Wikipedia:Reference desk/Archives/Mathematics/2009 June 13

= June 13 =

Paradox of the twots
Natural Numbers are positive integers, N={1,2,3,4,5,...}

A twot is a natural number which is the product of two unique primes.

For example:

6 is a twot because 2 * 3 = 6

10 is a twot because 2 * 5 = 10

14 is a twot because 2 * 7 = 14

15 is a twot because 3 * 5 = 15

Now as a twot is the product of two unique prime numbers, one of the two prime numbers must smaller than the other. Therefore we can call the smaller prime number the base prime and classify twots by their base primes

For base prime two

1st twot is  2 * 3 = 6

2nd twot is 2 * 5 = 10

3rd twot is 2 * 7 = 14

and so on

For base prime three

1st twot is 3 * 5 = 15

2nd twot is 3 * 7 = 21

3rd twot is 3 * 11 = 33

and so on

Because the number of prime numbers is infinity, we can match every twot of base prime two to a natural number. Therefore there are as many twots of base prime two as there are natural numbers.

Therefore we can say there are more twots than there are natural numbers because all the natural numbers can be matched to twots of base prime two and we still have twots of base prime three and other base primes leftover.

Yet, we know not all natural numbers are twots, in fact the set of twots is a subset of the set of Natural Numbers.

So the paradox of the twots. There are more twots than there are Natural Numbers and yet at the same time there are less twots than there are Natural Numbers.

PS: Originally, I did not call it by the name twot but by different name. Later I was told that the name I have chosen was an unmentionable word in the English language, so I had rename it to twot. 122.107.207.98 (talk) 06:16, 13 June 2009 (UTC)
 * Visit Hilbert's hotel for an analysis of the paradox. Bo Jacoby (talk) 07:29, 13 June 2009 (UTC).


 * There is no paradox. By showing there is a bijection between the set of natural numbers and one of its proper subsets, you have simply proved that the set of natural numbers is a Dedekind-infinite set. A simpler example is to consider the obvious bijection between the set of natural numbers and its proper subset of even numbers. Gandalf61 (talk) 08:16, 13 June 2009 (UTC)


 * (ec)The error lies in the statement " Therefore there are as many twots of base prime two as there are natural numbers." You are basically saying that these two infinities are equal just because they both are infinities. Infinity is not a number, and hence does not follow the rules of usual arithmetic. By the above logic i can devise hundreds of new paradoxes :
 * Infinity + Infinity = Infinity


 * But since infinity is equal to infinity, infinity - infinity is 0


 * Therefore Infinity = 0.


 * You see? You can't call two infinities the same, just because they are infinities. Limx->inf(x) is definitely smaller than Limx->inf(x2) Rkr1991 (talk) 08:22, 13 June 2009 (UTC)
 * Rkr1991, I am afraid your comment is incorrect. By showing a bijection from the set of 2-twots to the set of natural numbers, the OP has correctly demonstrated that there are as many 2-twots as natural numbers. The rest is questionable as well - $$\lim_{x \to +\infty}x = \lim_{x \to +\infty}(x^2)$$ in most common constructions. -- Meni Rosenfeld (talk) 17:47, 13 June 2009 (UTC)
 * I would not call it a paradox, at least not if you have a well-defined way to speak of when two sets have the same number of elements. For example, if I have the set {1,2,3}, I know that there are 3 elements in this set. If I have the set {a,b,c}, again there are three elements. "Without having a counting system," how do we know this? We know this because of the association:


 * 1 -> a


 * 2 -> b


 * 3 -> c


 * In fact, no matter whether we have three element sets {a,b,c}, {d,e,f}, {5,100,93045134314}, this "association method" allows us two decide whether the two sets have the same number of elements. However, as always in mathematics, this method must be formalized so we can check its validity in applications. I think the articles on countable set and bijection formalize this. -- PS T  08:25, 13 June 2009 (UTC)


 * This idea is originally due to Galileo (who used square numbers instead of nonsquare semiprimes). We have an article Galileo's paradox. Algebraist 11:13, 13 June 2009 (UTC)

Galileo called attention to this in the 17th century and Cantor addressed many similar problems in the 19th century.

Usage note: you should say "two distinct primes" if that's what you mean. "There is a unique even prime number" means there is exactly one of those. "The prime factors of 15 are distinct; whereas those of 18 are not distinct" is the right way to use that word. Michael Hardy (talk) 16:13, 14 June 2009 (UTC)

Twin prime version of Copeland-Erdos constant
Can anyone prove the constant created by concatenating prime p that p+2 is also a prime


 * 0.35111729...

is irrational?

Assume that there are infinitely many twin primes.

Thank you for your help.Motomuku (talk) 13:38, 13 June 2009 (UTC)


 * At the first glance, I'd say you need more information about the distribution of twins primes to conclude. Note that the statement about the irrationality of that Copeland-Erdős constant is just a consequence of a rough form of the prime number theorem, e.g. the number of primes with n digits diverges as n increases (therefore if by contradition that decimal expansion were periodic, some primes would be repeated more times in the list). Of course, any strictly increasing sequence of positive integer numbers which is 10o(n) gives rise to a non-periodic decimal expansion as well. --pma (talk) 20:54, 13 June 2009 (UTC)


 * Well 1,11,111,1111, etc or the prime in them which probably are infinite stuck together give a rational number. However for the original problem p and p+2 would eventually have repeating sequences in them - and if one had the repeats the other wouldn't. Perhaps the OP was wondering about trancendental numbers which are rather a bit harder? Dmcq (talk) 23:41, 13 June 2009 (UTC)
 * But he wants to put only "p" not "p+2" in the list... So the simplest reason for the irrationality seems that the sequence be 10o(n). --pma (talk) 23:59, 13 June 2009 (UTC)
 * Oops yes, quite right. I should have looked at his example more carefully. Dmcq (talk) 00:13, 14 June 2009 (UTC)
 * There is a bit of a problem in that even if there are infinitely many twin primes they may not be distributed as the Hardy–Littlewood conjecture on the Twin prime conjecture page says. It is possible, though extremely unlikely!, that above a certain level all the p of the type you want are of the form of a Repunit like 11111. Dmcq (talk) 09:06, 14 June 2009 (UTC)
 * On the other hand, if the integers (primes or not) of the sequence are eventually numbers of the form 11..1, the sequence has an exponential growth; that's why I was talking of 10o(n). However I must add that this way of making a decimal expansion out of an integer sequence by concatenation of digits gives me a sort of feeling of unhappiness... Isn't it just a way of restating some properties of an integer sequence in a quite involute and complicated way? I suspect that the previous constant with the primes has been introduced just as an exercise or a curiosity. In that case, I wonder what is the point of making variants with other sequences. Recreational mathematics suffers repetitions and generalizations. --pma (talk) 19:52, 14 June 2009 (UTC)
 * That's interesting, it hadn't occurred to me before, I guess a repunit in any base must be a repunit in any other base, barring bits being chopped off te beginning or end. Dmcq (talk) 11:56, 14 June 2009 (UTC)
 * Um sorry the Goormaghtigh conjecture says I'm very very wrong, I must have a good look at that. Dmcq (talk) 12:00, 14 June 2009 (UTC)

Need help with radical equation questions.
Question One: (y/y-1)^2 = 6(y/y-1)+7

Question Two: 3x(x^2 + 2x)^1/2 - 2(x^2 +2x)^3/2 = 0

Your help is much appreciated! Thanks! —Preceding unsigned comment added by 76.78.133.108 (talk) 18:11, 13 June 2009 (UTC)


 * For the first question: solve for (y/y-1) by factoring. Then expand and solve for y.
 * For the second one, factor out (x^2+2x)^1/2. The remaining part is a quadratic, and trivial to solve by factoring. --COVIZAPIBETEFOKY (talk) 18:25, 13 June 2009 (UTC)