Even Number, Odd Number ...

[khaled]

assume E is any even number, and O is any odd number

 

E = O + 1

 

O = E + 1

 

E + E = E

 

E + O = E + E + 1 = O

 

O + O = ( O + E ) + 1 = O + 1 = E

 

E * E = E

 

E * O = E

 

O * O = O

 

note: if a given (O = O * O) then that O can only

 

be factored by odd numbers.

 

what do you think ..?

 

I also think that,

 

 

given that A & B & O are odd, and C & D & E are even ...

 

 

C * D = E

 

(A+1) * (B+1) = E

 

AB + A + B + 1 = E

 

odd*odd + odd + odd + 1 = odd+1 = even

 

 

C * A = D, and C = B+1

 

(B+1) * A = D

 

AB + A = D

 

odd*odd + odd = odd+odd = even

 

 

A * B = O

 

(C+1) * (D+1) = O

 

CD + C + D + 1 = O

 

even*even + even + even + 1 = odd

Edited January 21, 2011 by khaled

[HamsterPower]

O + O = ( O + E ) + 1 = O + 1 = E

 

 

^^ for the equation above

isnt't it

o +o = (o+e) +(e+1) = e+e+ o+1 = e +e =e

not that it matter much

 

 

i think this very cool

where did you get the idea of think of such?

[khaled]

O + O = ( O + E ) + 1 = O + 1 = E

 

 

^^ for the equation above

isnt't it

o +o = (o+e) +(e+1) = e+e+ o+1 = e +e =e

not that it matter much

 

i think this very cool

where did you get the idea of think of such?

 

I didn't take the idea from anywhere, I thought it up, I was starting a numerical Analysis

to figure out some certain constraints of some numbers ...

 

Like, is there an odd perfect number ..?

 

I'm working in my free time on that question ...

[the tree]

assume E is any even number, and O is any odd number

Okay let's just stop you there.

 

The easiest way to denote every even number ever and every odd number ever is with the modulo classes [imath]\{ \bar{0},\bar{1} \}=\mathbb{Z}_2[/imath], and all those equations that you just listed are really quite obvious.

 

[a given odd number] can only be factored by odd numbers. what do you think ..?

Well straight from the schoolish definition of odd number, if it's odd then it cannot be divided by 2, so neither can any of it's factors so it can only be factored by odd numbers.

 

The argument from modular arithmetic is more subtle: since you are working in Z₂ and 2 is prime it is a known result that Z₂ is a field with no zero divisors. So there is no [imath]a \in \mathbb{Z}_2[/imath] such that [imath]a\cdot\bar{0}=\bar{1}[/imath] mod 2.

 

As it happens, 2 is also a really small number so you can also get that from inspection.

 

• odd*odd + odd + odd + 1 = odd+1 = even
  
• odd*odd + odd = odd+odd = even
  
• even*even + even + even + 1 = odd
  

• 1x1+1+1+1=4=0 mod 2, so yes.
  
• 1x1 + 1 = 2 = 0 mod 2, so yes
  
• 0*0 + 0 + 0 + 1 = 1 mod 2, so yes
  

Modular arithmetic: it really works.

[khaled]

So, it's Modular Arithmetic in [imath]\mathbb{Z}_2[/imath]

[Xerxes]

So, it's Modular Arithmetic in [imath]\mathbb{Z}_2[/imath]

Yes, but that is not all!

 

Try this party trick, amaze your friends....

 

1. Take any number at all with more than 1 digit and add the digits.

 

2. If the result has more than one digit, add again

 

3. Now subtract this single-digit number from your starting number (not added)

 

4. Add the resulting digits again.

 

5. I bet it will be the number 9

 

Can you see why?

 

Once you do, you will see that the restriction in (1) to more than a single digit is not required.

 

Hint in white follows (no peeking!!)

 

It is simply modulo 9 arithmetic where 0 = 9 mod 9. As it seems you peeked, your forfeit is to explain, using modulo 9 arithmetic, why this works

 

There are any number such "mind reading" tricks one can devise once you see what's going on

[the tree]

So, it's Modular Arithmetic in [imath]\mathbb{Z}_2[/imath]

Well everything that you said could be generalised to larger finite fields which can be described as [imath]\mathbb{Z}_p[/imath] for any prime [imath]p[/imath]. However the concept of parity does extend beyond just looking at the integers.

[DrFezza]

Xerxes that's cool!

So are you saying.... Take any 2 digit number and subtract it's single digits... Adding remainder down to single digit will always sum to nine!

As multiplying any number by nine and adding these digits to a single digit will sum to nine....

Any prior double digits are then factors of nine!!! Cool!