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/sci/ - Science & Math


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8982331 No.8982331 [Reply] [Original]

>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.

>> No.8982338

>>8982331
It's actually because of this we haven't solved the Riemann hypothesis.

>> No.8982342
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8982342

>>8982331
>Why don't we use this definition of prime numbers?
but that is the definition...?

>> No.8982344

>>8982331
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.

>> No.8982349

>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

>> No.8982351

>>8982342
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].

>> No.8982363
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8982363

>>8982351
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.

>> No.8982373

>>8982351
also this is wrong since it makes 1 a prime

>> No.8982374

>>8982331
>integer
>exactly two unique divisors: 1 and p
3
=1x3
=(-1)x(-3)
=(-1)x(-1)x1x3
=(-1)x(-1)x(-1)x(-3)

>> No.8982418
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8982418

>>8982331
>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).

>> No.8982420
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8982420

>>8982418
>the empty product
>beautiful

>> No.8982422

>>8982351
lol no

>> No.8982427

>>8982351
This is the correct definition. OP's definition refers to irreducible numbers not primes.

>> No.8982447

The point is so that each number has a unique prime factorization

>> No.8982503

>>8982447
1 still isn't a prime number by that definition.

>> No.8982529

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.

>> No.8982534

>>8982351
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

>> No.8982542

Did you guys know 7 is the only prime numbers that is a multiple of 7 ?

>> No.8982548

>>8982331
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.

>> No.8982556

>>8982542
Did you guys know that n is the only prime number that is a multiple of n?

>> No.8982558

>>8982351
a = 4, b = 14
2 divides a, b and ab. Is 2 not a prime?

>> No.8982560

>>8982558
t. brainlet

>> No.8982566

>>8982560
I think it's you who are the brainlet.
>for all a and b

>> No.8982568
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8982568

>>8982566
>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?

>> No.8982572

>>8982558
Yes, 2 factors out into (1 + i)(1 − i).

>> No.8982575

>>8982568
What issue brainlet? Prove 2 is a prime number.

>> No.8982577
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8982577

>> No.8982579

>>8982558
1*1 = 1, so 2 is prime

>> No.8982580

>>8982572
fuck outta here

>> No.8982583

>>8982579
You've proven 2 is irreducible in Z. Prove it is prime.

>> No.8982584

>>8982583
Z is a euclidean domain so primes and irreducibles are equivalent. Q.E.D.

>> No.8982585

>>8982580
What's the matter, is i too much for u?

>> No.8982587
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8982587

>>8982575
>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?

>> No.8982590

>>8982584
The proof for Z being euclidean assumes 2 is prime.

>> No.8982591

>>8982587
Congrats, you too have proven that 2 is an irreducible number. Now prove that it is prime.

>> No.8982592

>>8982590
https://proofwiki.org/wiki/Euclidean_Algorithm#Euclid.27s_Proof
No it doesn't. You lose.

>> No.8982595
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8982595

>>8982591
see
>>8982363

next?

>> No.8982599

>>8982592
I meant the poof that in an euclidean domain the two are the same assumes that 2 is prime.
>>8982595
see
>>8982351

>> No.8982601
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8982601

>>8982599
but that definition is wrong, it even turns 1 into a prime

next?

>> No.8982609

>>8982601
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.

>> No.8982614
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8982614

>>8982609
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?

>> No.8982618

>>8982614
Congratulations, you have proven that 2 is prime.
>next?
do it without appealing to the principle of the excluded middle.

>> No.8982638

>>8982618
don't hate on excluded middle

>> No.8982644

>>8982618
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

>> No.8982675

>>8982644
Congratulations, you have rewritten the proof in this post >>8982614
Now do it without appealing to the principle of the excluded middle.

>> No.8982709

>>8982587
>there's no integers between 1 and 2
Prove it buddy.

>> No.8982714

>>8982709
https://en.wikipedia.org/wiki/Peano_axioms

>> No.8982719

Prime numbers don't want to be divided into equal integers. The number 1 is a nigger for this reason.

>> No.8982724
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8982724

>>8982338
>we haven't solved the Riemann hypothesis
Speak for yourself.

>> No.8982728
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8982728

>>8982719
Lmao. I need to say this one in class. I bet all the students will be floored.

>> No.8982731

>>8982719
if
1 = 1
and
p > 1
or
1 < p
Then one is NOT prime

>> No.8982733

>>8982731
>he doesn't get it
BWAHAHAHAHAHA *breathes in* HAHAHAHAHA

>> No.8982881

>>8982675
can rational numbers be prime?

>> No.8982935

>>8982374
>Positive
Go back to middle school, you must be able to read in order to post here

>> No.8982942
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8982942

>>8982935
"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

>> No.8982952

>>8982881
All elements in fields are units.

>> No.8982954
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8982954

>>8982952
>zero is a unit

>> No.8982956

>>8982954
All non-zero* elements in fields are units.

>> No.8983094

>>8982952
>>8982956
duh, you are correct sir. brainfart

>> No.8983313
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8983313

>>8982942
Okay you double nigger. Two unique positive divisors.

>> No.8983320
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8983320

>>8983313
No need for racism.

>> No.8983436

>>8982675
>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 >>8982418.

>> No.8983502

>>8982609
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.

>> No.8984094

>>8982331
Both definitions are equivalent

>> No.8984105

>>8983502
Finally, another proper proof.

>> No.8984333

>>8982331
>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.

>> No.8984336

>>8984333
oops, I mean "positive integers", not "positive numbers" -- sorry.

>> No.8985092

>>8984333
Nobody said anything about considering 1 a prime. It still isn't a prime in OP's definition.