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Ramo
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Austin, Texas, USA
Oct 1999
time: 23:32
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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 .1k 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 1x 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?
Last edited by Ramo on 04-12-2003 at 11:00
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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...
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Enigma_Nova
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quote:
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/wilso...ite/8/t8-5.html
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East Street Trader
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London, United Kingdom
Jun 2000
time: 05:32
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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.)
Last edited by East Street Trader on 04-12-2003 at 22:37
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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.
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0 is not a prime, by my definition above
0 is even. it is also divisble by 3,
In facti it is divisble by every number.
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Peter Triggs
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Gone Fishin, Canada
Jan 2000
time: 05:32
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LulThyme and KrazyHorse are correct. There is nothing to argue about.
quote:
A slightly less illuminating but mathematically correct reason is noted by Tietze (1965, p. 2), who states "Why is the number 1 made an exception? This is a problem that schoolboys often argue about, but since it is a question of definition, it is not arguable."
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Well orange it depends on the context
usually if you want to exclude, you woudl say Strictly positive, usually positive includes 0 (as does negative btw)
Because 0 is the neutral element with respect to addition, so it makes the positive integers into a monoid. (another technical term)
The thing about math (at least this kind of math) si you cant argue it.
Primes are DEFINED as excluding units, that means 1.
That means that if you ask anybody who actually works with primes, there is no ARGUMENT about if 1 is a prime, you could say there is no reason either, it is just defined that way.
Just like if I DEFINE f(x)= 2x.
You cant argue that its not 2x.
Its the definition.
Now if you want to know why they define 1 as not being a prime, thats another story, with part of it on the first page already, its because it makes proving theorems easier, and makes them nicer (very important in mathematics).
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You have to understand that 1 is very special.
If we look at positive integers (thus including 0).
0 is very special because it is the additive neutral element, it is the multiplicative absorber (or whatever in english), and you can't divide anything by it.
Those 3 properties go hand in hand and 0 works like this in every ring.
Now 1 is also special, although only for multiplication (it has no special addition property).
First it is the neutral element.
And second because it is a neutral element it is a unit (although sometimes in other rings there are other units like in the ring of matrices)
This is the property that makes 1 special in our thread.
1 is a unit, so has a multiplicative inverse, so every number can be divided by it, and even more, since it is the neutral element, you get the same number.
So the very important fact is that
1*a=a for all a.
There is this very important theorem called Unique factorization. It means that there is only one way to factorize any strictly positive integer such that none of the factors can be further factorize.
So take any pos integer but 0, say 9.
it can be factorized as 3*3. none of these can be factorized further if we dont allow 1*3 as a factorization and so it is unique.
If we allow it, we can get 1*3*3 or 1*1*3*3 and so on
The key is here : the factorization process does not stop anymore, it is endless.
If 1 is not a prime and we count1*a as trivial, it is obvious that the factorization process will stop and nothing will be able to be further reduced.
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I thinkn adding this might help.
If we consider 1 a prime, no computer crashes, no algorithm stop working, no sky falls anywhere.
All math keeps working, we just have to add a lot of exceptions all over the place. Make exceptions in proofs and algorithms for 1. All primes except 1. etc...
So in a way it is an aesthetic choice.
Some of you (orange  ) might think that makes you right, but in fact it removes your last argument.
Even if you look at it from a "choice" point of you, the easiest and simplest choice is to exclude 1 from the list of prime, which I think even Euclid understood.
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04-12-2003 10:02
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