31 Aug '03 10:09>
Then i'm curious what proof you got. I still can't imagen it. But we'll see.
Originally posted by royalchickenThe set of rational numers is countable. We can map any rational on to an integer by the following operation (cribbed from http://www.cut-the-knot.org/do_you_know/countRats.shtml):
What is the probability that a randomly selected rational number (fraction in lowest terms) has an even denominator?
Originally posted by iamatigerYes, I was wrong - I think the proportion of Odd A's to even A's is 1/2 + 1/8 + 1/32 + 1/128... = 2/3. So the probability of an even denominator is 1/2 * 2/3 = 1/3
If X is even then its prime number composition will contain some power A (A>0) of 2. However a corresponding power of 2 will only appear in the prime number composition of N if A is ODD (if A is even then the power of two will contribute to the prime number decomposition of M).
So N is odd IF (X is Even) AND (A is Odd)
the probability is 1/2 * 1/2 = 1/ ...[text shortened]... may have gone wrong somewhere, possibly in the assumption that A is randomly ditributed...
Originally posted by TheMaster37Thanks for agreeing with me, but I have a few issues with your method : 🙂
I think you got it right the second time....
Pick a rational number. The numerator (N) is odd (NO) or even (NE), and the denominator (D) is odd (DO) or even (DE).
Case 1: NO+DO, Since the denominator is odd, there is no way that s ...[text shortened]... y case 3 has an even denominator. So the ratio even : odd is 1 : 3
Originally posted by FiathahelIts an interesting question - of course there is exactly one even number for every odd number - so I would have thought that the probability of getting an even number would be 1/2. But can anyone come up with any method at all of randomly (and fairly) selecting an integer when there is no upper or lower bound to the set of integers from which you are to select one?
Can someone proof for me that the probability of picking an even number out of all numbers is 1/2?
Cause this is what you all use, and this is what I doubt on.
Originally posted by iamatigercrap
Thanks for agreeing with me, but I have a few issues with your method : 🙂
a) One case out of 4 is a prob of 1/4, not 1/3
b) NE+DE could be something like 2/8 which simplifies to 1/4 (still an even denominator)
c) I think your method of generating a random number was perhaps biased - if you generate a random numerator and then a random denominator ...[text shortened]... use disproprtionately many of your generated numbers will simplify to simple fractions like 1/2.
Originally posted by royalchickenThere are some errors in your proof
THe answer is 1/3. Good going iamatiger!
Here's my proof. Let S(d) be the number of rationals in (0,N] with denominator d for some integer N.
Clearly, we are counting pairs (m,d) such that gcd(m,d)=1. If d is prime, then:
S(d) = N(1-1/d)
If d has k distinct prime factors, then:
S(d) = N(1-1/p1 - 1/p2 - ...-1/pk + 1/p1p2+...+1/p1p2p3p4 ...[text shortened]... t being even guarantees the presence of a prime factor, 2, while being odd guarantees nothing.