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Topic: INFINITE PRIMES? (Read 1626 times) |
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The_Fool
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Are there an infinite amount of primes? As the numbers get higher, so does the space between the primes. Therefore, as you approach infinity, you just can't get another prime. Am I right or wrong, and how do you approach this?
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towr
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Re: INFINITE PRIMES?
« Reply #1 on: Jan 12th, 2004, 10:30am » |
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You're wrong.
It's also allready on the site I think.. But I'll repeat the argument (hidden)
::Start by supposing there are finitely many primes, then there is a highest numbered prime pmax.
Now multiply all primes up to and including that prime and add one.
You now have a number N that isn't a multiple of any of the known primes, and thus has to have k distinct new prime factors (where 1 <= k < N/pmax).::
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| « Last Edit: Jan 12th, 2004, 10:47am by towr » |
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TenaliRaman
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Re: INFINITE PRIMES?
« Reply #2 on: Jan 12th, 2004, 10:51am » |
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Quote:| As the numbers get higher, so does the space between the primes. |
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Interestingly there is a conjecture which says that "there are infinitely many twin primes. This conjecture directly contradicts your statement.
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towr
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Re: INFINITE PRIMES?
« Reply #3 on: Jan 12th, 2004, 11:05am » |
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You have to consider, a conjecture is just that, conjecture, not theorem. (It would be nice if we could find a proof for it)
The prime density does seem to decrease (it approximately follow 1/ln(N), which for N -> [infty] is 0) So the average distance also decreases.
But all this just means the probability some number is prime is 0 for large numbers, not that they don't exist. 0 probability events do occur, every day even..
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| « Last Edit: Jan 12th, 2004, 11:06am by towr » |
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TenaliRaman
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Re: INFINITE PRIMES?
« Reply #4 on: Jan 12th, 2004, 11:13am » |
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towr,
I understand what you mean.
Its just the statistical data accumulated so far gives a strong support to this conjecture.(i don't mean to say that it is enough .. i hope u get what i mean).
If i get bored of life some day i will just sit and try to prove this conjecture and not to miss the riemann hypothesis as well. 
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James Lu
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towr,
What are examples of 0 probability events that occur?
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Sameer
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Re: INFINITE PRIMES?
« Reply #6 on: Jan 13th, 2004, 8:40am » |
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The event that you will ever post in this forum
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towr
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Re: INFINITE PRIMES?
« Reply #7 on: Jan 13th, 2004, 10:54am » |
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on Jan 13th, 2004, 8:30am, James Lu wrote:towr,
What are examples of 0 probability events that occur? |
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-Any event that only occurs once, for example. Or more generally any event that occurs a finite number of times, where there are an infinite amount to choose from.
-Selecting a real number with uniform distribution from an open interval. Or more generally with any continuous distribution.
-When moving from some point A to a point B, the probability you're at any specific point in between A and B.
There are also cases where two probabilities are 0, but when you divide them you get a very reasonable normal probability.. For instance with a conditional probability, where you look at a finite subset of an infinite set.
f.i. if you move through the open interval <0, 10>,
the probability of being on an integer, P(integer), is zero
the probability of being on an integer that is divisible by three, P(integer and div3), is also 0,
but the probability of being on a number divisible by three given that you are on an integer, P (div3|integer) = P(integer and div3)/P(integer), is 1/3.
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| « Last Edit: Jan 13th, 2004, 10:55am by towr » |
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Sir Col
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Re: INFINITE PRIMES?
« Reply #8 on: Jan 13th, 2004, 4:14pm » |
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In my understanding, the problem with probabilities is that they are only defined over exclusively countable (discrete) or exclusively measurable (continuous) domains for which the process of selection is clearly defined. It is meaningless to talk about the probability of selecting a discrete value from a continuous set. For example, P(selecting 2 from [0,4]) is undefined. Firstly, the process of selection is not possible, and secondly, 2 and [0,4] are incompatible elements. Whereas P(selecting [1.5,2.5] from [0,4])=1/4, and has meaning.
However, the greatest philosophical challenges lie with examples such as the probability of rolling 7 on an ordinary 6-sided die, or spinning a head or a tail with an ordinary coin. Is it reasonable to talk something in terms of probabilities if there is no measure of likelihood?
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SWF
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Re: INFINITE PRIMES?
« Reply #9 on: Jan 13th, 2004, 6:08pm » |
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Using The_Fool's reasoning, there are a finite number of integers that are pefect squares:
As the numbers get higher, so does the space between perfect squares. Therefore, as you approach infinity, you just can't get another perfect square.
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Icarus
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Re: INFINITE PRIMES?
« Reply #10 on: Jan 13th, 2004, 6:11pm » |
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The concept of probability lends itself easily to more general mixtures of discrete and continuous situations, but to do so adds another level of abstraction to the mathematics, which most students of probability are not ready for. So in courses discrete and continuous probabilities are kept chastely apart, lest virgin minds be polluted by their carnal mixing.
However, the probability of a discrete event in a continuum is well-defined in any case. For continuums unsoiled by the intrusions of discrete influences, the probability of any discrete event will always be zero. This means that any calculation involving them will either result in 0, or 0[cdot][infty], which is undefined. To define the values needed, you have to look at intervals. That is why in probability problems involving continuums you find everything expressed in terms of intervals or other sets with interior.
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John_Gaughan
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Re: INFINITE PRIMES?
« Reply #11 on: Jan 13th, 2004, 7:17pm » |
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on Jan 13th, 2004, 6:08pm, SWF wrote:Using The_Fool's reasoning, there are a finite number of integers that are pefect squares:
As the numbers get higher, so does the space between perfect squares. Therefore, as you approach infinity, you just can't get another perfect square.
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This reminds me of something I read about irrational numbers. I think it was a web page or book about random numbers, and it talked about [pi]. There are an infinite number of digits in [pi], and in those digits you will find every subsequence of digits. For example, somewhere in the digits of [pi] is a sequence of a million ones in a row. And this is to be expected. This is more random number theory but still is related to numbers and infinity so leave me alone 
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ThudnBlunder
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Re: INFINITE PRIMES?
« Reply #12 on: Jan 13th, 2004, 8:49pm » |
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Quote:| For example, somewhere in the digits of [pi] is a sequence of a million ones in a row. |
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Only one sequence? 
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| « Last Edit: Jan 13th, 2004, 11:28pm by ThudnBlunder » |
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Eigenray
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Re: INFINITE PRIMES?
« Reply #13 on: Jan 13th, 2004, 10:56pm » |
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Here's something to think about when picking real numbers "at random."
Define an equivalence relation on [0,1) by saying x is equivalent to y iff x-y is rational. Then the equivalence class of each x represents a countable set of real numbers, and there are therefore an uncountable number of equivalence classes.
Now, use the Axiom of Choice, and form a set S by taking one element from each equivalence class.
Pick a real number from [0,1) at random. What is the probability it lies in S?
Remember, we like probability measures to be
(1) Countably additive. If S1, S2, ..., is a countable collection of disjoint sets, then the probability that x lies in their union should be the sum of the probabilities of x lying in each set.
(2) Translation (rotation?) invariant. The probability that x lies in S should be the same as the probability that it lies in S+t = {x+t mod 1 | x in S}, for any t.
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towr
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Re: INFINITE PRIMES?
« Reply #14 on: Jan 14th, 2004, 12:59am » |
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on Jan 13th, 2004, 4:14pm, Sir Col wrote:| It is meaningless to talk about the probability of selecting a discrete value from a continuous set. |
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Not really, it is simply an integral of the probability density function over a domain containing one element, and thus zero.
It is the same as the probability of selecting a value within a certain interval, only the interval in the discrete case is just [v], rather than [a,b]. Integrating over [a,b] is equivalent to using <a,b>, and for [v] = <v> = {} it gives 0.
Using the distribution function (which is the primitive of the probability density function), for an interval [a,b] you get D(b)-D(a), and for [v] you get D(v)-D(v) = 0, perfectly well defined and logical.
Quote:| However, the greatest philosophical challenges lie with examples such as the probability of rolling 7 on an ordinary 6-sided die, or spinning a head or a tail with an ordinary coin. Is it reasonable to talk something in terms of probabilities if there is no measure of likelihood? |
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I don't see why that is a philosophical challenge. Since there isn't a 7 on an ordinary 6-sided die you logically can't role it. If somehow you do role a 7, then you need only remember that falsum implies everything, so you may make a wish.
Mathematics is very forgiving with the absurd in a way.. It's really not that judgemental..
(Of course there's a zero probability that due to quantummechanics one of the eyes from the other five sides displaces to the six eyed side, in which case you may role a seven, so it's not really impossible  )
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| « Last Edit: Jan 14th, 2004, 1:02am by towr » |
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John_Gaughan
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Re: INFINITE PRIMES?
« Reply #15 on: Jan 14th, 2004, 6:33am » |
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on Jan 13th, 2004, 8:49pm, THUDandBLUNDER wrote:| Only one sequence? |
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Actually I said "there is a sequence," as in [exists]. Not "one sequence." I suspect there are an infinite number of each sequence but I'm not a mathematician so I'll stop short of trying to prove it.
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Benoit_Mandelbrot
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Re: INFINITE PRIMES?
« Reply #16 on: Jan 14th, 2004, 8:35am » |
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Well, since there is no limit to numbers, you will eventually find another prime, because if you give me the highest prime you can come up with, I can find another number above that, until I find another prime, so there should be an infinite number of primes.
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Because of modulo, different bases, and significant digits, all numbers equal each other!
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John_Gaughan
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Re: INFINITE PRIMES?
« Reply #17 on: Jan 14th, 2004, 8:52am » |
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on Jan 14th, 2004, 8:35am, Benoit_Mandelbrot wrote:| Well, since there is no limit to numbers, you will eventually find another prime, because if you give me the highest prime you can come up with, I can find another number above that, until I find another prime, so there should be an infinite number of primes. |
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I remember seeing a proof somewhere that was either inductive or used a similar process that proved there are infinite primes. Of course with larger numbers primes are spread out more, but you can always find a larger one.
Someone mentioned here or in another thread that prime number distribution decreases logarithmically, so they get more sparse, but in a predictable (but not absolute) way. Since the logarithmic curves increase to infinity (however slowly), is there ever a prime number Pn with an infinite number of composites between it and the next prime Pn+1, or would this effectively say that Pn is the last prime (which is false)?
Maybe one of the math PhDs here could englighten us. This was the original question but thread got hijacked early on
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Dudidu
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Re: INFINITE PRIMES?
« Reply #18 on: Jan 14th, 2004, 9:10am » |
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on Jan 14th, 2004, 8:52am, John_Gaughan wrote:| Maybe one of the math PhDs here could englighten us. This was the original question but thread got hijacked early on |
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John_Gaughan hi,
Maybe this proof (from 300 BC) will help you.
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towr
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Re: INFINITE PRIMES?
« Reply #19 on: Jan 14th, 2004, 9:16am » |
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on Jan 14th, 2004, 8:52am, John_Gaughan wrote:| I remember seeing a proof somewhere that was either inductive or used a similar process that proved there are infinite primes. |
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You can read my version of it in the second post from the top.
Quote:| Someone mentioned here or in another thread that prime number distribution decreases logarithmically, so they get more sparse, but in a predictable (but not absolute) way. |
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in the fourth post :p
a better approximation is the li function. There are about li(n) primes among the first n postive whole numbers
Quote:| Since the logarithmic curves increase to infinity (however slowly), is there ever a prime number Pn with an infinite number of composites between it and the next prime Pn+1, or would this effectively say that Pn is the last prime (which is false)? |
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The latter, while not entirely meaningless (you should hear Icarus on this) transinfinite numbers are a hard concept to wrap you mind around, more importantly they aren't integers, so they can't be prime anyway.
Most of these things are also (better) explained in the following mathworld pages
http://mathworld.wolfram.com/PrimeNumberTheorem.html
http://mathworld.wolfram.com/PrimeNumber.html
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Icarus
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Re: INFINITE PRIMES?
« Reply #20 on: Jan 14th, 2004, 7:12pm » |
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To be a bit more explicit, transfinite numbers do not form a domain, so the concept of "prime" is not defined for them. I.e., multiplication on transfinite numbers does not allow the concept of "prime".
However, I don't think John meant to delve into the transfinite realm with his question, but rather was caught by a common misunderstanding of natural numbers.
Any natural number has only a finite set of other natural numbers less than it, so it is impossible for an infinite set of natural numbers to lie between any two. In particular, any two consecutive primes can only differ by a finite amount.
And to reiterate towr's argument in the 1st reply of this thread(which is a reiteration of Euclid's argument), there are infinitely many primes (each prime is finite, but there is always a next higher prime). Choose any finite set of primes. Take the product of all the primes in that set, then add 1. Finally factor the result back into primes. None of the primes in this second list is from the original set, since dividing the result by any of the primes in the original set always leaves a remainder of 1.
Hence, for any finite set of primes, there are other primes not in the set. So the set of all primes must be infinite.
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John_Gaughan
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Re: INFINITE PRIMES?
« Reply #21 on: Jan 14th, 2004, 7:35pm » |
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on Jan 14th, 2004, 7:12pm, Icarus wrote:| However, I don't think John meant to delve into the transfinite realm with his question, but rather was caught by a common misunderstanding of natural numbers. |
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Well of course I did not mean that since I don't know what transinfinite numbers are. But I figured there was something that might explain it in the higher mathematics.
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Icarus
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Re: INFINITE PRIMES?
« Reply #22 on: Jan 15th, 2004, 3:36pm » |
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on Jan 14th, 2004, 7:35pm, John_Gaughan wrote:
Well of course I did not mean that since I don't know what transinfinite numbers are. But I figured there was something that might explain it in the higher mathematics. |
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Transfinite (or [i] infinite[\i], but not "transinfinite") numbers are numbers greater than or equal to infinity. These numbers are not a part of the Natural numbers or the Reals, or the Complex numbers even. There are many ways in which such numbers can be defined. I described three of the most common in this post in the 0.999... thread. Another common one is a variation of the continuum infinities mentioned that attaches a single infinity to the complex plane which is the limit of anything whose magnitude increases without bound. This turns the complex plane into the "Riemann Sphere", a very useful tool in Complex Analysis. A better description can be found in the Complex Analysis forum.
However, I have never seen a set of infinite numbers with the appropriate properties for primes to exist.
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John_Gaughan
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Re: INFINITE PRIMES?
« Reply #23 on: Jan 15th, 2004, 9:38pm » |
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Ah, I misread the "transfinite" part. Either way it is a new concept to me. Thanks for the link. I don't understand half of that post but that's okay, I'm getting there. I really do love math. Between this forum and wikipedia I am learning more than I ever thought existed in math. Some day I hope to learn more about math. Hmm... I guess this is some day
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Barukh
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Re: INFINITE PRIMES?
« Reply #24 on: Jan 16th, 2004, 12:25am » |
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on Jan 14th, 2004, 8:52am, John_Gaughan wrote:| ...is there ever a prime number Pn with an infinite number of composites between it and the next prime Pn+1... |
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Interestingly enough, Euclid's argument about the infinitude of primes may be applied to show that for any finite n - no matter how big - there exist n consecutive composite numbers: just take (n+1)! + 2 to (n+1)! + (n+1).
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