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May 2[edit]

Solving Conditional Statements[edit]

I was given a deductive reasoning conditional proof in class from my textbook, but I can't seem to find a way to solve it. Could you help me find the starting assumed antecedent to get the ball rolling?

Premises:

$P\to Q\!$
$\neg P\to S\!$
$R\to \neg S\!$
$\neg Q\!$

Conclusion:

$\neg R\!$

Thanks. 66.76.125.76 01:50, 2 May 2007 (UTC)Caitlin C.[reply]

(Reformatted problem statement for clarity.) --KSmrq^T 07:54, 2 May 2007 (UTC)[reply]

Reformatted premises and conclusion using LaTex.--Kirby♥time 19:26, 3 May 2007 (UTC)[reply]

~Q seems like what you should start with. --Spoon! 03:56, 2 May 2007 (UTC)[reply]

Start by noting that since P-->Q, ~P-->~Q because if a statement is true, its contrapositive is also true. Foxjwill 04:10, 2 May 2007 (UTC)[reply]

"If it is Friday, then I am happy"; therefore, "since it is not Friday, then I am not happy."? I don't think so. As the linked article clearly states, the contrapositive of P→Q is ¬Q→¬P. Thus, "I am not happy, therefore it cannot be Friday."

In classical logic, the only way to contradict an implication P→Q is for P to be true and Q to be false. The contrapositive is logically equivalent, because it states that if Q is false then P cannot be true. It may help to note that P→Q is equivalent to (¬P)∨Q; thus ¬Q→¬P is equivalent to (¬¬Q)∨(¬P), in which we are allowed to cancel the double negation and to use commutativity. --KSmrq^T 04:54, 2 May 2007 (UTC)[reply]

In deductive reasoning, the conclusion of a conditional proof is always a formula of the form A → B. Since the conclusion you are asked to reach does not have that form, it is not clear why you should use that rule at all here. Does the assignment or exercise state that CP must be used? --Lambiam ^Talk 06:25, 2 May 2007 (UTC)[reply]

Probably the proper rule is modus tollens, not contrapositive. They do look very equivalent, but modus tollens is the one you would here. I.e., ~Q ^ (P->Q) => ~P (modus tollens). Going in the other direction is modus ponens. –Pakman044 07:19, 2 May 2007 (UTC)[reply]

The distinction makes little difference in classical logic. For, given P→Q we may assert the contrapositive, ¬Q→¬P. Then, given ¬Q we may deduce ¬P using modus ponens. Incidentally, I have been using the qualifying phrase "in classical logic" for a reason; other equally valid variations of logic exist, somewhat in analogy to non-Euclidean geometry. For example, in intuitionistic logic we are allowed to replace (¬A)∨B by A→B, but not the reverse; and we can replace A by ¬¬A, but again not the reverse. --KSmrq^T 07:54, 2 May 2007 (UTC)[reply]

I have a feeling my book just misplaced this problem into the conditional proof section instead of the direct proof section. As weird as that is, it seems like a common problem for my book. I'll talk to my teacher about it (not like he would actually know) and then I'll probably end up solving it the direct proof way. Seeing as I pretty much learned how to do conditional proofs yesterday by reading *one* example, I may just be completely off, but whatever. 66.76.125.76 11:46, 2 May 2007 (UTC)Caitlin C.[reply]

Well, you could rewrite the conclusion ~R as R->⊥, assume R, and deduce ⊥, but I personally don't see the point. Once you have equivalence of statement with contrapositive it's trivial however you do it. 131.111.8.96 14:14, 2 May 2007 (UTC)[reply]

01) P->Q
02) ~P->S
03) R->~S
04) ~Q
-
05) ~P (01,04)
06) S (02,05)
07) ~~S
08) ~R (03,06)--droptone 05:06, 3 May 2007 (UTC)[reply]

Sub-question[edit]

I found this question and it shows what I believe to be 'logic' and I would much like to one day learn this all properly - am I correct in assuming you can only formally study logic as part of a degree in philosophy? I'm a big fan of philosophy as well, but don't see myself finishing a degree in philosophy, and last I checked logic wasn't in the first year. Rfwoolf 19:25, 6 May 2007 (UTC)[reply]

Different professions, and their corresponding academia, use the word logic with different contexts, and other words like ethics that are related only in a particular context. I have studied logic in the context of computer logic which can be found both in the software set theory and binary math sense and the design of computer hardware where there is a mathematics to circuit design. User:AlMac|^(talk) 22:12, 6 May 2007 (UTC)[reply]

And when I took a course on logic it was run by the mathematics department at my university, and covered topics such as Turing machines, Peano arithmetic and Godel's incompleteness theorem, so it depends on what kind of logic you want to learn. Confusing Manifestation 22:47, 6 May 2007 (UTC)[reply]

Dyscalculia[edit]

How do I know if I have dyscalculia? I have the following symptoms:

Reliance on 'counting-on' strategies, e.g. using fingers, rather than any more efficient mental arithmetic strategies.

Difficulty with everyday tasks like checking change and reading analog clocks.

Inability to comprehend financial planning or budgeting, sometimes even at a basic level; for example, estimating the cost of the items in a shopping basket or balancing a checkbook.

Difficulty with times-tables, mental arithmetic, etc.

May do fairly well in subjects such as science and geometry, which require logic rather than formulas, until a higher level requiring calculations is needed.

Difficulty with conceptualizing time and judging the passing of time.

Problems differentiating between left and right.

Having a poor sense of direction (i.e. north, south, east, and west) and may also have trouble still even with a compass.

Difficulty navigating or mentally "turning" the map to face the current direction rather than the common North=Top usage.

Having difficulty mentally estimating the measurement of an object or distance (e.g., whether something is 10 or 20 feet away).

Inability to grasp and remember mathematical concepts, rules, formulae, and sequences.

Difficulty keeping score during games.

Also, how do I get tested for it and get it officially recognised (proof) that I have this disability? --Candy-Panda 08:45, 2 May 2007 (UTC)[reply]

This is not a mathematics question. I would expect your school to have a school psychologist who either can test you, or else can refer you to a good testing centre. I think you should speak to your parents or legal guardians, and they should contact the school's management. More important than "official recognition", in this stage, may be a proper diagnosis and remedial exercises that develop your abilities as much as possible. Although targeted training has limits, it can make a difference. 09:26, 2 May 2007 (UTC)

Probability[edit]

A round tabla has six seats, and six people are seated at it. Lets call the people a, b, c, d, e and f. If everyone are assigned random places, what is the possibility of e being placed next to f? 213.167.126.201 12:18, 2 May 2007 (UTC)[reply]

1-4C2/5C2

Otherwise known as 2/5. Once e is seated, there are five places f can go, all are equally likely, and exactly 2 are next to e. Algebraist 13:40, 2 May 2007 (UTC)[reply]

I understand your explanation but I don't get this 1-4C2/5C2-Czmtzc 19:19, 3 May 2007 (UTC)[reply]

Algebraist's 4C2 is a lazy way of writing ₄C₂, denoting the binomial coefficient four-choose-two. I prefer 4?2, but the most common form is awkward in-line:

{4 \choose 2}.

It is also tragically indistinguishable from a column matrix,

{\begin{pmatrix}4\\2\end{pmatrix}},

so we must hope that context saves us. The question mark notation suffers neither of those drawbacks, and extends nicely to multinomials, as in ?⟨2,3,1⟩ for the coefficient of x²y³z in the expansion of (x+y+z)⁶. --KSmrq^T 21:15, 3 May 2007 (UTC)[reply]

Covariance?!![edit]

Hello all. I've just been working through various problems for my statistics class and I've come across something which I just can't get my head around. I know its not a difficult question at all, and I've the answers final result, its just not explained correctly. Any help explaining the question would be appreciated: Ok, so I'm looking to get the correlation between two variables x and y. I know to get the correlation, I need the standard deviation of the variables which I have, and I also need the covariance of the two variables. Here is where I'm having the problem!

So the variables are: x = 1 3 4 5 7 y = 5 9 7 1 9

I've worked out the mean of x and y to be 4 and 6.2 respectively. Then the standard deviation to be 2 and 2.99 respectively. My answer states the covariance is 4/5 or 0.8, but I just can't figure out how to get that. Can anyone help me? --NotoriousTF 13:37, 2 May 2007 (UTC)[reply]

You know the definition of covariance, right? If so, I can't see the problem. The answer you give is correct, certainly. 131.111.8.102 13:48, 2 May 2007 (UTC)[reply]

Sorry, I know the answer is correct, but I'm just unsure how to calculate the answer. --NotoriousTF 13:53, 2 May 2007 (UTC)[reply]

I still don't see the problem. By definition, cov(x,y)=mean(x*y)-mean(x)*mean(y). You already have mean(x)=4, mean(y)=6.2, so all you nead is to calculate mean(x*y). Continuing your table, we have

x*y = 5 27 28 5 63

with mean 25.6. Algebraist 14:06, 2 May 2007 (UTC)[reply]

Oh I had just confused myself with the definition of covariance. Thank you for clearing it all up I've got it now. --NotoriousTF 14:34, 2 May 2007 (UTC)[reply]