What Is an “Almost Prime” Number?
blogs.scientificamerican.com
blogs.scientificamerican.com
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)
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...":
P1: "...a prime"
P2: "...a prime other than 1"
P3: "...an odd prime"
P1: "...1 or a prime"
P2: "...a prime"
P3: "...an odd prime"
P1: "...1, 2, or a prime"
P2: "...2 or a prime"
P3: "...a prime"
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.
And this from the same thread :)
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".
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.