Similar amusing math terms are "almost everywhere" and "almost surely"/"almost never". For example, if you pick a random real number between 0 and 1, it will almost never be exactly 0.5. It's possible that you'll get 0.5, but the probability of getting it is exactly 0.
I think the proper interpretation of that is that you can't "pick a random real number between 0 and 1". After all, you would have to choose an infinite number of digits.
My brain is fuzzy tonight, and I'm not good at math, but something about the distinction between countable and noncountable infinities makes me question the axiom of choice. I can't precisely state the issue, but something seems off and inconsistent.
You absolutely can "pick a random real number between 0 and 1", at least for the sake of mathematical argument. But for an interval [a, b] a random real number will be x s.t. a <= x <= b is zero. That's because the set {x} has measure 0.
An even more fun fact is that the probability that the randomly chosen number is rational is also zero, by the same argument.
However, once one starts to try to describe a distribution over the reals, e.g.: "let X be a real number chosen from a uniform probability distribution over [0, 1]", then you run into serious problems. That example begs the question because there is no uniform probability distribution over that set.
There definitely is a uniform probability distribution over the interval [0,1]. It is given by the probability density function f(x) = 1. There's no uniform probability distribution over the reals, or over the integers, but defining one over a closed real interval is trivially easy. That's what "pick a random real number between 0 and 1" means.
I don't believe there is a uniform probability distribution over any interval of the reals.
Suppose such a distribution existed over [0, 1]. Then as we have shown before, the probability of choosing 0 <= x <= 1 is zero.
The second axiom of probability states the probability of one of the numbers being chosen must be is one. The third axiom requires countable additivity.
We can easily see that for our distribution, we cannot satisfy both at the same time. If we strive for the sum of probabilities to equal 1, then we end up with the sum(P(x)) = 1, but P(x) is everywhere zero.
The countable additivity never enters into this, but for any uncountable set of events a uniform distribution is impossible.
> I don't believe there is a uniform probability distribution over any interval of the reals.
Then I'll observe that you probably shouldn't be trying to comment on this topic. I didn't just assert that there is one, I pointed out to you what it was.
The probability density function f(x) = 1, defined over the interval [0,1], is uniform (all values are equal, being 1) and covers an area of 1. That's all a uniform distribution is.
It satisfies the axiom which I assume you're referring to as the "second axiom of probability" in that the definite integral of the pdf over the entire interval is 1. It satisfies the requirement you're confused about in that the probability of choosing a number that falls into either of two disjoint subintervals is equal to the sum of the probabilities of (1) choosing a number falling into the first subinterval; + (2) choosing a number falling into the second subinterval.
> Suppose such a distribution existed over [0, 1]. Then as we have shown before, the probability of choosing 0 <= x <= 1 is zero.
For fixed x, the probability of choosing x from a uniform distribution over the reals is 0. That is a special case of the probability that a value chosen from a uniform distribution over the reals will lie within an interval. The probability of drawing a value from within the interval [a,b], for any distribution, is the definite integral of the distribution's pdf from a to b. As you can see, when a = b, this value is 0, but when a ≠ b, it isn't. The probability of choosing an x such that 0 <= x <= 1 from a uniform distribution over [0,1] is 1, not 0.
Much respect for confirming to the rest of us reading this thread that your interlocutor's argument is the better one / yours is wrong. Having determined the correct answer, we all walk away correct, regardless of who was corrected.
You won't be able to find a countable union of sets of just one point which covers [0,1], so the third axiom you state is not violated.
In fact, like the law says, if you take such a countable union, the measure of the set will be 0 (e.g. the rationals).
If you're going to hold that the sum of measure for all points is 1, you're not going to be able to make any continuous distributions at all. The thing is, the assignment of measure does not work like this, as in you don't just add up the measure of all the points to get the measure of the whole set. Instead, measure is also assigned to subsets of [0,1] (in particular too all the sets in the sigma algebra of your choosing). There are some laws which prevent you from whatever probabilities to everything (such as the one you mentioned about countable unions), but there is nothing that says it has to be the sum of the probability of the points.
Remember coming across this concept while reading about Measure theory. There are also concepts like "almost well-ordered" and "almost complete" in graph theory if my memory serves right.
Something applying to “almost all” elements in a (preferably infinite) set: Applies to all elements, with the exception of a finite number of elements.
Exactly! I can't remember where but I distinctly remember some theory quantifying this as almost-1, almost-2,... as a function of the number of finite (or countably infinite maybe) exceptions.
A warning: in general mathematical usage, this is not the definition of "almost all"; the notion of "almost all" is with respect to some notion of measure on a set. In the set of real numbers, for example, with the usual Lebesgue measure on it, if all but a /countable/ number of elements has a property X, then almost all the reals has that property X. (Note that uncountable sets in R exists that have zero measure; i.e. uncountable sets that "almost no" points in R lie in that uncountable subset of R)
Ah. "With the exception of finitely many" [1] was the definition given to me in University. The generalization makes sense, however.
[1] I don't think I'm doing a good job translating the original German definition I was given, which sounds much better: "Mit Ausnahme endlich vieler".
In case anyone reads the article and wonders why 1 is not a prime number I'll give a brief answer.
A prime number has to have the property that not every number is a multiple of it. In the language of abstract algebra this means that the ideal generated by that number is proper (not the entire set of numbers one is considering).
Another reason for discounting 1 as a prime number is that considering 1 to be a prime destroys the uniqueness of prime factorization. For instance, 24 is uniquely, up to order of powers of prime factors:
2^3 times 3
If we allow 1 to be prime then I can write 24 as
1^17 times 2^3 times 3
or I can write it as
1^2 times 2^3 times 3
We lose the property that representations of numbers as products of powers of primes is unique. Thus we have a good reason to discount 1 as being prime and there aren't any good reasons to count it as a prime.
The definition given in elementary school is not the correct one. It's a working definition that works well and is correct for all positive integers except 1. In my experience people remember the definition of: "only divisible by itself and 1". Then they ask, "Why isn't 1 considered prime then?" It's because the definition given isn't correct.
That reminds of the topic of groups, rings and fields that I had in either last year of high school or an early year of college (Math.). I thought that was a really cool topic. Not too hard (for the stuff we covered), and still some cool results/theorems in it. Also, they underlie a lot of other areas of mathematics.
> A prime number has to have the property that not every number is a multiple of it.
I understand why this observation is true given the definition that excludes 1 from being a prime number, but I don't follow why this is a necessary property of prime numbers (justifying the definition in the first place).
The definition of a prime number is arbitrary. We could have defined the term “prime number” to mean anything. The question is what concept do we wish to capture with the definition. If we allow numbers with the property that all other numbers in the number system are multiples of it then 1 is a prime number and this destroys uniqueness of prime factorization. Since we want to keep unique factorization it is convenient to exclude numbers that are called units (numbers in which all other numbers are multiples of it). Prime ideals need to be proper subsets of the number system or else we’d have to add the phrase
“Let P be a proper prime ideal”
to a vast number of theorems. This is inconvenient and allowing units to be prime doesn’t give any benefits. Only headaches so it’s best to just exclude them.
The referenced article mention this in passing without giving a reason. But didn't Pythgoras use musical whole number ratios like 2:1, 3:2 and 4:3?If so he must have considered 1 and 2 to be numbers.
Sorry to be lazy about citation. Here [1] is a peer reviewed article with better internal citations about Pythagorean mathematical perspectives on 1 and 2.
It has been argued [2] that the Pythagorean numbers were quite different from numbers as we think of them in arithmetic. The One (or Oneness), for instance, is more than the number one. However, because these views extended to at least 10, it isn't an argument for disincluding 1 & 2.
(BTW, [2] is probably the best thing I've ever downloaded from Kindle. Highly recommended. Be sure to read introduction.)
[1] Caldwell, C. K., & Xiong, Y. (2012). What is the smallest prime?. Journal of Integer Sequences, 15(2), 3.
[2] Guthrie, K. S., & Fideler, D. R. (Eds.). (1987). The Pythagorean sourcebook and library: an anthology of ancient writings which relate to Pythagoras and Pythagorean philosophy. Red Wheel/Weiser.
Adding to what amelius said. The formal term for the algebraic system we are talking about is ring. We want to consider all the elements of the ring to be numbers since they are all part of the same algebraic system. Also, let’s broaden the perspective a bit and think about rational numbers. These include all integers and within the system of rational numbers every nonzero element is a unit. So 2 which is prime within the ring of integers is no longer prime within the ring of rational numbers. Every rational number is a multiple of 2 in the sense of
2*(a/2) = a
The notion of primality is not a property of a number it is a property of a number within the structure of a ring. We don’t want to think of units as not being numbers.
Also, the construction of natural numbers is such that 1 is the successor of 0. It just happens that when considering the operation of multiplication 1 is a unit but this is happenstance and not a reason to exclude 1 as a number. It’s worth noting that under addition 0 is a “unit” (really identity but plays additive role that 1 does under multiplication).
To add to what others said about it being somewhat arbitrary, it should be noted that in different time and places other definitions have been used. Each of the following sets has been "the primes" to mathematicians at some time/place:
P1 = {1, 2, 3, 5, 7, 11, ...}
P2 = {2, 3, 5, 7, 11, ...}
P3 = {3, 5, 7, 11, ...}
The underlying mathematics is the same no matter which of those you call "the primes". All that really changes is what you then have to say when you want a specific one of those sets.
If P1 is "the primes", and your theorem needs a p that is a member of a specific one of those sets, you have to say "Let p be...":
P1: "...a prime"
P2: "...a prime other than 1"
P3: "...an odd prime"
If P2 is "the primes", it is:
P1: "...1 or a prime"
P2: "...a prime"
P3: "...an odd prime"
If P3 is "the primes" for you, it is:
P1: "...1, 2, or a prime"
P2: "...2 or a prime"
P3: "...a prime"
Given the state of mathematics since the 19th century, P2 as the "primes" probably results in the minimum verbiage.
I don't think anyone has used P3 in a very long time, so unless you are studying ancient math history you probably will never encounter anything using it.
P1 and P2 were used together up until at least the 18th century, with some mathematicians using P1 even longer, but pretty much everything you will encounter now will use P2.
Even is a book you might read now says it is presenting the historical proof of some theorem as it was originally proven, and that proof has done by someone who used P1 as "the primes" in the proof, the book will almost certainly reword it to be for P2 prime.
You will probably only ever have a chance of running into P1 primes if you actually go to original sources, finding copies the actual books or articles or papers of mathematicians from back when many used P1 primes.
Maybe it isn't so intuitive after all? Among the famous advocates of the primesness of 1, are [0] Goldbach, Legendre, Lehmer and Carl Sagan. The main discussion seems to come from 1 not being a composite number, but neither being welcomed in the home of the primes.
I prefer to think about it as there are three special types of numbers: zero, units (invertible elements), and prime numbers, which exclude zero and units. In the integers, the units are +/-1, so that’s why they are not prime. In the rationals, every non-zero element is a unit, which is why there are no prime numbers in the rationals.
I find it terrible naming. “Almost-x” is good but “2-almost-x” really sounds bad. “2-factor number” ? “2nd-order-prime”? Twime, Thrime? Even “2-quasi-prime” sounds better.
43 comments
[ 5.7 ms ] story [ 50.0 ms ] threadhttps://en.wikipedia.org/wiki/Almost_everywhere
https://en.wikipedia.org/wiki/Almost_surely
And this from the same thread :)
https://pbs.twimg.com/media/DqhjmPdU4AAb979.jpg
An even more fun fact is that the probability that the randomly chosen number is rational is also zero, by the same argument.
However, once one starts to try to describe a distribution over the reals, e.g.: "let X be a real number chosen from a uniform probability distribution over [0, 1]", then you run into serious problems. That example begs the question because there is no uniform probability distribution over that set.
Suppose such a distribution existed over [0, 1]. Then as we have shown before, the probability of choosing 0 <= x <= 1 is zero.
The second axiom of probability states the probability of one of the numbers being chosen must be is one. The third axiom requires countable additivity.
We can easily see that for our distribution, we cannot satisfy both at the same time. If we strive for the sum of probabilities to equal 1, then we end up with the sum(P(x)) = 1, but P(x) is everywhere zero.
The countable additivity never enters into this, but for any uncountable set of events a uniform distribution is impossible.
Then I'll observe that you probably shouldn't be trying to comment on this topic. I didn't just assert that there is one, I pointed out to you what it was.
The probability density function f(x) = 1, defined over the interval [0,1], is uniform (all values are equal, being 1) and covers an area of 1. That's all a uniform distribution is.
It satisfies the axiom which I assume you're referring to as the "second axiom of probability" in that the definite integral of the pdf over the entire interval is 1. It satisfies the requirement you're confused about in that the probability of choosing a number that falls into either of two disjoint subintervals is equal to the sum of the probabilities of (1) choosing a number falling into the first subinterval; + (2) choosing a number falling into the second subinterval.
> Suppose such a distribution existed over [0, 1]. Then as we have shown before, the probability of choosing 0 <= x <= 1 is zero.
For fixed x, the probability of choosing x from a uniform distribution over the reals is 0. That is a special case of the probability that a value chosen from a uniform distribution over the reals will lie within an interval. The probability of drawing a value from within the interval [a,b], for any distribution, is the definite integral of the distribution's pdf from a to b. As you can see, when a = b, this value is 0, but when a ≠ b, it isn't. The probability of choosing an x such that 0 <= x <= 1 from a uniform distribution over [0,1] is 1, not 0.
In fact, like the law says, if you take such a countable union, the measure of the set will be 0 (e.g. the rationals).
If you're going to hold that the sum of measure for all points is 1, you're not going to be able to make any continuous distributions at all. The thing is, the assignment of measure does not work like this, as in you don't just add up the measure of all the points to get the measure of the whole set. Instead, measure is also assigned to subsets of [0,1] (in particular too all the sets in the sigma algebra of your choosing). There are some laws which prevent you from whatever probabilities to everything (such as the one you mentioned about countable unions), but there is nothing that says it has to be the sum of the probability of the points.
For example, almost all prime numbers are odd.
[1] I don't think I'm doing a good job translating the original German definition I was given, which sounds much better: "Mit Ausnahme endlich vieler".
A prime number has to have the property that not every number is a multiple of it. In the language of abstract algebra this means that the ideal generated by that number is proper (not the entire set of numbers one is considering).
Another reason for discounting 1 as a prime number is that considering 1 to be a prime destroys the uniqueness of prime factorization. For instance, 24 is uniquely, up to order of powers of prime factors:
2^3 times 3
If we allow 1 to be prime then I can write 24 as
1^17 times 2^3 times 3
or I can write it as
1^2 times 2^3 times 3
We lose the property that representations of numbers as products of powers of primes is unique. Thus we have a good reason to discount 1 as being prime and there aren't any good reasons to count it as a prime.
The definition given in elementary school is not the correct one. It's a working definition that works well and is correct for all positive integers except 1. In my experience people remember the definition of: "only divisible by itself and 1". Then they ask, "Why isn't 1 considered prime then?" It's because the definition given isn't correct.
Equivalently, p is prime iff p | ab implies p | a or p | b.
The only issue with this definition is that you're forced to admit that 0 is prime in integral domains.
That reminds of the topic of groups, rings and fields that I had in either last year of high school or an early year of college (Math.). I thought that was a really cool topic. Not too hard (for the stuff we covered), and still some cool results/theorems in it. Also, they underlie a lot of other areas of mathematics.
https://en.wikipedia.org/wiki/Group_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Field_(mathematics)
I understand why this observation is true given the definition that excludes 1 from being a prime number, but I don't follow why this is a necessary property of prime numbers (justifying the definition in the first place).
“Let P be a proper prime ideal”
to a vast number of theorems. This is inconvenient and allowing units to be prime doesn’t give any benefits. Only headaches so it’s best to just exclude them.
Thanks for this explanation. I used to be puzzled about 1 being not a prime but now I feel better.
But can we also propose that 1 should not be considered a number? Because 1 is the unit with which all other numbers are measured.
That would (again) destroy a lot of useful properties without any benefit.
> Because 1 is the unit with which all other numbers are measured.
Not sure why this would make you question whether 1 is a number.
[0] http://www.math.tamu.edu/~dallen/history/pythag/pythag.html
It has been argued [2] that the Pythagorean numbers were quite different from numbers as we think of them in arithmetic. The One (or Oneness), for instance, is more than the number one. However, because these views extended to at least 10, it isn't an argument for disincluding 1 & 2.
(BTW, [2] is probably the best thing I've ever downloaded from Kindle. Highly recommended. Be sure to read introduction.)
[1] Caldwell, C. K., & Xiong, Y. (2012). What is the smallest prime?. Journal of Integer Sequences, 15(2), 3. [2] Guthrie, K. S., & Fideler, D. R. (Eds.). (1987). The Pythagorean sourcebook and library: an anthology of ancient writings which relate to Pythagoras and Pythagorean philosophy. Red Wheel/Weiser.
2*(a/2) = a
The notion of primality is not a property of a number it is a property of a number within the structure of a ring. We don’t want to think of units as not being numbers.
Also, the construction of natural numbers is such that 1 is the successor of 0. It just happens that when considering the operation of multiplication 1 is a unit but this is happenstance and not a reason to exclude 1 as a number. It’s worth noting that under addition 0 is a “unit” (really identity but plays additive role that 1 does under multiplication).
P1 = {1, 2, 3, 5, 7, 11, ...}
P2 = {2, 3, 5, 7, 11, ...}
P3 = {3, 5, 7, 11, ...}
The underlying mathematics is the same no matter which of those you call "the primes". All that really changes is what you then have to say when you want a specific one of those sets.
If P1 is "the primes", and your theorem needs a p that is a member of a specific one of those sets, you have to say "Let p be...":
If P2 is "the primes", it is: If P3 is "the primes" for you, it is: Given the state of mathematics since the 19th century, P2 as the "primes" probably results in the minimum verbiage.I don't think anyone has used P3 in a very long time, so unless you are studying ancient math history you probably will never encounter anything using it.
P1 and P2 were used together up until at least the 18th century, with some mathematicians using P1 even longer, but pretty much everything you will encounter now will use P2.
Even is a book you might read now says it is presenting the historical proof of some theorem as it was originally proven, and that proof has done by someone who used P1 as "the primes" in the proof, the book will almost certainly reword it to be for P2 prime.
You will probably only ever have a chance of running into P1 primes if you actually go to original sources, finding copies the actual books or articles or papers of mathematicians from back when many used P1 primes.
[0]https://cs.uwaterloo.ca/journals/JIS/VOL15/Caldwell2/cald6.h...
Some years I did something similar for the 'divisibility' of prime numbers and the result was pretty interesting:
http://www.gibney.de/does_anybody_know_this_fractal
The most common interpretation of a 'complex prime' is the gaussian prime. When rendered on the complex plane looks like this:
https://commons.wikimedia.org/wiki/File:Gauss-primes-768x768...
Looks rather random. Maybe going from 'is prime or not' to 'primeness' would reveal some more insight.