16 Aug '06 14:44

This is a thought that I've had for a time.

Every prime is odd. Odd, isn't it?

And I can prove it. Does anyone know how?

Every prime is odd. Odd, isn't it?

And I can prove it. Does anyone know how?

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mouse mouse mouse- Joined
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16 Aug '06 17:20Nono, unusual is odd enough but here I mean odd as not being even as matematicians do.

thespacemonkey says:

"Because all even numbers are divisable by 2...."

Well, either you give a glimpse that you know the proof or you're not. This I have to think over.

(*think, think, think*) What about two itself? (*think, think, think*)

Yes, you are right. The statement of yours that "all even nubers are evenly divisable with two" has the same kind of proof that the statement "all primes are odd" has. Well done.- Joined
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16 Aug '06 17:341 editI didn't think far enough here, sorry for that.

If thespacemonkey had said:

"All even numbers are divisable by 2 and still gives an even result"

then the proof of this would be very like the proof of that every prime is odd.

Please forgive me of this lapsus.

Let's stick to the statement that all primes are odd.- Joined
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16 Aug '06 20:40

Equally, no primes are divisible by 3 either.*Originally posted by FabianFnas***I didn't think far enough here, sorry for that.**

If thespacemonkey had said:

"All even numbers are divisable by 2 and still gives an even result"

then the proof of this would be very like the proof of that every prime is odd.

Please forgive me of this lapsus.

Let's stick to the statement that all primes are odd.- Joined
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p^2.sin(phi)17 Aug '06 00:241 edit

And now by induction I will expand this proof.*Originally posted by iamatiger***Equally, no primes are divisible by 3 either.**

A prime P will not be divisible by any n where n != 1 or p.

Already it has been shown for n=2 and n=3. Therefore all I have to show is that if the nth case is true then the n+1th case is true.

So if no A exists such that n*A = p then then no B must exist such that (n+1)*B = P. Utilizing the properties of primes we can show that no B exists that divides P except P and 1.

Therefore the n+1th case is true.

Therefore by induction we have shown that a prime number is only divisible by 1 and itself.

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p^2.sin(phi)17 Aug '06 07:48

A proof that 100% of primes are odd is not the same as proof that*Originally posted by FabianFnas***The first 10 primes is as follows: 2, 3, 5, 7, 11, 13, 17, 23, 29,31.**

Of theese 10 first primes only one is even, the rest is odd. Right?

The ratio of odd primes of the ten first is 9/10=90%. Right?

Do anyone se what I'm aiming at?**all**primes are odd.