And that you should consider euros to be a thing, and cents to be another thing, and not as cents being fractions of an euro.
Yes (as opposed to euros being an amount).
James, I agree with that; I made that same point earlier. But that applies to scientific inductions, i.e. taking a series of data points and extrapolating a general theory that fits them. It doesn’t apply to deductive methods, including mathematical induction (so it doesn’t apply to the reasoning used in the Blue Eye problem or the MI portion of the Master Logician problem).
And the point supports my SR argument: the fact that there are an infinite number of equations that fit any given set of points means that there is no certain argument that generalizes from a set of data points to find a non-given data point. In other words, if that were what the logician were expected to do, the problem would be impossible. And since it is a given that the problem is not impossible…
Your still not getting it. You first choose a theory concerning the puzzle. Then you find that the theory fits the puzzle. Then you declare that the theory is the answer merely because it fit the puzzle. What I have been telling you is that many theories might fit the puzzle. You have to prove that yours is the ONLY one, else yours might not be the one that the master is using. The video was expressing that just because something fits within given certain limited knowledge, doesn’t mean that it is the true answer. The puzzle requires that you prove your theory to be the only option.
Another effort to clarify/verify something on that puzzle:
[tab]if the proper entitlement is x,
e is the received euros, and
c is the received cents, then
2x + 5 = e+c
And if the received euros and received cents were confused then the proper entitlement is,
x = e/100 + c*100
That seems to be the stated situation. But is that the intent?[/tab]
Good luck!
Or should I give the whole solution?
Arminius,
[tab]
No. Just answer my question.
I was not asking about the solution. I was asking if the equations that I gave (in blue) represent the situation that you are trying to express. If not, why not?[/tab]
We’ve had this portion of the conversation before. The last time you made this point, I brought up the example of the Pythagorean Theorem, which has many proofs, none of which depend on or are threatened by the existence of any other.
If the solution is by way of a deductive proof, then you don’t have to prove that your solution is the only option. If it is deductively provable that, e.g., the blue-eyed islanders will know their eye color on day N and no sooner, then no other solution needs to be considered. Even if other syllogisms exist, they will arrive at the same conclusion.
I can’t believe that you are still saying that. I told you why your example is nonsense. It does NOT APPLY to what we are talking about. We know that the deduction works. That is merely the theory that you picked (“MY theory works!! My theory works!! My theory works!!!”). NOW you have to prove that the master is using THAT SAME ONE!!! But apparently that is beyond your grasp, so I’ll once again leave it alone. You are still incorrect for the exact same reason as before, merely completely blind to it, it seems.
I wish you success!
[tab]I think your next post will contain the right solution, James. 
[/tab]
Here’s my point, James. If the deduction works, then we don’t need to prove that everyone’s using it. That’s how deductions work.
So your argument here is at best question begging: If it’s a deduction, if it’s sound, then there’s nothing else to prove. Your argument here goes something like, “it isn’t a sound deduction, so you need to prove that everyone uses the theorem, so the deduction is flawed, so it isn’t a sound deduction.” That’s not a proper argument.
At worst, you’re just calling a syllogism a “theory” and then refusing to credit it with everything that follows from a valid logical deduction. That’s not a proper argument either.
That was a misquote of me. And you are wrong either way. You have at least two people who have to be “in sync” with the bells. If they are not using the same theory (making a different enabling assumption), they will possibly not be in sync.
The issue, as from the beginning, is that the premise assumption must be the same for everyone, else it is like saying because Pythagorean theorem works, it doesn’t matter how many sides the shape has.
Your THEORY is that your syllogism applies (not whether your syllogism is valid in itself). The master might have a better theory using a different enabling assumption and syllogism. You have to prove that the master cannot be doing that, because he controls who leaves on which bell.
[tab]No because those equations do not work out to sensible values for e and c. You end up with fractions of cents and also that;
Received back = e + c = 2.55102040816327
which obviously cannot be right And that is what has been holding us both up.[/tab]
Those equations you mentioned work, but note: they are merely abstract examples and not the solution for my concrete example. I hope you know that. You asked me to answer your question, then I answered your question. Now I hope you do not confuse my answer with the complete solution of the said task. Note: I merely answered your question.
[tab]As i said several times: You need to have both “e” and “c” on both sides of the equation, and then you have to find out which number (amount) the only correct one for the example is. Please note: You have both euros and cents, and your basis should be cents (just for the sake of convenience, because if you used euros as basis, then you would have to change the number “5” in your equation). Your equations work. You do not have the right numbers, James. In my concrete example are merely two whole numbers for “e” and “c” possible.[/tab]
Good luck!
No, they don’t and that is why I asked if that is what you really meant.[tab]Given those equations:
x = e/100 + 100c
2x + 5 = e + c
2(e/100 + 100c) + 5 = e + c
e/100 + 100c + 2.5 = e/2 + c/2
e(1/100 - 1/2) + 2.5 = c(1/2 - 1/100)
e(-49/50) + 2.5 = c(49/50)
2.5 - e(49/50) = c(49/50)
e(49/50) + c(49/50) = 2.5
e + c = 2.5(50/49) = 2.55102040816327
e is the number of euros in your hand and c is the number of cent coins in your hand, two integers. Yet they do not calculate out to be integers given the original equations.[/tab]
You used 100 twice and you should only use it once. Which one you use (/100 or 100*) depends on if the answer is in euros or cents.
James, please read what I wrote in my last post again:
[tab]What has been holding us both up is more the fact that you did not take my advice. For example this:
That must be a whole number.
[/tab]
This objection is overcome by the stipulations about how good a logician the participants are. To turn again the 2 Blue Eye problem as the simplest case of the logic, if there are 2 blues on the island, it seems clear that two perfect logicians, for whom it is common knowledge that they are perfect logicians, will know their eye color on the second day. There is no reliance on a everyone being “in sync” or using the same “theory”, there’s is pure deductive logic about what the islanders know and how they must behave and how they must deduce from their knowledge and behavior.
Returning to the case of the Master Logician, I take the same level of logical perfection to be a given, as the Master will enforce accurate deduction on the other logicians. It’s not whim or fancy that he’s enforcing, but what the logicians must deduce based on what they know and how they must behave and how they must deduce from their knowledge and behavior (or have such deduction forced on them by the master). The effect is the same.
Is that what you’re taking issue with here? I admit the premise is implicit; a more-clear statement of the problem would be explicit in the way the Blue Eyes problem is.
No… “e” is merely a number of objects received and so is “c”, although different objects.
The statement was that those two numbers got reversed: the e number was supposed to be the c number and vsvrsa.
But when translating those numbers into money values, the e number was supposed to be the number of cents, thus
e/100 = amount of cents (proper)
And also the c number was supposed to be the number of euros, thus
c*100 = amount of euros (proper)
So “e/100 + c*100” should be the amount of proper returned change (“entitlement”).
But if that was true, it would mean that the received NUMBER of objects, “e + c” had to be 2.55102040816327, which is obviously not possible.
Thus there is a miscommunication going on, as I said in the beginning.
The stipulation that the logicians were perfect demanded that they were in sync as a premise. This puzzle doesn’t have that stipulation. But even if it did, the puzzle would be unsolvable (or at least by you).
Well, that was not a given, but like I said, even if it was…
You didn’t solve the Blued problem either, so referring it doesn’t get you anywhere.
The fundamental problem is that you must PROVE that no other possible means for solving the problem can exist. And I have already provided you with several ways equal to yours (you merely deny that yours is just as good). One of those ways directly defeats your assumption by allowing you to make that assumption while also proving a means for everyone to leave by the second bell.
Again, again, and again, you merely keep saying the same thing over and over, that you think your assumed premise is the only possible valid assumption. You haven’t proven that. And repeating it over and over does not prove anything other than inference that there actually isn’t the actual proof that you need.
Please don’t say no when you don’t know.
If you have amount represented as e.c …
then the amount in cent is equal to 100e+c
and the amount in euros is equal to e+c/100
For example :
10.27
e=10, c=27
10.27 (in cents) = 1027 = 100*10 + 27
10.27 (in euros) = 10.27 = 10 + 27/100