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Contents 1 September 20 1.1 Words for the terms in an implication/entailment 1.2 San Juan 1.3 Wythoff's Game 1.4 Liquid Mixture Puzzle 1.5 Math problem

[edit] September 20

[edit] Words for the terms in an implication/entailment

What are the words for the terms in an entailment? For example, if , is A the "implicand" or the "implicant" or the "antecedent" of this entailment? Similarly, what's the word for B? I'm looking for something similar to addend. Having this word on hand would really help me think about the mathematical logic work I'm doing. I checked the entailment article, and asked on the talk page there, but I haven't gotten any responses. Thanks! -- Creidieki 12:43, 20 September 2006 (UTC)

I'd call them antecedent (for A) and consequent (for B). You could also use premise and conclusion. "Implicant" means literally: "that which implies" and might be used for A, while "implicand" means literally: "what is to be implied" and might be used to refer to B, but not to A. The similarity of these two words in English is confusing, and as, furthermore, entailment is not the same as implication, it is better to avoid them for use in relation to entailment. --Lambiam^Talk 15:43, 20 September 2006 (UTC)

Thank you! On a related note, are "implicant" and "implicand" the best terms to use when talking about an implication (rather than an entailment)? -- Creidieki 16:44, 20 September 2006 (UTC)

I would still use "antecedent" and "consequent", if only because of the confusing similarity between "implicant" and "implicand"; moreover, these terms are not widely used. Our Wikipedia article on material implication also uses "antecedent" and "consequent". --Lambiam^Talk 20:19, 20 September 2006 (UTC)

[edit] San Juan

The lowest temperature ever recorded in San Juan, Puerto Rico, is 60 degrees Fahrenheit. Write an inequality for T representing San Juan's recorded temperatures.

Do your own homework: if you need help with a specific part or concept of your homework, feel free to ask, but please do not post entire homework questions and expect us to give you the answers. Letting someone else do your homework makes you learn nothing in the process, nor does it allow us Wikipedians to fulfill our mission of ensuring that every person on Earth, such as you, has access to the total sum of human knowledge. -- Meni Rosenfeld (talk) 16:36, 20 September 2006 (UTC)

In the problem statement, it is the intention that T stands for any temperature (in degrees Fahrenheit) that may have been recorded in San Juan, Puerto Rico. While we do not know which temperatures have been recorded, we do know that 14 °F is not one of them. But, for all we know – if we only use what is given – it may have been 451 °F on a particularly hot day in 1898. The inequality involving T must therefore be such that if you replace T in it by 451, it becomes a true statement. That already rules out many things, such as T > 666 and T ≤ 2. A possible solution is T ≥ -459.67. But that solution does not take account of everything that is given in the problem statement. The information from the first sentence is not used at all. You want not just any inequality that is true for all recorded temperatures: you want it to be as strong as possible. In particular, you want the inequality to be such that it rules out all temperatures that cannot have been recorded, namely those that are lower than 60 °F. So if you substitute any lower value, for example 59, for T, you want the inequality to become a false statement. I hope this helps. --Lambiam^Talk 20:40, 20 September 2006 (UTC)

This is a joke right? Or do you really have problems understanding the concept of lowest and the concept of greater than? 202.168.50.40 00:23, 21 September 2006 (UTC)

The first two responses are appropriate; this third one is absolutely not. We can reasonably expect someone to read and abide by the guidelines at the top of this page, and chide them when they do not. We can offer guidance in a subject area without providing homework answers.

We want people to ask questions here, and to feel comfortable revealing their areas of ignorance in order to learn. Mocking and insulting responses will not achieve these goals. Please do not do this again. --KSmrq^T 13:16, 21 September 2006 (UTC)

[edit] Wythoff's Game

Does anyone know where I can find the solution to Wythoff's Game, with a proof? There are a few sites scattered around that claim staunchly that it's all about the Golden Mean and the Fibonacci numbers, and as far as I can tell they're right, but nobody bothers to really explain it. Black Carrot 17:13, 20 September 2006 (UTC)

Whoa! I just read the page. Rounding down to integers?! That sounds like things Ramanujan came up with. Such occult phenomena! Surprising number sequence, in any case. Do you have more link tips? —Bromskloss 20:00, 20 September 2006 (UTC)

I am sure I read a pretty full analysis of the game in John Conway's book "On Numbers and Games" - but that was nearly 30 years ago, so I might have mis-remembered ... but I still have the book somewhere, so if you are really desperate, I will try and find it :-) (unless you have access to a copy) Madmath789 20:16, 20 September 2006 (UTC)

Sure. Iirc, a complete proof is given in this book:

Csákány Béla, Diszkrét matematikai játékok. Polygon, Szeged, 2nd ed, 2005.

This is a nice easily understandable book that requires little prior knowledge. I recommend it heartily. – b_jonas 11:31, 21 September 2006 (UTC)

Searching for variations on -wythoff nim- gives quite a bit. If it helps, I can easily prove the claim at [1], which gives a vague link to Fibonacci-eque numbers. And the whole rounding-down thing comes pretty naturally (I thought of it too, just didn't have a number to plug in) from writing the numbers out in order. They aren't going by intervals of 1, or 2, or 1.5, or 1.6, but seem to be jittering around in that area. Yeah, I'd appreciate it if you could find that book. Black Carrot 13:34, 21 September 2006 (UTC)

[edit] Liquid Mixture Puzzle

A while ago I found this puzzle that I cannot solve. It states that:

• There is a tank of water, with a capacity of 100 litres.
• This tank contains 100 litres of LiquidA
• This tank has 2 pipes connected to it, one at the top and one at the bottom
• From the top pipe, LiquidB flows into the tank at 1 litre per minute
• From the bottom pipe, the contents of the tank (both LiquidA and LiquidB - assume they mix perfectly throughout the tank) both flow out at 1 litre per minute

What is the ratio of LiquidA to LiquidB in the tank after one hour?

I do not seem to be able to solve this problem using algebra and after some searching, it seems like I need to know some calculus in order to solve it (which I don't). Can someone tell me the answer please, and how to work it out? If the answer requires knowledge of calculus, can someone please point me to a good reference that is easy to understand in order to learn it (if I wait for school to teach me it, I will have to wait around 3 to 4 years - I am currently 14)? Thanks for any help you can provide.

P.S. Side question: Does anyone know why in England, you have to wait until Year 13 to do logarithms? I have had a look at them and they don't seem particularly difficult. Also, they seem quite useful, but if you not take Maths for A Level, you will not learn about. Strange... -80.229.152.246 21:06, 20 September 2006 (UTC)

The exact symbolic solution to the problem is a differential equation. You can get an approximate numerical solution using repeated algebraic computations: see numerical analysis. --Serie 21:41, 20 September 2006 (UTC)

Here is how you solve it.

(1) Change the unit of time to minute. Having a consistent unit of time helps a lot.

(2) Have two variables VolA and VolB which represent the volume of each type of liquid within the Tank

(3) Write out the equation which represents the volume of each type of liquid in the Tank.

VolA(t) = input_VolA(t) - output_VolA(t)

VolB(t) = input_VolB(t) - output_VolB(t)

 determine output_VolA(t) and output_VolB(t)

(4) input_VolA(t) = 0

(5) input_VolB(t) = int(t=T0,t=T,1,dt)

(6) output_VolA(t) = int(t=T0,t=T,ratio_A(t),dt)

(7) output_VolB(t) = int(t=T0,t=T,ratio_B(t),dt)

(8) ratio_A(t) = VolA(t)/( VolA(t) + VolB(t) )

(9) ratio_B(t) = VolB(t)/( VolA(t) + VolB(t) )

(10) Solve the differential equation.

202.168.50.40 00:30, 21 September 2006 (UTC)

Let's begin with some bounds and a crude guess. If liquid B only was removed, after 60 minutes (1 hour) the ratio would be 0:100. If liquid A only was removed, the ratio would be 60:40. That is, B must be somewhere between 0% and 60% of the total volume, and a crude guess might be 30% (a ratio of 30:70). Now let's try to be more precise.

We can easily state exactly how much of B has been added after t minutes, namely t litres. We also know that the amount of B present is this minus the amount removed. And the rate at which B is removed is the amount present divided by 100. Of course, at the beginning the amount removed, r, is zero. Thus, with a little rearranging, we wish to solve the ordinary differential equation

with initial condition

The final amount of B can be recovered as 60−r(60), the amount added minus the amount removed after 60 minutes.

Here we have a first order linear differential equation, which can be solved quite easily, as our article explains. It turns out that after 60 minutes we will have just over 45 litres of B, and a B-to-A ratio of approximately 23:28. --KSmrq^T 00:08, 22 September 2006 (UTC)

Thanks very much for the answer everyone. It looks like I'll have to wait until I do calculus (3 years and counting) before I completely understand exactly what you are saying, but I think I get the gist. Thanks. --80.229.152.246 21:21, 22 September 2006 (UTC)

[edit] Math problem

I know you guys don't do homework, but I've got this question and our teacher just told us to use trial and error or the Pythagorean triplets to solve it because he said that the algebraic solution is too difficult. But im just curious, how do you solve this algebraically?:

a) Express the volume of a cone to the surface area of the same cone (includes the bottom)

b) Using the answer from part (a), make the ratio of volume to surface area equal to 1. What are the values of the radius, height and slant height?

The answer is radius=6, height=8 and slant height =10. But can someone post the solutions to part b? Jamesino 22:47, 20 September 2006 (UTC)

This requires that we work with dimensionless units. Equating the formulas for volume and surface area for a cone with radius r and height h gives us:

Divide both sides by πr/3:

Subtract 3r from both sides, and square:

Multiply out and bring everything to one side:

Divide by h, and then solve for h:

.

Any value of r > 3 will give you now a value for h such that together they are a solution. For example, r = 21 gives h = 49/8, which results in a slant height of 175/8 and a volume and surface area of 7203π/8. If you want both r and h to be whole numbers, note that h − 6 = 54 / (r² − 9) must also be an integer, so r² − 9 must be an integer divisor of 54, leaving for r² only the possibilities 10, 11, 12, 15, 18, 27, 36, and 63. The only square is 36, giving r = 6 and h = 8. --Lambiam^Talk 00:11, 21 September 2006 (UTC)

Niggling note: The step "Divide both sides by πr/3" should be accompanied by a test that . This test would show that r=0 is a solution (but not an interesting one). -- Fuzzyeric 15:47, 21 September 2006 (UTC)

Good point. It would also be necessary, because of the fifth step, to test h=0, but that's not a solution. Black Carrot 16:56, 21 September 2006 (UTC)

There is no need to test for r ≠ 0. In fact, how could we "test" for that? – we are trying to find the value of r here. What I'm doing is deriving a sufficient condition for the equation to hold. By reducing something of the form r·X = 0 to X = 0, I discarded the solution r = 0. However, since a cone is constrained to have a positive radius, it is harmless to drop that case, which I did tacitly. Likewise, I did not explicitly mention my throwing overboard the algebraic solutions with negative h. --Lambiam^Talk 19:53, 21 September 2006 (UTC)

Because any step that reads "divide both sides by X" had better ensure that X is not zero, or the step is invalid. Division by zero is undefined. Also, since you did not check for this, you did not discover that r=0 is also a solution for all values of h. -- Fuzzyeric 04:06, 22 September 2006 (UTC)

You do not understand. Suppose you have to solve X³−X² = 0 for X. Clearly, that follows from X²−X = 0, which has solutions X = 0 and X = 1. So I have divided here by X, replacing the equation to be solved by a simpler equation all of whose solutions will be valid solutions to the original equation. That is the meaning of "sufficient condition". One of the solutions of X²−X = 0 is X = 0, so I have as it were divided by 0, but that does not invalidate the step. In a logic formula, XY = 0 ← Y = 0 (where "←" stands for "follows from"). That is also true if X happens to be equal to 0. Of course I saw that algebraically r = 0 was a solution for all h; I just assumed the side condition (constraint on the solutions) r > 0 and h > 0 to be so obvious that it could be dealt with implicitly. --Lambiam^Talk 15:14, 22 September 2006 (UTC)

Would it somehow be clearer to point out that not checking for division by zero is inherently wrong, that it is incorrect method, and that it will permit you to prove that 1 = 2. It's a common enough error that there's an entire talk page about it at Talk:Division_by_zero. The examples I'm quoting are present at Talk:Division_by_zero#Wow.2C_2.3D1.3F -- Fuzzyeric 04:15, 23 September 2006 (UTC)

Oh, and your example is an excellent one for demonstrating why it is wrong to not check. You justify the step by "dividing by x". (Repeating this step will eliminate the solution x=0 completely, which is incorrect.) However, the valid implication is "by the definition of integral domain". Integral domains are distinguished by the property that you can divide by sides of an equality by any nonzero element you want. R[x,y] is an integral domain under addition and multiplication and therefore in your demonstration, it is required to verify that no divisor can be zero. -- Fuzzyeric 04:30, 23 September 2006 (UTC)

I do not justify the step because I do not take that step. I do not derive a NECESSARY condition but a SUFFICIENT condition. Sorry for the shouting, but I said this twice already and you just seem to ignore it. --Lambiam^Talk 15:45, 23 September 2006 (UTC)

Just in case Lambiam's explanation is still unclear: The step he takes is , which is obviously legitimate (via multiplication, not division, by x, which is always valid). -- Meni Rosenfeld (talk) 15:57, 23 September 2006 (UTC)

So which part of Lambiam's "Divide both sides by πr/3: " is a multiplication? -- Fuzzyeric 22:34, 23 September 2006 (UTC)

Is this a homework question? What about ? --Lambiam^Talk 23:04, 23 September 2006 (UTC)

However, the demonstration you wrote was ! -- Fuzzyeric 04:43, 24 September 2006 (UTC)

This is turning into an Achilles–tortoise dialogue. I did not "write" that. I showed the operation by which I obtained one equation from the other. I could instead have just presented the solution, for which it is easy enough to verify it satisfies the original equation, but I also wanted to sketch the steps that brought me there. Maybe the presentation was confusing, but it is you who made the mental leap to an entailment. --Lambiam^Talk 07:59, 24 September 2006 (UTC)

Thanks alot =) Jamesino 21:04, 21 September 2006 (UTC)