Supposedly, drinking some of the yeast helps you to avoid a hangover, according Charlie Papazian. I haven't made any wine; but, I have produced some very nice beers and meads. I don't bother with killing the yeast or filtering it out, but I do rack the stuff a few times before bottling to remove most of the spent yeast.
Re:Ethanol
on
Hacking Vodka
·
· Score: 4, Informative
No, by percentage of effect, its the impurities that give you the hangover.
Care to back that up? According to this ethanol causes dehydration, electrolyte imbalance and low blood sugar. Further, it states that pure ethanol can cause hangovers, and that it is unknown whether ethanol or the impurities have the greater effect.
That's why they are shooting for 100% pure, in theory no hangover...
Who is they? Care to back this up? You do realize that it's impossible to get 100% pure ethanol, right? Although one could probably produce 99.999% pure ethanol, as soon as the bottle was opened, it would begin absorbing water from the atmosphere until it reached the azeotropic composition, about 95% purity, if I remember correctly.
Re:Does this work for Rubbing Alcohol?
on
Hacking Vodka
·
· Score: 3, Funny
I don't know; but, methanol seems to taste better after processing with one of those filters. I... hey, waitaminute... someone's been fiddling with the brightness on my monitor. Who turned out the lights???
Re:Common knowledge?
on
Hacking Vodka
·
· Score: 2, Funny
Where does vomit rank on this scale? At the bottom under "all of the above"?
It is probably good for a first reference --- provided the entries give bibliographic information. You cannot become an expert on a topic by reading an encyclopedia, and that includes Britannica. There is no substitute for true research... unless you just want a new topic to discuss for your next party.
and this is the attitude that is creating a third-world economy in the US and shaping much of our foreign policy; "I don't care if something is made well, i just want it cheap and in bulk so i can stuff more crap into my life."
You sound like a rich apple loving fanboy. I take it then that only rich people should stuff more (expensive and well-made) crap into their lives. Everyone else can just do without.
The positive integers are countable because f(x)=x is a bijection from the positive integers to the positive integers. Therefore, they are not uncountable.
As for the Cantor argument you present, I take it you mean the entries in the list are sequences of integers with only finitely many nonzero entries --- for example, (...,0,0,3,2,7) would represent the number 327. Now, these sequences are in bijective correspondence with the positive integers. Here, diagonalization will produce a "number" that will have infinitely many nonzero entries, and therefore does not correspond to an integer since there are no integers with infinitely many digits. This "number" was not in the list, but that doesn't violate the assumption that all naturals were enumerated, since the result doesn't belong in the list anyway.
A little notation is in order: A bijectionf from a set A to a set B is a function that is onto, meaning every element b of B has a pre-image a in A, such that f(a) = b; and is one-to-one, meaning that distinct elements of A get mapped to distinct elements of B.
Now, I see from several posts "blah blah blah is/is not infinite by definition". But, the poster(s) supply no definition. In mathematics, the language is very precise. There are (at least) two working definitions of finite, which are equivalent provided we assume the Axiom of Choice:
1) A set is finite if there exists a bijection from the set to the set of counting numbers {1,...,n} for some counting number n.
2) A set is finite if there exists no bijection from the set to a proper subset of itself.
Now, a set is said to be infinite if it is not finite. That is, for an infinite set, there does exist a bijection from the set to a proper subset of itself.
Therefore, all questions regarding whether some set is finite or infinite boil down to the existence or nonexistence of certain bijections.
Now, it turns out that bijections exist between:
The set of Natural numbers and the set of Integers.
The set of Integers and the set of Rationals.
But, no bijection exists between any three of these sets and sets of the form {1,2,...,n} where n is a counting number. So, the sets of Natural numbers, Integers, and Rationals are indeed infinite.
Now, we need another definition. Two sets are said to be cardinally equivalent if there exists a bijection between them. Furthermore, if a set is finite, then the number of elements of the set is called the cardinality of the set. So, the cardinality of the set {8,7,-4,175,111} is 5. So, the finite cardinals are exactly the natural or counting numbers, along with zero.
Now, by a famous proof of Cantor, which I will not repeat here, there exists no bijection from the set of Natural numbers to the set of Real numbers. To fully appreciate the meaning of this statement, we need to talk about ordinal numbers --- truly the foundation of all number systems.
I will not get into the technical details of what makes a set an ordinal. Suffice it to say that the empty set {} is an ordinal. When treated as an ordinal number, we will call the empty set 0, or zero. Now, given any ordinal s, which is a set, we define its successor to be the set s union {s}, which is itself an ordinal. So, the successor to 0 is 0 union {0} or simply {0} which we will call 1, the successor to 1 is the set 1 union {1}, or {0} union {1} or, {0,1} which we will call 2. Likewise, 3={0,1,2}, 4={0,1,2,3}, and so on.
A few nice things happen here, 1 is a subset of 2 is a subset of 3, etc. so the usual order is built in. Also, the ordinal 4 cannot be put in bijection with 3 or 2 or 1 or 0, thus is finite, by the second definition of finite above. However, one of the axioms of set theory, namely the Axiom of Infinity, says that an infinite set exists that includes 0 and the successor of every element of the set. This axiom is what essentially "bounds" infinity, by regarding an infinite collection as a completed thing.
But, just because every successor belongs to this infinite set (the set of natural numbers or finite ordinals) does not mean every ordinal belongs to it. Indeed, this infinite set satisfies the definition of ordinal number, we call it omega. Now, omega has a successor, called "omega + 1" which is omega union {omega} per the previous definition of successor. "Omega + 1" has a successor, "Omega + 2", and so on. None of these ordinals can be put in bijection with any finite ordinal. But they are cardinally equivalent to omega itself. The process continues.
Now, if one assumes the Axiom of Choice, which in turn implies that the real numbers can be well-ordered, then there exists an ordinal that can be put
There are not a series of larger and smaller infinities - by definition they are NOT infinite!
Here is where the issue lies. What is your working definition of infinity?
If you don't like the use of the word "infinity", use "transfinite cardinals" instead. They are indeed numbers --- cardinal numbers. They are well-ordered by less than. Each transfinite cardinal is greater than any finite cardinal. They can be added and multiplied. The set of all cardinals does not exist, but there are sets of distinct transfinite cardinals of any cardinality you like.
At the very least, 9 years in the pen is 9 years this guy won't be spamming. That sounds like a pretty effective way to stop the behavior --- at least for 9 years.
As for murder --- if anyone kills a family member of mine, and if I know who it is, then I will retaliate with extreme prejudice. Second chances and morality be damned.
I didn't say anything about revenge. The ultimate objective of punishment is to STOP THE BEHAVIOR, right? And if fines don't STOP THE BEHAVIOR then harsher punishment is warranted.
Your phraseology is a little ambiguous. You are not talking about Z_n except for the case where n is p^1 for some prime p. Also, the SET of powers of a prime is not finite, hence not a Galois Field. More general Galois Fields are extensions fields of Z_p for prime p, and it can be shown that they have p^n elements for some positive integer n. However, a discussion of the details of these fields is a little more involved.
I mean the standard lemniscate belonging to the extended real number system commonly used in Calculus and Real Analysis. I was not referring to transfinite cardinal arithmetic.
I don't know how interesting this will be to you, but for each positive integer n there is a finite ring called Z_n or the integers mod n that consists of the integers 0 through n-1 inclusive. Addition, subtraction, and multiplication work the same as in the set of integers, EXCEPT all results are modulo n.
For example, in Z_8, 15*3 = 45 = 5 since the remainder of 45 upon division by the modulus 8 is 5. Strange things can happen in some of these rings. For example, in Z_8 we have 2*4 = 0. Here, 2 and 4 are called zero divisors. Neither 2 nor 4 have reciprocals in this ring, so we cannot form the quotient 3/2 for instance.
Here is the interesting part having to do with primes. If, and only if, p is prime, then Z_p is actually a FIELD, meaning that every nonzero element has a reciprocal element. This implies that there can be no zero divisors in Z_p, and that the ring is closed under division (except for dividing by zero, of course). I believe there are some computer applications of finite fields.
The proof that the sum of all the POSITIVE integers equals the product of all the primes is quite trivial: Their common value is infinity.
Re:I haven't taken anything like this...
on
IT Literacy Test
·
· Score: 1
Because we all know a degree from Brown is the same as one from Mississippi State, right? And someone who got a 3.5 from Yale must surely not be as talented as someone who has a 3.9 from Syracuse.
Well, if that's the case, then why not just drop the pretense and say on the application that those without degrees from Brown or Yale need not apply?
How are we to compare them if not for these tests?
As I implied earlier, accreditation is either valid, in which case additional testing is redundant, or it isn't, in which case it should be abandoned.
I have serious doubts about the validity of these tests. What precisely is being measured?
Furthering the field of testing. As we all know, testing provides an invaluable service to mankind: more food on the table, more peace treaties, more architectural masterpieces, more rocket ships to Alpha-Centauri. Tests, how I miss them. I knew I should have continued with a Ph.D. --- not for the money, not for the self-improvement, but for the EXAMS! God, I, I,... I'm going into withdrawal.
Re:I haven't taken anything like this...
on
IT Literacy Test
·
· Score: 1
I never have understood the idea behind testing graduates with degrees from universities with accredited programs. If the program is accredited, then why the need for the student to prove themselves with yet another exam? This would seem to imply that the universities have no faith in the evaluation process.
You've never been to the South if you haven't heard of Piggly Wiggly.
Here is a better one: Ding Dongs vs. King Dons. I grew up in Arkansas where those little chocolate snack cakes were called Ding Dongs. Later I moved to Colorado only to discover that the same cakes were called King Dons. What the Hell???
Slashdot Mirror
Supposedly, drinking some of the yeast helps you to avoid a hangover, according Charlie Papazian. I haven't made any wine; but, I have produced some very nice beers and meads. I don't bother with killing the yeast or filtering it out, but I do rack the stuff a few times before bottling to remove most of the spent yeast.
No, by percentage of effect, its the impurities that give you the hangover.
Care to back that up? According to this ethanol causes dehydration, electrolyte imbalance and low blood sugar. Further, it states that pure ethanol can cause hangovers, and that it is unknown whether ethanol or the impurities have the greater effect.
That's why they are shooting for 100% pure, in theory no hangover...
Who is they? Care to back this up? You do realize that it's impossible to get 100% pure ethanol, right? Although one could probably produce 99.999% pure ethanol, as soon as the bottle was opened, it would begin absorbing water from the atmosphere until it reached the azeotropic composition, about 95% purity, if I remember correctly.
I don't know; but, methanol seems to taste better after processing with one of those filters. I ... hey, waitaminute ... someone's been fiddling with the brightness on my monitor. Who turned out the lights???
Where does vomit rank on this scale? At the bottom under "all of the above"?
Since when are the sewer pipes under high pressure? The sewerometer reads 130,000 psi (poops / square inch).
Matters not to me one way or the other. I quit watching T.V. years ago.
It is probably good for a first reference --- provided the entries give bibliographic information. You cannot become an expert on a topic by reading an encyclopedia, and that includes Britannica. There is no substitute for true research ... unless you just want a new topic to discuss for your next party.
and this is the attitude that is creating a third-world economy in the US and shaping much of our foreign policy; "I don't care if something is made well, i just want it cheap and in bulk so i can stuff more crap into my life."
You sound like a rich apple loving fanboy. I take it then that only rich people should stuff more (expensive and well-made) crap into their lives. Everyone else can just do without.
The positive integers are countable because f(x)=x is a bijection from the positive integers to the positive integers. Therefore, they are not uncountable.
As for the Cantor argument you present, I take it you mean the entries in the list are sequences of integers with only finitely many nonzero entries --- for example, (...,0,0,3,2,7) would represent the number 327. Now, these sequences are in bijective correspondence with the positive integers. Here, diagonalization will produce a "number" that will have infinitely many nonzero entries, and therefore does not correspond to an integer since there are no integers with infinitely many digits. This "number" was not in the list, but that doesn't violate the assumption that all naturals were enumerated, since the result doesn't belong in the list anyway.
It is truly amazing how little mathematics people understand. The extended reals are not the same as the hyperreals. I said what I meant. Try again.
Okay, here we go (I love this topic).
A little notation is in order: A bijection f from a set A to a set B is a function that is onto, meaning every element b of B has a pre-image a in A, such that f(a) = b; and is one-to-one, meaning that distinct elements of A get mapped to distinct elements of B.
Now, I see from several posts "blah blah blah is/is not infinite by definition". But, the poster(s) supply no definition. In mathematics, the language is very precise. There are (at least) two working definitions of finite, which are equivalent provided we assume the Axiom of Choice:
1) A set is finite if there exists a bijection from the set to the set of counting numbers {1,...,n} for some counting number n.
2) A set is finite if there exists no bijection from the set to a proper subset of itself.
Now, a set is said to be infinite if it is not finite. That is, for an infinite set, there does exist a bijection from the set to a proper subset of itself.
Therefore, all questions regarding whether some set is finite or infinite boil down to the existence or nonexistence of certain bijections.
Now, it turns out that bijections exist between:
The set of Natural numbers and the set of Integers.
The set of Integers and the set of Rationals.
But, no bijection exists between any three of these sets and sets of the form {1,2,...,n} where n is a counting number. So, the sets of Natural numbers, Integers, and Rationals are indeed infinite.
Now, we need another definition. Two sets are said to be cardinally equivalent if there exists a bijection between them. Furthermore, if a set is finite, then the number of elements of the set is called the cardinality of the set. So, the cardinality of the set {8,7,-4,175,111} is 5. So, the finite cardinals are exactly the natural or counting numbers, along with zero.
Now, by a famous proof of Cantor, which I will not repeat here, there exists no bijection from the set of Natural numbers to the set of Real numbers. To fully appreciate the meaning of this statement, we need to talk about ordinal numbers --- truly the foundation of all number systems.
I will not get into the technical details of what makes a set an ordinal. Suffice it to say that the empty set {} is an ordinal. When treated as an ordinal number, we will call the empty set 0, or zero. Now, given any ordinal s, which is a set, we define its successor to be the set s union {s}, which is itself an ordinal. So, the successor to 0 is 0 union {0} or simply {0} which we will call 1, the successor to 1 is the set 1 union {1}, or {0} union {1} or, {0,1} which we will call 2. Likewise, 3={0,1,2}, 4={0,1,2,3}, and so on.
A few nice things happen here, 1 is a subset of 2 is a subset of 3, etc. so the usual order is built in. Also, the ordinal 4 cannot be put in bijection with 3 or 2 or 1 or 0, thus is finite, by the second definition of finite above. However, one of the axioms of set theory, namely the Axiom of Infinity, says that an infinite set exists that includes 0 and the successor of every element of the set. This axiom is what essentially "bounds" infinity, by regarding an infinite collection as a completed thing.
But, just because every successor belongs to this infinite set (the set of natural numbers or finite ordinals) does not mean every ordinal belongs to it. Indeed, this infinite set satisfies the definition of ordinal number, we call it omega. Now, omega has a successor, called "omega + 1" which is omega union {omega} per the previous definition of successor. "Omega + 1" has a successor, "Omega + 2", and so on. None of these ordinals can be put in bijection with any finite ordinal. But they are cardinally equivalent to omega itself. The process continues.
Now, if one assumes the Axiom of Choice, which in turn implies that the real numbers can be well-ordered, then there exists an ordinal that can be put
There are not a series of larger and smaller infinities - by definition they are NOT infinite!
Here is where the issue lies. What is your working definition of infinity?
If you don't like the use of the word "infinity", use "transfinite cardinals" instead. They are indeed numbers --- cardinal numbers. They are well-ordered by less than. Each transfinite cardinal is greater than any finite cardinal. They can be added and multiplied. The set of all cardinals does not exist, but there are sets of distinct transfinite cardinals of any cardinality you like.
At the very least, 9 years in the pen is 9 years this guy won't be spamming. That sounds like a pretty effective way to stop the behavior --- at least for 9 years.
As for murder --- if anyone kills a family member of mine, and if I know who it is, then I will retaliate with extreme prejudice. Second chances and morality be damned.
I didn't say anything about revenge. The ultimate objective of punishment is to STOP THE BEHAVIOR, right? And if fines don't STOP THE BEHAVIOR then harsher punishment is warranted.
Because he'll do it again. As Steve said, fines aren't going to do the trick.
Your phraseology is a little ambiguous. You are not talking about Z_n except for the case where n is p^1 for some prime p. Also, the SET of powers of a prime is not finite, hence not a Galois Field. More general Galois Fields are extensions fields of Z_p for prime p, and it can be shown that they have p^n elements for some positive integer n. However, a discussion of the details of these fields is a little more involved.
I mean the standard lemniscate belonging to the extended real number system commonly used in Calculus and Real Analysis. I was not referring to transfinite cardinal arithmetic.
I don't know how interesting this will be to you, but for each positive integer n there is a finite ring called Z_n or the integers mod n that consists of the integers 0 through n-1 inclusive. Addition, subtraction, and multiplication work the same as in the set of integers, EXCEPT all results are modulo n.
For example, in Z_8, 15*3 = 45 = 5 since the remainder of 45 upon division by the modulus 8 is 5. Strange things can happen in some of these rings. For example, in Z_8 we have 2*4 = 0. Here, 2 and 4 are called zero divisors. Neither 2 nor 4 have reciprocals in this ring, so we cannot form the quotient 3/2 for instance.
Here is the interesting part having to do with primes. If, and only if, p is prime, then Z_p is actually a FIELD, meaning that every nonzero element has a reciprocal element. This implies that there can be no zero divisors in Z_p, and that the ring is closed under division (except for dividing by zero, of course). I believe there are some computer applications of finite fields.
The proof that the sum of all the POSITIVE integers equals the product of all the primes is quite trivial: Their common value is infinity.
Because we all know a degree from Brown is the same as one from Mississippi State, right? And someone who got a 3.5 from Yale must surely not be as talented as someone who has a 3.9 from Syracuse.
Well, if that's the case, then why not just drop the pretense and say on the application that those without degrees from Brown or Yale need not apply?
How are we to compare them if not for these tests?
As I implied earlier, accreditation is either valid, in which case additional testing is redundant, or it isn't, in which case it should be abandoned.
I have serious doubts about the validity of these tests. What precisely is being measured?
Furthering the field of testing. As we all know, testing provides an invaluable service to mankind: more food on the table, more peace treaties, more architectural masterpieces, more rocket ships to Alpha-Centauri. Tests, how I miss them. I knew I should have continued with a Ph.D. --- not for the money, not for the self-improvement, but for the EXAMS! God, I, I, ... I'm going into withdrawal.
I never have understood the idea behind testing graduates with degrees from universities with accredited programs. If the program is accredited, then why the need for the student to prove themselves with yet another exam? This would seem to imply that the universities have no faith in the evaluation process.
You've never been to the South if you haven't heard of Piggly Wiggly.
Here is a better one: Ding Dongs vs. King Dons. I grew up in Arkansas where those little chocolate snack cakes were called Ding Dongs. Later I moved to Colorado only to discover that the same cakes were called King Dons. What the Hell???
$500+??? Damnation, that's expensive. I can't imagine how long it takes the bag lady to save up for one of those.
implementations be compliant with the technical documentation?!?! I hear Bugs Bunny: "Ha ha ha, what a moroon! Ha ha ha!!!"