4 ms·

It comes from how we define infinity/the number line. Any number divided by infinity is 0, that's easy. But between any two points there is an infinite number

It comes from how we define infinity/the number line. Any number divided by infinity is 0, that's easy. But between any two points there is an infinite number of rational numbers and between any two rational numbers you can derive an infinite number of irrational numbers. I know this isn't kosher, but it is pretty much infinity/(infinity ^ infinity). That leads us back to the probability of picking a rational number as 1/infinity, because for every rational number there are infinity irrational numbers that we also could have chosen. David Foster Wallace had a really good book called Everything and More where he explored the history of infinity. Might be of interest to you.

SomeStupidPoint 10y ago
That's a really bad argument, since between any two irrational numbers, there's an infinite number of rational numbers. Ed: For an explicit construction of such numbers -- Suppose we have a<b, two irrational numbers. Let b-a = d, the difference between them. Then there exists N, an integer, such that 10^(-N) < d. Let b1 be b truncated at the (N+1)th digit past the decimal point. Then b1 is rational (finitely many digits) and b-b1 < d (and b1 < b), so b1 is in (a,b). You can then add back one digit of b at a time, to get b2, b3, etc which form an infinite sequence of rational numbers converging to b from below. (We get infinite unique numbers from the fact b is irrational. Not all the b_n need be distinct.)
posterboy 10y ago
> Any number divided by infinity is 0, that's easy That's a dirty hack for the sake of argument, but it isn't really true. The convergence limit of division with infinitely increasing divisors is zero. Corollary division by zero isn't infinity, it's simply not defined.
thanatropism 10y ago
I wish there was a third alternative to upvote/downvote -- "sounds like a good chap acting in good faith, but knows less than nothing of this subject".

SomeStupidPoint 10y ago

That's a really bad argument, since between any two irrational numbers, there's an infinite number of rational numbers. Ed: For an explicit construction of such numbers -- Suppose we have a<b, two irrational numbers. Let b-a = d, the difference between them. Then there exists N, an integer, such that 10^(-N) < d. Let b1 be b truncated at the (N+1)th digit past the decimal point. Then b1 is rational (finitely many digits) and b-b1 < d (and b1 < b), so b1 is in (a,b). You can then add back one digit of b at a time, to get b2, b3, etc which form an infinite sequence of rational numbers converging to b from below. (We get infinite unique numbers from the fact b is irrational. Not all the b_n need be distinct.)

posterboy 10y ago

> Any number divided by infinity is 0, that's easy That's a dirty hack for the sake of argument, but it isn't really true. The convergence limit of division with infinitely increasing divisors is zero. Corollary division by zero isn't infinity, it's simply not defined.

thanatropism 10y ago

I wish there was a third alternative to upvote/downvote -- "sounds like a good chap acting in good faith, but knows less than nothing of this subject".