Nono, 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.
Well as the definition of an even number is a number that is divisible by 2, and the definition of a prime number is a number only divisible by 1 and itself, if a number is not 2 (as you decided to take it out of the prime numbers for some reason) it has to be odd to be prime.
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.
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Originally posted by FabianFnasEqually, no primes are divisible by 3 either.
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.
Originally posted by iamatigerAnd now by induction I will expand this proof.
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.
(Seriously what's going on in this thread?)
Originally posted by FabianFnasA proof that 100% of primes are odd is not the same as proof that all primes are odd.
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?