Wikipedia:Reference desk/Archives/Mathematics/2009 April 28

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April 28

Łukasiewicz notation for propositional functions

Jan Łukasiewicz used C to denote implication, K to denote conjunction, A for disjunction, and E for logical equivalence, as noted at Polish_notation#Polish_notation_for_logic. Why these letters? Are they the initial letters of some relevant Polish words? If so, what words? —Dominus (talk) 04:15, 28 April 2009 (UTC)

Hmm, I always had a vague impression that the letters were based on Latin, but Polish actually makes more sense now that you mention it. Koniunkcja, alternatywa, ekwiwalencja (though równoważność appears to be the more common name), negacja, możliwość and dysjunkcja (which, strangely enough, does not mean disjunction in Polish, but Sheffer stroke) are transparent. I do not understand the source of C for implication (implikacja) and L for necessity (konieczność). — Emil J. 11:47, 28 April 2009 (UTC)

I suppose C might have come from czyni (makes), but I can't figure out the source for L, either. --CiaPan (talk) 14:56, 28 April 2009 (UTC)

Thanks. The article is probably wrong when it says that Łukasiewicz originated the use of L and M for modal operators. It is certainly wrong when it says that he originated the use of Σ and Π for quantifiers. —Dominus (talk) 15:04, 28 April 2009 (UTC)

Well, the article does not actually claim that Łukasiewicz originated all the notation, so it is not wrong. You may be right that L and M may come from a different source. But then the question remains, what is the source and what does it mean. According to modal logic#Axiomatic Systems, the usual $\Diamond$ and $\Box$ notation was already used by the founder of modern modal logic, C. I. Lewis. J. J. Zeman confirms it in the case of $\Diamond$, but he notes that $\Box$ is a later addition. Either way, M and L must have been introduced when $\Diamond$ was already in use, which seems to suggest that they indeed originated in the context of the Polish prefix notation, even though nowadays they are also used in infix notation. — Emil J. 15:42, 28 April 2009 (UTC)

Easy probability question

• Event occurs: probability 1
• Result 1 occurs: probability 1/x
• Result 2 occurs: probability (x-1)/x

What are the chances of result 1 happening if the event occurs x times? Vimescarrot (talk) 15:51, 28 April 2009 (UTC)

Sounds a bit like homework. Assuming the trials are independent, you can calculate the probability that only event 2 ever occurs using the multiplication rule, from which you can derive the result you want. You may also have a look at e to turn it into a neatly-looking approximation for large x. — Emil J. 16:08, 28 April 2009 (UTC)

And have a look at binomial distribution in case the number of occurrences of Result 1 is of interest.81.132.236.12 (talk) 16:14, 28 April 2009 (UTC)

I realised I forgot to specify "once or more", but never mind. It's not homework, it's just that...Well, this applys in computer games a lot (if the chances of this monster dropping this item are 1/100, what are the chances of getting it after kiling it 100 times?) Anyways, thanks very much for the help. Vimescarrot (talk) 18:03, 28 April 2009 (UTC)

What event occurs x times?? Could it be that you meant that their are x trials, and on each trial the probability of success is 1/x? It gets confusing when you don't use terminology in a standard way? And why do you mention "Result 2" if it has nothing to do with your question? Michael Hardy (talk) 19:33, 28 April 2009 (UTC)

I think the question was quite clear. A event occurs and can have one of two outcomes, the probability of outcome 1 is 1/x, the probability of outcome two is 1-1/x (=(x-1)/x). What is the probability of outcome one occurring at least once in x (independent) trials? The question has been answered by EmilJ, and the OP seems to be happy, so another question successfully resolved! --Tango (talk) 19:45, 28 April 2009 (UTC)

...OK, further guesses: What you meant was that "Result 2" was the complement of Result 1, i.e. to say that "Result 2" happens just means "Result 1" doesn't happen. Really, it wouldn't have hurt to say so, but even better would have been not to mention Result 2 at all. At any rate, if my guesses are right then the probability that "Result 1" never occurs in x trials is (1 − 1/x)^x. That number approaches 1/e as x grows (where e is the base of natural logarithems). So the probability that "Result 1" occurs at least once is 1 minus that. Michael Hardy (talk) 19:48, 28 April 2009 (UTC)

I didn't use standard terminology because I don't know standard terminology. Vimescarrot (talk) 21:54, 28 April 2009 (UTC)

Your terminology was fine, I don't know what Michael is complaining about. Perhaps he missed the fact that (x-1)/x=1-1/x, which makes it clear that one is the complement of the other? --Tango (talk) 11:09, 29 April 2009 (UTC)

You're mistaken, Tango. If the sum of the two probabilities is 1, that does not mean they are complements of each other. The probability of getting a "1" when rolling a die is 1/6. The probability of getting a number no more than 5 is 5/6. The sum of those two is 1. But they are not complements. And the poster used the word "event" where he probably meant "trial". Michael Hardy (talk) 01:55, 30 April 2009 (UTC)

It is a matter of common sense English that two different results cannot happen at the same time. — Emil J. 13:04, 1 May 2009 (UTC)

Putting it more simply, isnt the answer 1 minus the Xth power of (1/x)? 89.242.97.56 (talk) 14:02, 3 May 2009 (UTC)

The set of complex numbers is the largest possible set of numbers?

I've once seen a proof that claimed that the set of all complex numbers was the largest possible set of numbers that can be conceived of, but do not remember the details of that proof. Does anyone know if this is true, and if so, have a link to a proof? JIP | Talk 18:50, 28 April 2009 (UTC)

Probably you are thinking of the fundamental theorem of algebra, which says that the only proper algebraic field extension of the field of real numbers is the field of complex numbers. Similar results for systems of numbers larger than the complex numbers are described in the articles Frobenius theorem (real division algebras) and Hurwitz's theorem#Hurwitz's theorem for composition algebras. JackSchmidt (talk) 19:06, 28 April 2009 (UTC)

The definition of "number" is not fully standard. Sometimes things like non-standard real numbers are considered numbers. Transfinite cardinal and ordinal numbers are called "numbers". Sometimes things like quaternions or members of finite fields are considered "numbers".

Maybe Jack Schmidt's guess as to what you remember is right. The term "fundamental theorem of algebra" is something of a misnomer. It says you don't need to extend your set of "numbers" beyond the complex numbers in order to have solutions of all algebraic (i.e. polynomial) equations. Michael Hardy (talk) 19:36, 28 April 2009 (UTC)