The Prime Debate: Is 1 prime?

[Kuciwalker]

Originally posted by SnowFire It may be harder to understand why 1^infinity is the same. But basically, when you multiply a number infinite times, things get wonky. Don't forget that a number infinitely close to 1 is still "1" (just as .9999... = 1), but if you do infinite multiplications on it, you might be able to "undo" that closeness and bring out some larger number.

Just one quibble. .9 repeated is in fact exactly equal to one, because it is the sum from k=1 to infinity of (1/10)^k, which is one (by one of the rules about sums). 1^infinity isn't a number because infinity isn't a number. However, the limit as x goes to infinity of 1^x is 1.

[Ramo]

What Snow is getting at is the basic reason why .9.... = 1, namely that the sequence of .9, .99, .999, etc. converge to 1 (look up epsilon-delta proofs). Summing up an infinite series of .1^k adds some unneccessary baggage wrt small, but very important analysis details (not to mention the hassle of proving the theorem you're using).

And I think you miss his point: the limit as x approaches infinity of 1^x isn't well-defined, for the reason he pointed out. Think of it this way: let's say you're looking at .999999999999999 and take it's 1000000000000'th power. Not as close to 1 as .999999999999, right? Now, convergence is defined such that when there's an arbitrarily small difference between the two numbers, they're the "same" number. So how much of distortion do you think there would be if you took this arbitrarily small perturbation from 1 to an arbitrarily large power (infinity)? Think about it for a minute. Is it large? Is it small? Is it either?

[Guest]

Skywalker is right
If you stick to the definition
"a prime is a number with only itself and 1 as factors"
Then 1 is indeed prime.
The THING is, that as he said, that is not the right definition.
the right definition in terms of rings (abstract algebra) in general, excludes units and reduces to the one above in the case of the ring of integers for the non-units.
So in high school, they teach you the one above because it is much easier to remember and tell USUALLY tell you that 1 is not a prime by convention...

[TheStinger]

Shouldn't you guys be doing something more useful like
watching paint dry

[Enigma_Nova]

Originally posted by geeslaka
1 being prime does not eliminate unique factorizations.

16 = 2^4 * 1^(anything) * (every other prime)^0

If you wish to include all the factors then indeed there is only one unique factorisation.
This happens to include an infinite number of terms.
Thus, for 16 as for all numbers, there are an infinite number of factors.

If all numbers have an infinite number of factors, how do you determine which is prime?
Do you consider only the non-one factors, and if there is only one of these, the number is prime?

In that definition, one ceases to be prime, as it has NO non-one factors.

Pardon me while I get the FTA's definition up on-screen

http://www.sonoma.edu/users/w/wilsonst/Papers/finite/8/t8-5.html

[East Street Trader]

The definition can be re-stated,

"For an integer to be a prime number it must be divisible by precisely two numbers, by one and by itself."

If you ask which two numbers one is divisible by the answer is that it is not divisible by two numbers but by one number only.

So as a matter of semantics the integer one does not meet the definition.

Which is why Euclid used the word "and". (The express exclusion in the definition is otiose.)

[Japher]

Shouldn't you guys be doing something more useful like
watching paint dry

did that... watching it fade now

[DinoDoc]

Originally posted by Japher
did that... watching it fade now

Is it more entertaining?

[orange]

A number is prime when it has two factors: itself and 1. I believe 1 qualifies.

Here's the definition of a prime number from the dictionary:

A positive integer not divisible without a remainder by any positive integer other than itself and one.

Positive Integer: Yup
Not divisible without a remainder by any positive integer other than itself and one: Yup

[KrazyHorse]

There's no debate. By convention 1 is not a prime. The motivation for this is the uniqueness of prime factorization.

[Sava]

1 is prime... the definition a prime number is:

a number divisble by only itself and 1... is 1 divisible by any other number other than itself (1) or 1? no...

every number can be infinitely divided by 1, so that argument is meaningless...

EDIT: Damn you orange!

[Guest]

Lol
Who would you listen to : a dictionnary? or mathematicians on the matter.
a primer number is a technical definition in the field of math. what the dictionnary says is way less important to how mathematicians define it.
Crash Course in Ring Algebra :
Rings are sets with 2 binary operations defined on them that act sort of like addition and multiplication.
For example, the ring of integers with addition and multiplication, but also the ring of polynomials, ring of continuous fonctions etc...
all ring must have a 0, which is the neutral element with respect to addition, that is when you add 0, you get 0.
(the ring of integers has one : 0).
Although you can always define substraction in a ring by the opposite of addition, it is NOT always possible to define Division as the inverse of multiplication, because some elements of the ring may not have multiplicative inverses.
For example, if we look at integers, the multiplicative inverse of 2 is not an integer. Now the multiplicative inverse of 1 is 1 and is an integer. Elements which have multiplicative inverses are called UNITS.
For example, if you know something about matrices, you will know that some of them (those with det = 0) are not inversible, while the others are.
So those which are inversible and units and the others not.
In some rings, for example the rational numbers (the fractions of integers), all elements but 0 (which never has an inverse) are invertible. Such rings are called fields (if multiplication is commutative) or division rings.
But we are interested in rings which have units and non-units , like the ring of integers, which has 0 (which is special) 1 (which is a unit) and all the other integers (non-units).
The last important thing is this , if , in a ring, a*b= 0 implies a= 0 or b=0, then it is called a domain.
That is a ring is a domain if no two non-zero elements multiplied together makes 0.
So the ring of integers is a domain (an integral domain actually).
Now comes the part we were waiting for.
Factorization of elements in and integral domain!!
We say that an element a of a ring R is factored if it equal to a product of two or more elements of R.
If we have units in the ring(there is always at least 1 : 1, the neutral element of multiplication), we have some trivial factorization : a= a * u /u (remember that u is a unit, so as I said earlier, we are allowed to divide by u, but this is not true in general).
A factorization a = ub where u is a unit is called a TRIVIAL factorization.
in the case of integers this means that a=1*a is the only trivial factorization (since 1 is our only unit).
These factorizations are of no interest obviously.
AND NOW FINALLY :
an element p in an integral domain is called an irreducible element if it satisfies both :
(1) p not equal to 0,
(2) P NOT A UNIT
(3) If p=a*b then a or b is a unit.
All these are very important to keep on going in the theory of Domains. From here we could go on, and get to Unique Factorization, and the Euclidean Algorithm and so on.
Now in the case of integers :
(1) irreducibles are not 0
(2) irreducibles are not 1
(3) any factorization of irreducibles, is trivial, meaning the only factorization of p is p = p*1*1*1..
THERE are NO non trivial factorization.
There is a convention not to count trivial factorization.
So then the true definition of irreducibles for integers is any number,except 0 and 1, which has no non-trivial factorizations.
These are the positive and negative primes, or if we restrict ouself to positive integers, the primes.

So yeah if you use the dictionnary definition (which I believe they jsut wanted to make short and simple) 1 is a prime.

But if you want to use primes for anything except talking about aroudn a cofee table or on Apolyton , like for proving theorems and such, then 1 is NOT a prime, because, as a unit, it is already excluded.
If you want to know why the definition is that way, I could explain but then we would get in a bit heavier theory.

[Kuciwalker]

Originally posted by TheStinger
Shouldn't you guys be doing something more useful like
watching paint dry

I fail to see how this is any less useful than any other thread in the OT

[Japher]

Is it more entertaining?

Not really. It's raining now so I hope it gets better!

Is 0 prime, no it can be devided by everything!!! Even two! So, is it even? odd? What the heck is Zero!!! Actually, where is Zero?

[Kuciwalker]

Japher - see the post above mine.