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pibbur who
Guest
THEN I WON'T ASK YOU, OK?
Eh… Sorry about that.
Interesting. Mine is a combination of the Egyptian and the Greek style.
pibbur who still won't ask the eye who doesn't look like an eye anything. Even kisses. Or hugs. The wife's coming home today after spending two days in Copenhagen visiting graveyards.
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Eye
Guest
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pibbur who
Guest
She's working at the graveyard authority (don't know if that's the right word) here in Bergen. Administration, not digging.
pibbur who likes the new avatar of the eye, but who admits she can choose any avatar she wants to.
pibbur who likes the new avatar of the eye, but who admits she can choose any avatar she wants to.
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pibbur who
Guest
Kangawoo season!!!
Pibbbur who wonders when it's koawa season.
PS. Did you know that you can configuwe Googwe to use Ewmew Fudd as wanguage? Along with Klingon, Pirate and Bork! Bork! Bork! Sadly not Sindarinf, Quenya or Blackspeech. PS.
Ripper
Зичу Вам успіхів
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Yes, the other day someone asked what the Watch was worth, and I attempted to show that its value is undetermined, since the answer must be x divided by the membership price (zero). The Watch seemed unmoved by my theorem.
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Case (1)
Let be x <> 0
Lets assume a definite number y exist, that
x/0 = y
then y*0 must be x.
But y*0 is always 0. x = y*0 = 0.
A contradiction to our precondition x <> 0.
Case (2)
Let be x = 0
Lets assume a definite number y exist, that
x/0 = 0/0 = y
Then y*0 = 0.
But this is true fo every y.
A contradiction to our precondition that y is a definite number.
Ergo:
y= x/0 is not defined
Let be x <> 0
Lets assume a definite number y exist, that
x/0 = y
then y*0 must be x.
But y*0 is always 0. x = y*0 = 0.
A contradiction to our precondition x <> 0.
Case (2)
Let be x = 0
Lets assume a definite number y exist, that
x/0 = 0/0 = y
Then y*0 = 0.
But this is true fo every y.
A contradiction to our precondition that y is a definite number.
Ergo:
y= x/0 is not defined
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pibbur who
Guest
And for the question whether x/0 is infinity or not:
We cold use limits and demonstrate that y/x->infinity as x->0. But this assumes that x >0. Because if x<0 we get y/x->- infinty as x->0. Therefore we cannot say that y/0 is infinity, instead x/y approaches (we can make |x/y| as large as we want) + or - inifnity depending on whether x> or x<0.
pibbur who has an upper limit to the number of quarks and electrons in his body.
We cold use limits and demonstrate that y/x->infinity as x->0. But this assumes that x >0. Because if x<0 we get y/x->- infinty as x->0. Therefore we cannot say that y/0 is infinity, instead x/y approaches (we can make |x/y| as large as we want) + or - inifnity depending on whether x> or x<0.
pibbur who has an upper limit to the number of quarks and electrons in his body.
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pibbur who
Guest
Now for a quick question: Why do we define 1 as not prime? It surely isn't divisible by any integer value other than itself.
pibbur who knows the answer (and he's not talking about 42).
pibbur who knows the answer (and he's not talking about 42).
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pibbur who
Guest
Nitpicking: x/0 is undefined, not undetermined. There is a subtle difference between those two terms.
Assume 1/0=y exists. Then we have 1=0*y. There is no value of y that multiplied by 0 equals 1. Therefore x/0 is undefined.
On the other hand, using a similar argument for 0/0=y. Then as HiddenX pointed out 0=0*y, which holds for any value of y. So 0/0 is undetermined.
pibbur who thinks he's both defined and determined.
Ripper
Зичу Вам успіхів
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Infinity is a fairly troublesome concept in itself, and is also not well-defined, as it means slightly different things in different cases. Some infinities can be shown to be larger than others, for example.
A number of philosophers and mathematicians are not happy about this sort of thing, but most people just shut up and calculate.
A number of philosophers and mathematicians are not happy about this sort of thing, but most people just shut up and calculate.
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pibbur who
Guest
Yep, there is the question of whether an infinite set is countable or not. In a countable set, every element can be associated with one integer. The set of integers is of course countable. So is the set of all even numbers. Or numbers divisible by 7. And the set of rational numbers (fractions). All of these are of the same cardinality (one way of saying that their "equally large").
But the set of real numbers (pi, e,square root of 2…) can be shown to be not countable and therefore of a higher cardinality.
But it is to some degree a question of how we define things.
Interesting.
pibbur who insists that the set P={pibbur} is countable. He will however not be held accountable for what he writes.
PS.
For those confused by number sets, here are the important sets, each of them includes the above mentioned sets.
N is the set of natural numbers, integers > 0.
Z is the set of postive and negative integers and 0. The set includes N
Q is the set of rational numbers, positve or negative fractions. The set includes Z
R is the set of real numbers, whch includes the rational numbers and numbers like pi, e and so on. Every number that can be represented on a number line belongs to R. Other numbers do not belong to R, those belong to the next set, the complec numbers.
C is the complex numbes, of the form a+b*i, where a and b are real numbers and i (for imaginary number) is the square root of -1. i is not a real number, since no real number satisfies x^2=-1. C contains R, we get R by setting b=0.
DS.
Edit: Corrected a nasty typo. Complex numbers are of the form a+b*i, not a=b*i, which doesn't make sense. *shivers*
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pibbur who
Guest
Well, nobody came up with an answer, so here it is.
We could (and we did for a long time) consider 1 a prime. But if we do, we're making things a bit more complicated than necessary.
One example, the fundamental theorem of arithmetic which says: Every integer >1 is either a prime, or a unique product of primes. As the name suggests, this is a very important theorem in mathematics.
We have for instance 42=7*3*2. No other product of primes give this number (sorting the factors differently doesn't count). Now, if 1 is prime we would also get:
42=7*3*2*1
42=7*3*2*1*1
and so on. This invalidates the uniqueness part of the theorem.
We could solve this by reformulating the theorem: Every integer >1 is either a prime, or a unique product of primes larger than 1. But there are many other theorems where we would have to do the same. So, the most practical solution is to exclude 1 from the set of primes. After all 1 is a very special number with fundamental properties not shared by primes >=2.
Among those we have:
- a*1=a, which actually is one of the fundamental axioms (multiplicative identity axiom) defining the set of integers (and the rational, the real and the complex).
- a*b=1, which is one of the axioms defining the rational numbers (the reals and complex as well), the existence of the multiplicative inverse.
Now for another question about primes. 2,3,5,7,11,13,17,19,23,29,31,37 are all primes, they come up quite often in the example (the primes in integers< 40). Now, is there a sequence of 1 000 000 consecutive numbers where none of them is a prime? If so, how to prove it?
pibbur who doesn't think "which number has the unique prime product 2*3*7?" is the question.
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1000000!+2 ... 1000000!+1000000
gives you 1000000-1 numbers that are not prime.
by definition n! is divisible by all numbers from 1...n
and
n!+2 is divisible by 2
n!+3 is divisible by 3
...
n!+n is divisible by n
gives you 1000000-1 numbers that are not prime.
by definition n! is divisible by all numbers from 1...n
and
n!+2 is divisible by 2
n!+3 is divisible by 3
...
n!+n is divisible by n
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pibbur who
Guest
Exactly! (not the faculty)
This also means that there is an infinite sequence of consecutive numbers where none is prime. And of course, being countable, this sequence is just as large as all the integers.
pibbur who, being divisible by for instance b (giving piur), is not prime, although being the only one, is the prime of pibburs.
This also means that there is an infinite sequence of consecutive numbers where none is prime. And of course, being countable, this sequence is just as large as all the integers.
pibbur who, being divisible by for instance b (giving piur), is not prime, although being the only one, is the prime of pibburs.
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Unless both a and b are 1.
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