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TheDeath
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posted March 22, 2009 08:43 PM |
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Regarding the logic problem beforehand...
sorry for bringing this back up, I've been thinking about this for a while, and I found what you are talking about. It's called Material Conditional but notice that it has many flaws, for example (in that link):
Quote:
So, although a material conditional from a contradiction is always true, in natural language, "If there are three hydrogen atoms in H2O then the government will lose the next election" is interpreted as false by most speakers, since assertions from chemistry are considered irrelevant conditions for proposing political consequences.
Also: Indicative conditional which "solves" this but isn't clearly defined. Here's my opinion on it:
in my opinion, regardless of condition (even if it is false), obtaining an answer to the statement's validity should always require evaluation of the 'consequence'. Of course you don't have to evaluate the consequence if you do not need the answer "is this statement true?"
so in the example with the government above, I would actually say it is unknown -- unless, that is, we can "test" the preposition when H20 has 3 hydrogen atoms (somehow weirdly), that could just as well be impossible since after all, a lot of things are unknown (and some are irrelevant anyway).
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Ecoris
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posted March 23, 2009 10:30 AM |
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The example you quoted has an emprirical nature. Reason alone is not enough to decide whether it is true or not. In mathematics there is always a connection between the concepts involved. There is no flaw in the use of such statements in mathematics.
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TheDeath
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posted March 23, 2009 05:56 PM |
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Edited by TheDeath at 17:57, 23 Mar 2009.
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Moon made of cheese and superman are different than that?
Or "1=0" (false condition) and anything else are different than that?
To be honest I think the following rule would solve them all:
A statement is only true if you verified the cases. Of course you don't need to evaluate anything if the condition is false, but it doesn't say anything about the statement's validity, unless you verify the statement when the condition is true -- if it ever is 
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dimis
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posted March 23, 2009 06:39 PM |
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Edited by dimis at 18:39, 23 Mar 2009.
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To me it seems that it is still not clear why I made the post about "gates". If you want to discuss more about it, you have to "attack" the arguments made there. Since these arguments form our intuition and in the end give the definitions for the various connectives. And these truth tables give you the answer. For example you don't have to evaluate the "then ... " part when the condition is false, because regardless of the outcome the statement is characterized by a "T". So, is it that you don't agree with these "T's"? And what do you mean by verifying cases? And then you also say that "it doesn't say anything about the statement's validity, unless you verify the statement when the condition is true". What is that supposed to mean? If the condition is true (satisfied), then you have to evaluate the "then" part to decide whether the statement is true or false. What's wrong with that? And by the way, this is what is said about statement's validity in these cases.
My opinion is try to attack the arguments made on gates. I believe this will clear everything out. For example, in the "A <=> B" (equivalence relation), the only output value which is susceptible is when both A and B are false and I have a "T" in the output. But, if I had an "F" right there in the output, then "A <=> B", would be identical with the "A and B". Is this really what we mean when we say that A is equivalent to B? No way! And by setting this to True, then you get the rest. Really. Forget about wikipedia, or anyone else. Try to counter those arguments. If you accept the truth tables, then everything should follow naturally.
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zamfir
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posted March 24, 2009 05:13 PM |
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Let's leave this for a while and return to true maths. Let's see if you solve this problem (ok, easy one for start).
Let's say logx(y) means the base x logarithm with the argument y(I wish HC had an equation editor ).
We have a, b and c natural numbers, a,b,c>0.
If loga(b),logb(c) and logc(a) are natural numbers, find a, b and c.
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dimis
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posted March 25, 2009 05:45 AM |
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Edited by dimis at 05:52, 25 Mar 2009.
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Sorry, but we are talking about true math. 
And you should also give that a, b, c are different from 1. Not that it matters that much, but anyway ...
edit: Similarly, a, b, c don't have to be natural numbers, but any a, b, c > 0 with a, b, c different from 1 is enough.
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dimis
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posted March 26, 2009 10:00 PM |
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Edited by dimis at 22:05, 26 Mar 2009.
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logarithms problem
I feel like I have to answer this.
Say
loga(b) = x = natural number
logb(c) = y = natural number
logc(a) = z = natural number
Then,
a^x = b
b^y = c
c^z = a
and by combining the first two we get
(a^x)^y = c ==> a^(x*y) = c
Substituting this last one on the c^z = a we get
a^(x*y*z) = a
Since a, b, c are different from 1 we can take the logarithms in base a on the last one, giving us (edit: by monotonicity of the logarithm if this is not clear why /end of edit)
x*y*z = 1.
This implies that none of the x, y, z is zero. Moreover, since x, y, z are natural numbers, this means x, y, z are greater than or equal to 1. If at least one of them is strictly more than 1, then the product is strictly more than 1. Hence all of them are equal to 1; i.e.
x = y = z = 1.
Substituting back we get
a^1 = b ==> a = b
b^1 = c ==> b = c
c^1 = a ==> c = a
which implies that any triple a = b = c is a solution to the problem.
Of course this triple has to respect some properties posed by the problem; i.e. all a, b, c > 0 and all of them different from 1.
It was a really nice problem Zamfir, and I think instructive on solving it. Give us your next one.
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Ecoris
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posted March 26, 2009 10:45 PM |
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Edited by Ecoris at 22:47, 26 Mar 2009.
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New problem: Which part of dimis' solution above incomplete/slightly incorrect?
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dimis
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posted March 26, 2009 11:33 PM |
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Edited by dimis at 23:44, 26 Mar 2009.
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I fail to see what it is. Meanwhile you made me think about it, and I came up with another solution which leads to the same answer. lol Here it is:
Clearly none of them is zero (given + you can not define the logarithm).
One of them is 1: This is problematic on defining the logarithm in that base. That's why I take all of them bigger than 1.
a, b, c >= 2:
Think about the first logarithm loga(b) = x. What can x be? Either 0, 1, or at least 2.
* If it is 0, then, b = 1, contradiction.
* If it is at least 2, then b > a (since a^(something) increases since a >= 2). In this case consider the logarithm logb(c) = y. Then c = b^y. Again y is not zero, since this implies c = 1. But any other value (natural number) for y, implies that c is at least as big as b. So we also have c >= b > a (i.e. c > a). Now look at the last logarithm logc(a) = z. Again z can not be zero, and any other value (natural number) for z implies that a >= c. This is a contradiction. Hence x can not be >= 2.
This leaves only one case for inspection; namely x = 1, which implies a = b. This implies that both loga(c) and logc(a) are natural numbers. Again excluding the case where they are equal to 0, this forces them to be equal to 1 so that they hold simultaneously. And this is ok, since it implies a = b = c >= 2.
edit: I know that I am using the fact that a, b, c are natural numbers this time, but it is not really needed. The proof only becomes bigger if this is not assumed.
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posted April 13, 2009 09:23 PM |
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*bump*
I was sorry to see this thread go into disuse. So, let's have another fun problem, shall we?
A race car is traveling around a square track, with each side measuring exactly 1 mile. Along the first two legs, the car travels at 60 miles per hour, then along the third leg it travels at 30 miles per hour. In order to average 60 miles per hour around the entire route, how fast must the driver drive along the final leg of the course?
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zamfir
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posted April 13, 2009 09:32 PM |
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Edited by zamfir at 21:38, 13 Apr 2009.
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@Dimis: I'm glad you enjoyed my problem. Let's see another one (I give problems I solved, so if anything is not clear you can ask me):
Solve the ecuation: 2^(x^2+x) + log2 (x)=2^(x+1) (log2 = base 2 logarithm).
EDIT: BTW, your solution is correct. a=b=c, as long as they don't break the restrictions.
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dimis
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posted April 13, 2009 09:40 PM |
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Edited by dimis at 05:09, 24 Apr 2009.
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My threads tend to remind chicks.
Either many people come up with ideas simultaneously, or there is a period of ... inactivity
From now on try to have some order on the problems guys; otherwise it's gonna be a mess and we have no referee. 
Thanks for both problems. I'll let others try first.
Best
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TheDeath
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posted April 13, 2009 09:50 PM |
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Edited by TheDeath at 21:55, 13 Apr 2009.
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Quote:
A race car is traveling around a square track, with each side measuring exactly 1 mile. Along the first two legs, the car travels at 60 miles per hour, then along the third leg it travels at 30 miles per hour. In order to average 60 miles per hour around the entire route, how fast must the driver drive along the final leg of the course?
120mph?
EDIT: Sorry to bring this back in again but I thought I should better clarify myself:Quote:
If the condition is true (satisfied), then you have to evaluate the "then" part to decide whether the statement is true or false. What's wrong with that? And by the way, this is what is said about statement's validity in these cases.
There's nothing wrong with that at all. What I was talking about was when the condition is false -- that is, if it is false, you DO NOT have to evaluate the "then" part and I agree.
However this says nothing about the validity of the statement. For that, in my viewpoint, is unknown at this point. I.e "if the moon is made of cheese, then everyone will be transformed into a bear" you have NO WAY of knowing whether this STATEMENT (not the condition!) is true or not, unless let's say, you teleport into an alternative Universe where the Moon IS made of cheese and then everyone transforms into a bear -- in which case, it was true.
But for common sense purposes, IMO it is unknown. Sure it may be twisted logic, but at least it solves the "natural problem" with 'inconsistencies' (I gave a link before). False condition means unknown statement validity -- you do not have to know the statement validity to ignore the "then" part or do anything with it at all anyway, that's the condition's job.
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posted April 13, 2009 10:08 PM |
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Quote:
120mph?
Nope.
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TheDeath
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posted April 13, 2009 10:11 PM |
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Oh sorry my bad, I meant 150 (made a silly typo when I calculated it ). If that isn't true either, at least I'd like to know the solution, or should I explain it and you'll figure out the error?
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posted April 13, 2009 10:14 PM |
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That's not the answer either, but explain away.
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TheDeath
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posted April 13, 2009 10:18 PM |
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Well, to keep it simple, I took another 'imagined' car that goes with 60mph average. In 5 minutes, it moves 5 miles.
The real one, excluding the last leg, does 3 miles in 4 minutes. So to catch up with the average one it would have to go with 120mph, to get 2 miles in 1 minute.
Oops it seems I'm back at 120, sorry for rushing and confusing this before
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posted April 13, 2009 11:47 PM |
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Well that's certainly a convoluted way to try to solve the problem, but ultimately, no, it didn't give you the right answer.
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TheDeath
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posted April 14, 2009 12:01 AM |
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Right and I think I know the problem, it is because it is measured as the average in 5 miles, but I read that it has to be measured over 4 miles only! (i.e just one course)
So he would have to make 4 miles in 4 minutes...? (generally speaking, x miles in x minutes) That means infinite speed?
I know I'm missing something. Can anyone else help?
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posted April 14, 2009 12:50 AM |
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Corribus:
I don't know. It depends on if it's the speed or the velocity that we're looking for.
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This thread is pages long: 
