>A prime number is a positive integer p that has exactly two unique divisors: 1 and p.
Why don't we use this definition of prime numbers? It seems less arbitrary than constraining the definition to numbers greater than 1. I thought mathematicians liked beauty and all that.
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It's actually because of this we haven't solved the Riemann hypothesis.
>Why don't we use this definition of prime numbers?
but that is the definition...?
Mathematicians like definitions that can be generalized to work for several algebraic structures.
Talking about numbers greater than 1 requires your ring to have some sort of order relation which only few rings have.
>Not defining prime numbers based on dubs
>A prime number is a positive integer p such that it isn't a 2-digit dubs number in any non trivial base (p-1) except 6
brainlets
The definition of a prime is that [math]p [/math] is a prime iff
[math]p [/math] divides [math]ab [/math] implies [math]p [/math] divides [math]a [/math] or [math]p [/math] divides [math]b [/math] for all [math]a [/math] and [math]b [/math].
I don't think you'll find many number theory texts using that definition, or at least I opened 3 random ones and none of them use that. Wikipedia doesn't either.
also this is wrong since it makes 1 a prime
>integer
>exactly two unique divisors: 1 and p
3
=1x3
=(-1)x(-3)
=(-1)x(-1)x1x3
=(-1)x(-1)x(-1)x(-3)
>I thought mathematicians liked beauty and all that.
"Two unique divisors" is ugly though.
If you want to see a beautiful definition, try this:
>A prime number is a positive integer p such that for any factorization of p into a product of finitely many positive integers [math]p = \prod_{i=1}^k n_i[/math] there is an i such that [math]p = n_i[/math].
This covers the case p=1 naturally, because there is a factorization of p that doesn't contain p itself as one of the factors: namely, the empty product (k=0).
As a bonus this definition of primality generalizes to any other type of mathematical structure for which a product is meaningful (which is just about everything).
>the empty product
>beautiful
lol no
This is the correct definition. OP's definition refers to irreducible numbers not primes.
The point is so that each number has a unique prime factorization
1 still isn't a prime number by that definition.
Definition 1:
A chocolate bar is a set of n rows and m collumn of squares of chocolate with n and m >1. (otherwise it's a chocolate stick)
Definition 2:
The set of prime numbers are all quantities of square of chocolate you can't obtain with a rectangular chocolate bar.
This is a more general definition for a general R, and p in R, where p is not a unit
prime always implies irreducible
in PID, like Z, the converse is also true
Did you guys know 7 is the only prime numbers that is a multiple of 7 ?
Because what you've defined there is an irreducible number, not a prime number. The two notions only happen to coincide in unique factorisation domains.
Did you guys know that n is the only prime number that is a multiple of n?
a = 4, b = 14
2 divides a, b and ab. Is 2 not a prime?
t. brainlet
I think it's you who are the brainlet.
>for all a and b
>I think it's you who are the brainlet.
>2 divides ab=4*14 and 2 divides 4
>doesn't contradict 2 being prime
what's the issue exactly brainlet?
Yes, 2 factors out into (1 + i)(1 − i).
What issue brainlet? Prove 2 is a prime number.
...
1*1 = 1, so 2 is prime
fuck outta here
You've proven 2 is irreducible in Z. Prove it is prime.
Z is a euclidean domain so primes and irreducibles are equivalent. Q.E.D.
What's the matter, is i too much for u?
>Prove 2 is a prime number.
>look for divisors of 2 between 1 and 2
>there's no integers between 1 and 2
>therefore 2 is prime
next?
The proof for Z being euclidean assumes 2 is prime.
Congrats, you too have proven that 2 is an irreducible number. Now prove that it is prime.
proofwiki.org
No it doesn't. You lose.
see
next?
I meant the poof that in an euclidean domain the two are the same assumes that 2 is prime.
see
but that definition is wrong, it even turns 1 into a prime
next?
That definition is incomplete. not wrong.
A prime is not a zero and not a unity such that whenever it divides a product ab it divides a or it divides b.
Prove 2 is prime.
assume 2 divides ab, so ab=2c for some c
if 2 divides a or b we're done, so assume otherwise
then a=2k+1 and b=2l+1 for some k,l
so 2c=ab=(2k+1)(2l+1)=4kl+2(k+l)+1
so 1= 2c-4kl-2(k+l)=2(c-kl-(k+l)), a contradiction
therefore 2 is prime
next?
Congratulations, you have proven that 2 is prime.
>next?
do it without appealing to the principle of the excluded middle.
don't hate on excluded middle
suppose 2 divides ab
in the ring Z/2Z we have (a+2Z)(b+2Z)=ab+2Z=2Z
Z/2Z has multiplication rule
(2Z)(2Z)=2Z
(2Z)(1+2Z)=2Z
(1+2Z)(1+2Z)=1+2Z
therefore one of a+2Z or b+2Z is 2Z
therefore 2 divides a or b
Congratulations, you have rewritten the proof in this post Now do it without appealing to the principle of the excluded middle.
>there's no integers between 1 and 2
Prove it buddy.
Prime numbers don't want to be divided into equal integers. The number 1 is a nigger for this reason.
>we haven't solved the Riemann hypothesis
Speak for yourself.
Lmao. I need to say this one in class. I bet all the students will be floored.
if
1 = 1
and
p > 1
or
1 < p
Then one is NOT prime
>he doesn't get it
BWAHAHAHAHAHA *breathes in* HAHAHAHAHA
can rational numbers be prime?
>Positive
Go back to middle school, you must be able to read in order to post here
"two unique divisors" doesn't have the word positive in it
Go back to middle school, you must be able to read in order to post here
All elements in fields are units.
>zero is a unit
All non-zero* elements in fields are units.
duh, you are correct sir. brainfart
Okay you double nigger. Two unique positive divisors.
No need for racism.
>do it without appealing to the principle of the excluded middle
Same proof but with binary products replaced by finite products, and using the definition in .
To start, 2 is not a zero or a unity.
Suppose 2 divides ab.
If 2 divides a and b, we are done.
WLOG, suppose 2 does not divide a. We have 2c = ab and a = 2k + 1 for some integers c and k. Therefore
2c = 2bk + b
and
2(c - bk) = b.
It follows that 2 must divide b.
QED.
Both definitions are equivalent
Finally, another proper proof.
>I thought mathematicians liked beauty and all that.
They do. And "beauty" is precisely why they excluded 1 from being a prime number.
It all comes down to the unique prime factorization theorem:
"All positive numbers have a unique ordered prime factorization."
For example, 6 has only one unique ordered prime factorization: 2 * 3. (The "ordered" requirement prevents you from claiming that 3 * 2 is also a factorization, because the prime factors must always be listed in non-descending order.)
If you allowed 1 to be considered "prime", then the unique prime factorization theorem would be false. Example:
6 = 2 * 3
6 = 1 * 2 * 3
6 = 1 * 1 * 2 * 3
etc.
So in order to expose the beauty and elegance of the unique ordered prime factorization theorem, mathematicians had to exclude 1 from being prime.
(This is only one of many examples where 1 would muck up various formulas and theorems if it was allowed to be prime. Once you see a dozen cases of this, it becomes really clear that mathematicians did the right thing by excluding 1 from being prime.)
This also explains why negative integers are not considered prime. Example:
6 = 2 * 3
6 = -2 * -3
Since "negative primes" would also destroy the theorem, they are also excluded from being prime.
oops, I mean "positive integers", not "positive numbers" -- sorry.
Nobody said anything about considering 1 a prime. It still isn't a prime in OP's definition.