Pondering my navel regarding Number Theory

[maddog]

I must admit the "Cartesian product" had me a bit confused as well.

 

In my thinking of the problem my mapping was Z -> Z and not Z x Z -> Z.

I wasn't using ordered pairs to indicate the operation. I suppose you

could. My question is why bother (?)

 

In any cases, another point -

 

Corrallary to the initial post:

 

1. for any x, y in Z st x * y is even then either x or y must be even.

2. for any x, y in Z st x * y is odd the both x and y must be odd.

 

maddog

 

ps: I would love it if I just took the time to make my posts all look like

the pretty print here by using the LaTex tools that Alexander provided.

Oh well.... :-( who has time...

[maddog]

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.

 

This might imply that primes are the most "hip" of numbers being they are the class

that are farthest from being "square" and ARE Definitely "with it". They are Kewl!

 

In addition primes are all Odd (with the exception of 2 which is even).

 

So we have a new number than One being the loneliest of numbers. With respect to

primes two also holds that slot -- yet I digress... :P

 

maddog

[Ben]

I wasn't using ordered pairs to indicate the operation. I suppose you could. My question is why bother (?)

Answer: Because you have no choice. That is the definition of the domain of a binary operation (like multiplication. Or addition, if it comes to that). In spite of what else you might have read here......

 

ps: I would love it if I just took the time to make my posts all look like

the pretty print here by using the LaTex tools that Alexander provided.

Oh well.... :-( who has time...

Yes, it does take a little time, but not that much (with practice). Plus it makes one's posts sooooo much more readable

[Qfwfq]

This might imply that primes are the most "hip" of numbers being they are the class

that are farthest from being "square" and ARE Definitely "with it". They are Kewl!

Absolutely!

 

Indeed, it was a reason behind one of my choices last week:

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

Wherin the acute observer might notice that the odd number is prime and the even one has a factor 2 raised to a pretty good exponent. Clearly these give good contributions to the difference in densities. After all, 2 is the smallest of all prime factors that a natural number can have.

 

I must admit the "Cartesian product" had me a bit confused as well.

Cartesian product - Wikipedia, the free encyclopedia and, in particular:

Cartesian product - Wikipedia, the free encyclopedia

Look, you can mock me all you want - anyone can, I don't mind

Especially when you've been rilly rilly go'n' lookin' for it. :shrug:

 

C'mon, Ben. :beer:

 

I wasn't using ordered pairs to indicate the operation. I suppose you could. My question is why bother (?)

Because it is what your OP was really all about.

 

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]?

If you prefer, we can say that those of which the image is odd are "of the grapefruit type" and those of which the image is even are "of the banana type". But I think that calling them, respectively, odd and even is a less ambiguous use of language.

 

Bizarre.

It's exactly what the OP says, so why do you keep bleming me for it?

 

BTW, since 0 is commonly considered an even number, I disagree with the disequality in your definition of even numbers in [imath]\mathbb{Z}[/imath].

[Ben]

Blooped.