Can a complex number be prime?

The intent of the post is unclear, but at any event the Gaussian integers are sure a subset of the complexes which happen to contain primes.

You do not understand primes. The idea of prime only makes sense in the context of an integral domain. For example: is the number 3 prime?

That question does not make sense. But the question: Is the number 3 prime as an element of the integers? The answer is yes.
Is the number 3 prime as an element of the reals? No. Because 5 * 3/5 = 3.

So no, there are no primes in the complex numbers. The complex numbers are an integral domain without primes, but if you restrict the complex numbers down to the gaussian integers you then get primes.

>being this much of a brainlet

Prime is a relative term. You have to specify the ring you're working with. 2 is prime in [math]\mathbb{Z}[/math] but not in [math]\mathbb{Z}[i][/math] for example.

On the contrary, you do not understand grammar.

The OP's question can be re-read at your convenience. He asked for an example of a complex number which is a prime number. I correctly replied that any and all primes will do.

It is fairly clear by context that the OP was really asking for "generalizations" of the notion, but he phrased his question poorly (and yet, what he ended up writing makes perfect sense in a way that he probably did not intend). That's what I've been pointing out with my cute insistence on my correct point about what was actually written.

I also understand that you're trying to salvage your thing by emphasizing "with respect to x or y", but that doesn't even work. And the reason why this is so is because you can have a number in thingy y (the set of complex numbers) which also has this other property with respect to thingy x (it is a prime number, see: the conventional, well-understood definition). And that's what the language of the OP actually asks for, but you fail to appreciate this.

What a convoluted ass backwards explanation of everything.

Also, it is pretty cringe that you call yourself cute. I hope you are a roastie because if you are a man I recommend you chop off those balls.

...

>quickmeme

oh God

Clearly the second sentence of this post is false as the initial notion of a prime number arose with regard to the naturals, which are not even a ring, much less an integral domain. This suffices for "the idea of prime", clearly, which is later generalized elsewhere.

The fundamental disconnect is that you fail to relate and yet distinguish the abstract algebraic notions of primality to the (distinct!) commonly understood arithmetic one, which is helpfully expounded here

en.wikipedia.org/wiki/Prime_element

And this goes back to the subtly confused language of the OP. The OP is making the same basic problem in his own language. This is what I point out.

I take you to mean in your latter sentence that /prime elements/ are what you have in mind, yet this other read is still worth doing. With regard to your statement "The complex numbers are an integral domain without primes", Let [math] \mathbb{C} - \mathbb{P} = \mathbb{C} - \{2,3,5,7,...\} [/math] , and since an integral domain is by definition a commutative ring (with unity) in which the cancellation property holds, please let me know what element of [math] \mathbb{C} - \mathbb{P} [/math] may be added to its element -3, to give 0 (integral domains are rings, and so must have additive inverses).

You're going to whine that I'm missing the point but I'm not. It's the same point that I made above.

There is possibly slight wiggle room from author to author about the definition of integral domains, so you might use that.

You misunderstood me. I did not mean that if you take out the primes, the complexes become an integral domain.

I meant that the complex numbers are an integral domain, but as one it has no prime numbers.

I gave the example: 3 is not prime because 3 = 5 * 3/5

Maybe you could build an arithmetic in [math]\mathbf{Z}\,+\,\mathrm{i}\,\mathbf{Z}[/math], but I bet it's useless.

GAUSSIAN INTEGERS
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>[math]\mathbf Z\left[\mathrm i\right][/math]
lmao I forgot about them

every commutative ring has a notion of prime. you can say that an element p is prime if p|ab implies that p|a or p|b.

is referring to primes in Z[i], the ring of numbers of the form a+bi, where a,b are integers.

is very misleading.
primality isn't really a useful concept in fields (where there is division), like R or C. every nonzero number is invertible, and so the only prime element is 0.

primality is a relative notion having to do with divisibility, which will depend on how many other elements are lying around.

the pills won't work unless you take them user

All prime numbers are integers
All integers are real numbers
All real numbers are Complex numbers
So all prime are Complex numbers.

>No. Because 5 * 3/5 = 3.
This only makes sense in a computer

Yes. Let p be a prime such that p=4k+3 where k is a natural number. Then p is prime Gaussian Integer.

>This only makes sense in a computer
[math] \frac{3}{5} 5 = 3 [/math]

>integers = naturals
kys my man
you tried to sound really smart

[math]2.122398 \cdot \frac{4}{2.122398}[/math]
>Hurr 4 is prime
once again, kys my man
your weak definition of prime applies only where integer division is implicitly floored

>0
let k = 3, element of N
p = 4(3) + 3 = 15
then 15 is prime

when the fuck are the captchas coming to save Veeky Forums from the poltard influx?

>once again, kys my man
>your weak definition of prime applies only where integer division is implicitly floored

Are you absolutely retarded? Do you understand integral domains. Let me rewind this shit because you seem to be brain dead.

Theorem: [math] \mathbb{R} is an integrla domain [/math].

The proof is trivial and left as an exercise for the reader.

So now that you have an integral domain you can find prime numbers in it. But what happens in the real numbers? That fractions exist. So technically every number can be divided by any other (exluding 0). Therefore there are no primes in the real numbers. And that was my point.

>>Hurr 4 is prime
No, my point is that while 3 is a prime in the integers, it is not a prime in the rationals, reals or complex numbers.

STOP doing the REDDIT SPACING!

Calling it "REDDIT SPACING" is actually more r/cancerous than it. Why? Because for some reason you care about that shit. Go back to plebbit where they can upvote you for posts like this because here nobody gives

a

fuck

you

faggot

Reported for reddit spacing kys fag

Cancerous.

No, 5*3/5=3.00000001

This.
You only recognize it as "reddit spacing" because that's where you're from. Nobody here cares

Whoops, meant to quote

how did you get past the captcha?

Guys complex numbers are numbers like the sqrt(-1) or sqrt(-2) or sqrt(-3)

7 is prime but is not complex

You are mistaken and you do not understand the definition of a complex number. You furthermore do not understand the general notion of subsets and supersets as it relates to the simple number systems, naturals, integers, rationals, real and complexes.

No, I didn't. This is the whole point. It is incumbent on the user to specify in what sense they mean "primes", especially in view of the commonly understood definition.

What's really happening is that you're misunderstanding me, because mine has primacy.

The latter rebuttal in this post misses the same point which has been at this point wilfully missed by the other user. In order for an algebraist to be in a position to explain to a general audience about "primality" in a general way, one has to define one's terms very precisely. The OP didn't, and that's exactly the point. Hence it is not me who is misleading, but the OP with the confusion of the question itself. This is what I've pointed out and correctly argued throughout the thread.

>implying prime numbers aren't complex
???

The set of real numbers is a subset of the complex numbers, so every real number is a complex number. Real numbers are complex numbers with an imaginary part equal to 0

Yes. For example, 2.

I think you're out of your league here, son. The concept of an irreducible is irrelevant unless you specify how your multiplication operation works.

Sopa de macaco

2 is not prime in the complex numbers. 2 is prime in the integets, but the definition of prime requires the element to not be a unit. 2 is most certainly a unit in the field of complex numbers as fractions exist.

Basically- once you have a field, you have no more primes relative to that field.

If you want you could construct it such that 2, 2i, and 2sqrt2 + i2sqrt2 are all prime, but I don't really know why you would.

Still not getting it, after all this time. Another user correctly made the same observation, and cue the broken record...