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Guest
Guest
Is there anyone from Oregon that would like to meet others from Oregon ?
fishingjeff
Virgin
- Joined
- Jan 15, 2002
- Posts
- 10
I am here too
I am in Portland and would love to meet others
I am in Portland and would love to meet others
Peekaboo69
Experienced
- Joined
- Aug 1, 2002
- Posts
- 49
Well I am from Central Oregon and we like to meet new people too ....anuny1mous
Virgin
- Joined
- Aug 3, 2002
- Posts
- 1
eugene oregon here. new to this, so forgive me any faux paus.
Peekaboo69
Experienced
- Joined
- Aug 1, 2002
- Posts
- 49
nice to meet u
hope i do too still new to all this
but love the excitement and the fun
hope i do too still new to all this
but love the excitement and the fun
Devil_d0ll
I'm Always Wet
- Joined
- Feb 25, 2002
- Posts
- 3,453
hey... we have a wa/oregon thread in the playground.. with lots of ppl from around the area...
come chat with us there
27 f/close to portland/astoria but in WA
come chat with us there
27 f/close to portland/astoria but in WA
Hay i have not been in here for a while...
looks like Oregon is now in the House...
38 m Eugene ore here, would like to get to know ya!!!
wally9235@hotmail.com
looks like Oregon is now in the House...
38 m Eugene ore here, would like to get to know ya!!!
wally9235@hotmail.com
retired guy
Really Experienced
- Joined
- Jul 28, 2002
- Posts
- 144
hello
anyone from oregon?
hi all salem, mwm 56 looking for more than i get at home.
retired guy
anyone from oregon?
hi all salem, mwm 56 looking for more than i get at home.
retired guy
Peekaboo69
Experienced
- Joined
- Aug 1, 2002
- Posts
- 49
well everyone seems about 2-3 hrs away from me
maybe i should throw a party and oregonians
can get together
woo hoo
tell me your thoughts please
maybe i should throw a party and oregonians
can get together
woo hoo
tell me your thoughts please
Overdrivenat40
Virgin
- Joined
- Mar 15, 2002
- Posts
- 28
40 MWM...........east of Portland...........
Peekaboo69
Experienced
- Joined
- Aug 1, 2002
- Posts
- 49
hmmmm sounds good hun
G
Guest
Guest
In Oregon
Hey,
I'm from Oregon City, looking for females, moms, etc. pm me please..
Cory
Hey,
I'm from Oregon City, looking for females, moms, etc. pm me please..
Cory
Infinite Wombat
Experienced
- Joined
- Aug 6, 2002
- Posts
- 65
Eugene, Oregon is the place I call home.
Peekaboo69
Experienced
- Joined
- Aug 1, 2002
- Posts
- 49
lived in sspringfield before we moved here
Infinite Wombat
Experienced
- Joined
- Aug 6, 2002
- Posts
- 65
Sexy
Here's an interesting proof that the sqrt(3) isn't a rational number.
Assume that we can express sqrt(3) = p/q, for some p,q in Z (The field of integers), with q not equal to 0. Further, we ensure that p and q have no common factors. This is really no problem, because if p and q both have some common factor, say r, then we can just divide p and q by r. We can continue to do this until we have removed all common factors.
sqrt(3) = p/q
3 = p^2/q^2
3*q^2 = p^2
This implies that p^2 has 3 as a factor (p^2 = 3*q^2).
Any integer can be expressed as the product of prime numbers raised to powers, so multiplying an integer by itself has the result of multiplying each prime factor by itself. For example,
15 = 3*5, so 15^2 = 3^2*5^2 = 225. The important point is that no new primes are introduced when we square an integer.
Therefore, since p^2 has 3 as a factor, p must also have 3 as a factor, since 3 is a prime number. Thus we can write p = 3*k, for some k in Z.
Given our new expression for p, we can rewrite our original equation, sqrt(3) = p/q, as sqrt(3) = (3* k)/q
3 = 9k^2/q^2
3 * q^2 = 9k^2
q^2 = 3k^2
So we find that q^2 has 3 a factor, which, using the preceding argument, implies that q also has 3 as a factor. Therefore, both p and q are evenly divisible by 3, but we made sure at the start that p and q had no common factors. Thus, we have a contradiction, so it must be that our initial assumption that sqrt(3) = p/q for some p,q in Z was incorrect.
As you might have noticed, it's possible to use a logically equivalent proof for all prime numbers.
Pretty sexy, eh?
Here's an interesting proof that the sqrt(3) isn't a rational number.
Assume that we can express sqrt(3) = p/q, for some p,q in Z (The field of integers), with q not equal to 0. Further, we ensure that p and q have no common factors. This is really no problem, because if p and q both have some common factor, say r, then we can just divide p and q by r. We can continue to do this until we have removed all common factors.
sqrt(3) = p/q
3 = p^2/q^2
3*q^2 = p^2
This implies that p^2 has 3 as a factor (p^2 = 3*q^2).
Any integer can be expressed as the product of prime numbers raised to powers, so multiplying an integer by itself has the result of multiplying each prime factor by itself. For example,
15 = 3*5, so 15^2 = 3^2*5^2 = 225. The important point is that no new primes are introduced when we square an integer.
Therefore, since p^2 has 3 as a factor, p must also have 3 as a factor, since 3 is a prime number. Thus we can write p = 3*k, for some k in Z.
Given our new expression for p, we can rewrite our original equation, sqrt(3) = p/q, as sqrt(3) = (3* k)/q
3 = 9k^2/q^2
3 * q^2 = 9k^2
q^2 = 3k^2
So we find that q^2 has 3 a factor, which, using the preceding argument, implies that q also has 3 as a factor. Therefore, both p and q are evenly divisible by 3, but we made sure at the start that p and q had no common factors. Thus, we have a contradiction, so it must be that our initial assumption that sqrt(3) = p/q for some p,q in Z was incorrect.
As you might have noticed, it's possible to use a logically equivalent proof for all prime numbers.
Pretty sexy, eh?
Master rogue
Virgin
- Joined
- Aug 11, 2002
- Posts
- 9
new to the board here............. in Salem,
G
Guest
Guest
hey Im from the gorge area!
