannot possibly induce a well-defined partition on the domain.

 

I much prefer equivalence classes

Me too, but they are imposed on the domain, not by the codomain, like this: suppose we denote by [math]2\mathbb{Z}[/math] those integers that are exactly divisible by 2. Now form the quotient set [math]\mathbb{Z}/2\mathbb{Z} \equiv \mathbb{Z}_2[/math].

 

This induces 2 equivalence classes in the integers, the evens and the odds. The elements in [math]2\mathbb{Z}[/math], the evens, are [math]\{....-4,-2,0,2,4,....\}[/math] from wich we "choose" a class representative, say 0, and write this class as [math][0] [/math].

 

Elements in the residuum of [math]\mathbb{Z}_2[/math] are therefore the odds [math]\{....-3,-1,1,3....\}[/math] and, again choosing a class representative form the equivalence class [math][1][/math].

 

Then the rest follows without difficulty: for all [math]x \in [0],\,\, y \in [1][/math] we have that [math]x*x \in [0],\,\,y*y \in [1], x*y \in [0][/math] which, in terms of my equivalence classes becomes

 

[math][0]*[0] = [0],\,\,[1]*[1]=[1],\,\,[0]*[1]= [0][/math].

 

A very reassuring result, I hope all will agree, and is no more than my pompous way of restating maddog's aphorisms, but in a way that, I hope, makes them less mysterious.

 

 

Ben, is 25 a square?

A square WHAT?

[Qfwfq]

I don't believe it does.

Of course iI induces the partition onto the domain; just say that all elements in the anti-image of odd numbers are odd, those in the anti-image of even numbers are even.

Believe it or not... it works!

 

Now choose completely arbitrary elements [math] x,\,\,y \in \mathbb{Z}[/math]. If [math]x^* y[/math] is even, how do we know whether these are both even or one is even and one odd? This is the crux of the aphorisms.

No, the crux of the aphorisms is what I said above, in the sense that the above is the meaning of the is in the aohorism.

 

I contend this fact shows that the image set cannot possibly induce a well-defined partition on the domain.

Only because you are considering [imath]\mathbb{Z}[/imath] as a domain instead of [imath]\mathbb{Z}\times\mathbb{Z}[/imath] (or [imath]\mathbb{N}_0\times\mathbb{N}_0[/imath]). The very fact that the product (result) doesn't determine the two factors, i. e. the fact that [imath]*[/imath] is not injective, is what underlies the density conundrum.

 

Me too, but they are imposed on the domain, not by the codomain, like this:

Except that I was talking about bananas and you're talking about grapefruit.

 

I meant as a construction of [imath]\mathbb{Q}[/imath] out of [imath]\mathbb{Z}[/imath], without needing to define the reciprocals of primes as a further postulate.

 

What you are doing there is toe exact same thing as I do, only using [imath]\mathbb{Z}_2\times\mathbb{Z}_2[/imath] (and reducing it due to commutativity). This collapses it down to two anti-images having only a few elements, one of which has only one single element: [imath][1]*[1]=[1][/imath]

 

I agree that it simplifies the matter, so perhaps it does make it less mysterious, at least for those that find my rants to complicated to follow. :)

 

A square WHAT?

Ever since around the sixties (or the fifties, was it?), a square is someone that just ain't with it, who is regarded as dull, rigidly conventional, and out of touch with current trends.

 

Now everybody knows that the number 25 is just completely not with it, at all. So totally and completely not with it indeed, that folks often say 25 is a perfect square. It will never be a social success with the younger generations; it stands a snowball's chance in hell.

 

Actually, :doh: sorry, I should have said, everybody knows 25 is a perfect square, but not everybody knows the true reason stated above. There is an urban legend that confuses a lot of people. Sure enough, it's due to an absolutely horrid notation, a ghastly notation in fact: many folks say that [imath]25=5^2[/imath] and they read it as "Twenty-five is five squared." and, what even ghastlier, they often abbreviate to "Twenty-five is five square."

 

Now, I know that such a ghastly horrid notation just can't be the real reason why 25 is a square, don't panic and fret, I'm aware of it. I know the real reason is that 25 just ain't with it; just that I should not have said everybody knows this. They do all know it's a perfect square but some folks believe it's due to that absolutely vile, horrid notation. In fact some nerds even go further in their delirium, they phrase it as: "Twenty-five is the square of five."

[Ben]

Of course iI induces the partition onto the domain; just say that all elements in the anti-image of odd numbers are odd, those in the anti-image of even numbers are even.

Believe it or not... it works!

Lemme get this straight. Given [math]*:\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z},\,\, 12 \in \mathbb{Z}[/math] you want that both [math](2,6)[/math] and [math](3,4)[/math] as pre-images [math]*^{-1}(12)[/math] to be defined as even in [math]\mathbb{Z} \times \mathbb{Z}[/math]?

 

Bizarre. The usual definition of an even number [math]x[/math] is that there is some umber [math]y \ne x[/math] such that [math]x = 2y[/math] (or some variant of this). I repeat, in spite of your definition of the evens in the Cartesian product, I don't think the notion of oddness and evenness has any real meaning for ordered pairs.

 

 

Ever since around the sixties (or the fifties, was it?), a square is someone that just ain't with it, who is regarded as dull, rigidly conventional, and out of touch with current trends.

Look, you can mock me all you want - anyone can, I don't mind - but your humour doesn't exempt you from the obligation to use language unambiguously.

 

Had you not earlier referred to something called "the Cartesian square"? Is 25 a Cartesian square? Is there some number [math]n \in \mathbb{Z}[/math] such that [math]n^2[/math] is a Cartesian square?

 

Clearly this is not what you meant, but I don't read minds, just words.