Wikipedia:Reference desk/Archives/Mathematics/2007 October 5

= October 5 =

Logic
I have to prove the following tautology:

$$ (P \to Q) \land (Q \to R) \to (P \to R) \equiv T $$

Here's my logic:

$$ (\neg P \lor Q) \land (\neg Q \lor R) \to (\neg P \lor R) $$

$$ \neg (\neg P \lor Q) \lor \neg(\neg Q \lor R) \lor (\neg P \lor R)$$

$$ (P \land \neg Q) \lor (Q \land \neg R) \lor (\neg P \lor R) $$

Can't see how to get out of that last statement. I'm thinking that maybe I cocked it up at the second line where I flipped the AND to OR. But $$ \neg (P \land Q) \equiv \neg P \lor \neg Q $$, right?


 * Wrong, $$ \neg (P \land Q) \equiv \neg P \lor \neg Q $$ — see De Morgan's laws.


 * Woops, typo on my part. That's what I meant. Fixed.


 * Anyway, open last parentheses, reorder terms and try to reduce $$ (P \land \neg Q) \lor \neg P$$ and $$(Q \land \neg R) \lor R$$
 * CiaPan 09:15, 5 October 2007 (UTC)
 * (contd.)
 * $$\begin{align}

(P\land\neg Q)\lor\neg P & \equiv (P\land\neg Q)\lor\neg P\lor\neg P \\ & \equiv (P\land\neg Q)\lor(\neg P\land 1)\lor(\neg P\land 1) \\ & \equiv (P\land\neg Q)\lor(\neg P\land (\neg Q\lor Q))\lor(\neg P\land 1) \\ & \equiv (P\land\neg Q)\lor(\neg P\land \neg Q)\lor(\neg P\land Q)\lor(\neg P\land 1) \\ & \equiv ((P\lor\neg P)\land\neg Q)\lor(\neg P\land (Q\lor 1)) \\ & \equiv (1\land\neg Q)\lor(\neg P\land 1) \\ & \equiv \neg Q\lor\neg P \end{align} $$
 * Reduce the other part $$(Q \land \neg R) \lor R$$ in similar way, and you get a logical sum with some $$X\dots\lor\neg X\dots$$ terms, which will give you, by law of excluded middle, the final result of true, i.e. the tautology proof. --CiaPan 15:35, 8 October 2007 (UTC)


 * To list all combinations of truth values for P, Q, and R will require a table of only eight entries. If the expression is true for all combinations, that's a proof.
 * {| class="wikitable" style="background-color:white;text-align:center"

! P || Q || R || P→Q || Q→R || P→R || &Sigma; ! F || F || F ! T ! F || F || T ! T ! F || T || F ! T !  || ⋮ || ! ⋮ ! T || T || T ! T
 * T || T || T
 * T || T || T
 * T || F || T
 * T || T || T
 * }
 * Or, perhaps it's a proof; that depends on the rules. Formal study of logic introduces many variations, including some in which we must distinguish between P→Q and &not;P&or;Q. See Classical logic and its links. --KSmrqT 11:56, 5 October 2007 (UTC)

Mass-Energy Equivalence
In the famous equation e = mc2, this article says that e measured in Joules; m measured in kilograms, and c measured in meters per second. My question is: does this apply to any kind of matter? Would the energy in a kg of feathers be the same as the energy in a kg of lead?

David A. Johns (e-mail address removed) Doghall 11:35, 5 October 2007 (UTC)


 * Yes, and it is independent of the specific units of measurement used to express the physical quantities involved. A more appropriate place for this question would have been Reference desk/Science. --Lambiam 12:09, 5 October 2007 (UTC)


 * (edit conflict) Both of Einstein's relativity theories assert that we should detect no difference, as do most theories. However, the universe is not obliged to respect our theories, so this is an experimental question. I believe tests of this sort have been made, and none have been accepted as falsifying the theory. But mathematicians do proofs, not science experiments, so why are you asking us? This question belongs on the science reference desk, not here. --KSmrqT 12:22, 5 October 2007 (UTC)

MATHEMATICS (Interest homework)
Moses invested a certain sum of money in a housing bank for one year. For the first three months he earned 18% per annum simple interest and for the remaining time he earned 16% per annum compounded interest quaterly.The total interest earned for one year period was 5616.Find the sum invested.Please i beg your help!Georgekalusanga 14:50, 5 October 2007 (UTC)


 * I can never quite figure out what they mean by "interest", is that the nominal yield not including compounding or effective annual yield ? Let's assume it's the nominal yield not including compounding.  Therefore, the quarterly rates based on 18% and 16% are 18/4 and 16/4 or 4.5% and 4%.  That gives us:

P × 1.045 × 1.04 × 1.04 × 1.04 = 5616


 * Or:

P × 1.045 × 1.043 = 5616


 * On the other hand, if we assume "interest" means effective annual yeild (including compounding), then the quarterly rates are 1.181/4 and 1.161/4. They might even mean to mix and match the two, with quarterly interest rates of 18/4 and 1.161/4.  This actually gives us the answer closest to a whole number, so that's my guess.  That would give us the formula:

P × 1.045 × 1.161/4 × 1.161/4 × 1.161/4 = 5616


 * Or:

P × 1.045 × 1.163/4 = 5616


 * You do the math to finish it at let us know if you have any trouble. StuRat 15:27, 5 October 2007 (UTC)

opposite of disjoint sets
Is there a terms for a pair of distinct sets that share at least one element? 71.29.100.133 17:28, 5 October 2007 (UTC)


 * I'd call them "overlapping sets", a term that is used, for example. in Distance, Connection form, and Seifert–van Kampen theorem. --Lambiam 17:38, 5 October 2007 (UTC)


 * I think the term I've heard most often is just 'sets with nonempty intersection'. Algebraist 18:11, 5 October 2007 (UTC)

Parabola/Catenary
What is the difference between the two? Btw please don't just give the differences in their equations, I would like an explanation about what is physically different. Thanks 172.159.49.31 17:54, 5 October 2007 (UTC)


 * They're totally different curves? I'm not sure what you want here. If you mean 'how do they look different to the eye', the most obvious difference is apparent on a zoomed-out plot: the catenary is massively steeper for large x (exponential rather than quadratic). Algebraist 18:15, 5 October 2007 (UTC)


 * Maybe our articles on parabola and catenary can be of help. Both articles mention relations to the physical world and not just equations. PrimeHunter 20:21, 5 October 2007 (UTC)


 * I would especially recommend the Hanging With Galileo page mentioned in the catenary article. And why are people now asking mathematicians about physics? If we were clever enough to deal with the "real world" we would be engineers, with endless job opportunities and unbounded income potential. ;-) --KSmrqT 21:24, 5 October 2007 (UTC)


 * What's wrong with merchant banking and management consultancy? Algebraist 13:21, 6 October 2007 (UTC)


 * I other words as the link "Kmsqr" supplies says above - the parabola typically arises when a flat heavy thing is supported by a wire (eg humber bridge), whereas the cantenary arises when the majority of the mass is in the wire. eg overhead line
 * [[Image:PylonHorizon.jpg|200px|catenaries]][[Image:Humber Bridge.png|200px|parabola]]


 * Just out of curiosity can anyone tell the difference between a graph of either - just by looking??87.102.23.214 14:11, 6 October 2007 (UTC)

The difference is that the wire's graph has no bounds for it's derivative (because it's the derivative of a Parabola), versus the Bridge has a bounds of +/-n where is a finite positive constant, because it's the derivative of the upper half of a Hyberbola) em I right?? —Preceding unsigned comment added by 69.54.140.201 (talk) 22:25, 6 October 2007 (UTC)
 * Did you read the Hanging With Galileo page? --Lambiam 23:59, 6 October 2007 (UTC)

1 equals to -1

 * $$\frac{1^n}{x^n}=\left(\frac{1}{x}\right)^n$$
 * Let x == -1
 * $$\frac{1^n}{(-1)^n}=\left(\frac{1}{-1}\right)^n\,$$
 * $$\frac{1^n}{(-1)^n}= -1^n$$
 * Let n == 1/2
 * $$\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{-1}$$
 * $$\frac{1}{i}=i$$
 * Multiply both side by i
 * $$1=-1\,$$
 * $$\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{-1}$$
 * $$\frac{1}{i}=i$$
 * Multiply both side by i
 * $$1=-1\,$$
 * Multiply both side by i
 * $$1=-1\,$$
 * $$1=-1\,$$
 * $$1=-1\,$$

211.28.129.8 22:12, 5 October 2007 (UTC)


 * See invalid proof. --Lambiam 22:22, 5 October 2007 (UTC)


 * Such is the peril of asserting that $$i = \sqrt{-1}$$. It's better to say that i is a solution to the equation x2 = –1. Strad 23:37, 5 October 2007 (UTC)


 * Not to mention that when you take non-integer roots in the complex numbers, you really have to consider the impact of multiple solutions. Confusing Manifestation 07:16, 6 October 2007 (UTC)

The flaw in your proof is assuming that a square root and the power of 1/2 are the same thing, I did a proof Sophomore year of high school in Algebra 2 that 4throot(x^2) and (4throot(x))^2 are not equal for negative x. In your case, the fact that square roots have two solutions versus 1/2 power having only one solution has caused you to make an error. If you don't fully understand how this is the case, try looking up De'Moivre's formula. Hope that clarifys things. —Preceding unsigned comment added by 69.54.140.201 (talk) 09:56, 6 October 2007 (UTC)


 * To be more specific, let's define $$\forall x \in \mathbb{C}\quad R(x):=\{y\in\mathbb{C}|y^2=x\}$$. Then


 * $$\forall x \in \mathbb{C}\setminus\{0\}\quad\frac{\sqrt{1}}{\sqrt{x}}=\sqrt{\frac{1}{x}}$$


 * doesn't make sense since you do not explain how you choose the roots. In particular, as your example demonstrates, it is not true in general that an arbitrary choice for each root would satisfy the equation. What you can say however is:


 * $$\forall x \in \mathbb{C}\setminus\{0\}\quad\frac{R(1)}{R(x)}=R\left(\frac{1}{x}\right)$$


 * In particular:


 * $$\frac{R(1)}{R(-1)}=R(-1)$$


 * $$R(1)={R(-1)}^2$$


 * $$\{-1;1\}={\{-i;i\}}^2$$


 * $$\{-1;1\}=\{-1;1\}$$


 * And there is no contradiction. Ceroklis 09:59, 6 October 2007 (UTC)
 * Hm? $$(-i)^2=i^2=-1$$, so it seems that your last set should be the singleton $$\{1\}$$.  Or are you using some non-element-wise definition of exponentiation?  Of course, it makes no sense to divide two sets element-wise anyway: sets have no order, so you can permute one of them and change the answer. --Tardis 16:40, 9 October 2007 (UTC)
 * Denoting $$AB = \{ab|a \in A,b \in B\}$$ is fairly standard, and likewise for $$\frac{A}{B}$$. $$\{-i,i\}^2$$ was meant as a shorthand for $$\{-i,i\}\{-i,i\}$$ which is indeed $$\{-1,1\}$$. -- Meni Rosenfeld (talk) 17:12, 9 October 2007 (UTC)
 * Ah; my mistake. I didn't realize that a cartesian product (but with the infix operator used, rather than "pair") was the intended understanding.  Of course, it's a bit odd to say that $$A^2\,\not\!:=\{a^2|a\in A\}$$, but it's comprehensible.  Thanks.  --Tardis 16:01, 11 October 2007 (UTC)
 * The product $$AB=A*B$$ is not at all the same thing as the cartesian product $$A \times B$$. For instance
 * $$\{1;2\} \times \{2;4\}=\{(1,2);(1,4);(2,2);(2,4)\}$$
 * $$\{1;2\}*\{2;4\}=\{2;4;8\}$$
 * And there is in general no bijection between the two. Ceroklis 19:02, 11 October 2007 (UTC)
 * Tardis did say, with the infix operator used, rather than "pair", the infix operator being in this case multiplication. -- Meni Rosenfeld (talk) 12:38, 12 October 2007 (UTC)