Thread: How Good Are You At Reasoning?

Originally Posted by Xei

The case being trivial doesn't somehow make the assertion incorrect... I was careful to be conscious of the case that Hawaii is a single group when I wrote it.

I must be missing something then. If we consider the trivial group a valid communicating group then why isn't it a member of every set? In that case the proof goes:

The empty set is a member of every set. There is no one in the empty set. Hence the empty set constitutes a group in which no one can contact anybody outside of that group.

That's a correct statement, there's just no meat to figuring it out.

I'm having to paraphrase technical mathematical terms in terms of intuitive things so it's a bit rough around the edges (as I'm sure you know I'm basically talking about a kind of equivalence relation partitioning a set, but that would be to much jargon for most people). I'll try to fix it.

Yeah, this is tough stuff. Jargon exists for a reason. Consider the fact that my proof of question one really goes:

Suppose N is square. Decompose the set of divisors, S, into orbits under the involution f(d) = N/d. Then each orbit has at most two elements. If N = n^2 then the orbit of n has only one element. But all other elements lie in an orbit with two elements, hence the order of S is k*2 + 1 for some k and must be odd. Conversely suppose that S is of odd order. Splitting it into orbits under f, we can count it as K*2 + j*1 with j odd and j > 0. So there is at least one orbit with one element, n. Then f(n) = N/n = n and N = n^2.

Jargon allows precision and brevity.

There's a slight blip because person 1 is a closed communicating class here, but yeah I know what you mean. I suppose you could have labelled the islanders with the integers.

Yeah, I'm trying to keep stuff short but I should have at least included the n >= 0 part. Or labeled with the integers. So person 1 would be in a class with person 0 who could contact person 2. Person 3 would be in a class with person 2 who could contact person 4, etc.

My answer was even simpler though kind of trivial.

I'd like to see it. Your proof of #7 was very nice. You're good at writing short proofs. It makes me feel like I overindulge in formality...

It wasn't much simpler; it was just 1 can contact 2, 2 can contact 3, etcetera. Yours is just as good at illustrating the point at hand.

Originally Posted by PhilosopherStoned

I must be missing something then. If we consider the trivial group a valid communicating group then why isn't it a member of every set? In that case the proof goes:

The empty set is a member of every set. There is no one in the empty set. Hence the empty set constitutes a group in which no one can contact anybody outside of that group.

That's a correct statement, there's just no meat to figuring it out.

Well, the statement is about any size group. The fact that this encompasses a trivial case where the group is the set just isn't a problem. And it happens all the time... the fundamental theorem of algebra includes the trivial case of polynomials of degree 1, rank nullity includes the case of vector spaces of dimension 0, etc. etc.

Originally Posted by Xei

It wasn't much simpler; it was just 1 can contact 2, 2 can contact 3, etcetera. Yours is just as good at illustrating the point at hand.


Well, the statement is about any size group. The fact that this encompasses a trivial case where the group is the set just isn't a problem. And it happens all the time... the fundamental theorem of algebra includes the trivial case of polynomials of degree 1, rank nullity includes the case of vector spaces of dimension 0, etc. etc.

Yeah, yeah. We're talking past each other cause I was confused about something. Trivial case as in zero size is no problem. The trivial case with only one group is trivially true though because they can't contact anyone outside the group by virtue of there being no one outside the group. I was somehow missing that. Simplicity wise, I was thinking you had a simpler proof of the main part. I feel like there is one.

1. This is an interesting question. I'm going to say the answer is that every coin with an odd number of multiples is heads.

2. 22

3. This one is kind of a doozy. I'm not sure of my answer. If you test positive for having cancer, you have a 7.69% chance of having cancer.

4. 7,920 out of 104,900. Or 792 out of 10,490

5. The second and fourth card.

6. Given the information, you have to check if the 16 year old is drinking beer and you have to check if the person drinking beer is underage.

7. True. If there are two friends together, there must be a third friend for one of the two friends to have one more friend than the other. But with the addition of a third friend, there are now two people with only one friend in the group. There must be two more friends added to the group to give one of the single-friended friends more friends than the friend with two friends. But this means yet more friends must be added to the the group to give these new additions a different number of friends than that guy with only a single friend. This goes on for infinity. Thus, a group of two people or more containing a finite number of people must have at least two people that have the same number of friends.


Edit: I see where I went wrong with number 4. I added the twenty people who were tree huggers but would not admit it.

80/1030

Found a few nice questions whilst browsing one of my tutor's websites for revision stuff and remembered this thread. Again, none of them require maths knowledge.


1. I draw N circles on a piece of paper. Prove that they divide the paper into no more than N(N - 1) + 2 regions.


2. x is any number on the number line such that x + 1/x is a whole number. Prove x^n + 1/x^n is also a whole number, for any positive whole number n.


3. We have the set of consecutive numbers from 1 to some even number, for instance, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Now we take one more than half of the numbers from this set, for instance, {2, 3, 4, 7, 9, 10}.

a) Prove that in this new set, there will be two numbers with no common factors (for instance, 4 and 9).

b) Prove that in this new set, one number will divide another (for instance, 3 and 9).


4. You wish to drive around a mountain. There are fuel stops at various places on the road, and the total amount of fuel in them is exactly enough to make the journey once. Prove that it's possible to make the journey in a car with an empty fuel tank from some fuel stop.


5. There are six towns. Between each pair of towns is either a bus service or a train service, but never both. Prove that you can travel through three towns in a loop (without visiting any others) with only one type of transport.

Does this have to be true for five towns?

my thouoghts, without looking at anyone else's answers:

1. every coin is flipped except the first one

2. 24

3. 90.4% chance

4.7.77%

5. I would only need to flip the card with 2 on it

6. what the 16 year old is drinking, and the age of the guy drinking beer

7. true

probably all wrong :p

Alright, let's see if we can pull the 'ol modus ponens and find some answers.

Originally Posted by Xei

1. I draw N circles on a piece of paper. Prove that they divide the paper into no more than N(N - 1) + 2 regions.

Originally Posted by Xei

2. x is any number on the number line such that x + 1/x is a whole number. Prove x^n + 1/x^n is also a whole number, for any positive whole number n.

I interpret this as,

x ∈ ℤ, where ƒ(x) = [x + (1/x)], and ƒ(x) ∈ ℤ.

ƒ₂(x) = [x^n + (1/x^n)], where ƒ₂(x) ∈ ℕ only when n ⊂ x.

I'm not sure the first function should look like [(x+1)/x] or [x+(1/x)].

If I were to take the sum of the geometric series which results from expanding ƒ₂(x) while keeping x static at some value, namely 2, then the series doesn't have a common ratio and does diverge to an arbitrary whole number.

Originally Posted by Xei

3. We have the set of consecutive numbers from 1 to some even number, for instance, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Now we take one more than half of the numbers from this set, for instance, {2, 3, 4, 7, 9, 10}.

a) Prove that in this new set, there will be two numbers with no common factors (for instance, 4 and 9).

b) Prove that in this new set, one number will divide another (for instance, 3 and 9).

B ⊂ A and A ∪ B and B ∩ A. If both sets are communicative with respect to the other, and each set contains elements of the other, then every valid operation for the set A will be valid for the set B. The number of valid operations depends on the total number of elements taken from the consecutive sequence set to the subset B and is proportional to limiting value of ½+1 number of elements in the subset.

Originally Posted by Xei

4. You wish to drive around a mountain. There are fuel stops at various places on the road, and the total amount of fuel in them is exactly enough to make the journey once. Prove that it's possible to make the journey in a car with an empty fuel tank from some fuel stop.

Do you mean the car or the fuel stations when you say "them"? The gas stations would need to be at equivalent distances around the mountain and the surface of the terrain would need to be consistent for this to be possible.

Originally Posted by Xei

5. There are six towns. Between each pair of towns is either a bus service or a train service, but never both. Prove that you can travel through three towns in a loop (without visiting any others) with only one type of transport.

Does this have to be true for five towns?

There will always be at least one more bus than train station or vice versa between all the towns, and no that doesn't have to be true for five towns.

OK. Time to start knocking them out.

Originally Posted by Xei

4. You wish to drive around a mountain. There are fuel stops at various places on the road, and the total amount of fuel in them is exactly enough to make the journey once. Prove that it's possible to make the journey in a car with an empty fuel tank from some fuel stop.

For a contradiction, supose that some arrangement of depots exists for which it is not possible. Choose a direction of traversal. Each depot, T, defines a segment, S(T) of the road which is reachable by starting from T, traveling in the given direction, and stopping at all reachable depots along the way. There is at least one depot, T₁, which defines a longest segment. Start at T₁ and travel until you run out of gas. Push the car to the next depot. The car hence traverses some distance D₁ of roadway with no fuel. Fill the car with fuel and procede in this manner until you arrive at the starting point. You hence completed one circuit of the road and put enough enough fuel in the car for precisely one circuit. However the car was pushed for distance D = D₁ + ... + D_n and so there must be enough gas left in the tank to travel D units of distance. This means that there is some depot, T₂, such that T₁ is reachable from it. Specifically, it will be the last depot into which the car had to be pushed. But then S(T₂) is longer than S(T₁) because it properly contains the latter segment and this contradicts our assumption that S(T₁) is maximal. Hence there must always be an arrangement of depots such that there is at least depot, T, with S(T) being the entire road.

Hope that's reasonably clear.

I assume that by whole, one means integral. Otherwise, there's more work to do.

Originally Posted by Xei

2. x is any number on the number line such that x + 1/x is a whole number. Prove x^n + 1/x^n is also a whole number, for any positive whole number n.

Using strong induction, the one case is given to us. Specifically, x¹ + 1/x¹ is whole by the terms of the problem.

For the inductive case, note that (x^n-1 + 1/x^n-1)(x + 1/x) - (x^n-2 + 1/x^n-2) is whole as, by the inductive hypothesis, each term in brackets is whole. But trivial algebraic simplification reduces this to xⁿ + 1/xⁿ. Hence we've proven it for n as well.

Mathematician circlejerk....

Technically it's an n-gon jerk. But if you don't want to play could you please bawww elsewhere, it's a little embarrassing to watch and you're killing the mood.

Originally Posted by Xei

But if you don't want to play could you please bawww elsewhere, it's a little embarrassing to watch and you're killing the mood.

Originally Posted by Xei

Technically it's an n-gon jerk. But if you don't want to play could you please bawww elsewhere, it's a little embarrassing to watch and you're killing the mood.

In reality, circles don't have points, therefore you cannot classify it as n-gon. I wasn't bawwwing either, I was simply pointing out being good at maths does not mean you are good at reasoning. And vice versa.

Originally Posted by tommo

In reality, circles don't have points, therefore you cannot classify it as n-gon

I think that was Xei points exactly, as each person would represent a point and you'd therefore not be circle jerking.

Originally Posted by khh

I think that was Xei points exactly, as each person would represent a point and you'd therefore not be circle jerking.

Oh, right.

But yeah I might try them out just coz I like to test my intelligence.

Originally Posted by tommo

I was simply pointing out being good at maths does not mean you are good at reasoning. And vice versa.

I think we would all be surprised to know how much our structure of logic and mathematics are actually related.

Originally Posted by Wayfaerer

I think we would all be surprised to know how much our structure of logic and mathematics are actually related.

Oh, no doubt. But one doesn't imply the other.
But definitely. You need logic if you want to figure out any problem with already knowing the method , not just maths.

I'm terrible at reasoning. I failed basic logic, I overgeneralize all the time, I believe in the gambler's fallacy (stats just don't make sense to me at all), and I tend to believe eyewitness testimonies over objective evidence. When I start an argument, I never take the time to define premises clearly, or ambiguous terms. It just takes too much effort, y'know?

Oh wait. That's not me at all. I'm the opposite of that. How could I forget?

And in reality you can't form a circle out of a chain of masturbators.

Like I said, the questions are not intended to require mathematical knowledge. If I wanted to pleasure the mathematicians here I would have done some tricky integrals or some groups questions which would have looked like Greek to everybody else. Though to be honest I think reasoning and mathematics (the action rather than the body of knowledge) are pretty synonymous, and indeed the whole point of the questions was to allow everybody to have a go at the kind of questions that a mathematician would typically tackle, but specially picked so that you didn't have to know anything at all about maths per se.

That was the goal, anyway. I suppose that question 1 requires a modicum of algebra to complete, but that's not really important. If you want, exchange N circles with 6 circles and N(N - 1) + 2 regions with 32 regions. Then the only knowledge you need is what 6 and 32 are and what circles are.

Question 2 is the most algebra heavy, but it's the kind of very basic stuff you learn at school. And question 3 only requires a basic familiarity with numbers, if you know for instance how to split something into factors you're fine. But if you really hated maths at school then I'd leave those two out.

And then questions 4 and 5 really are pure reasoning questions.

Please add your own, that was the point of the thread in the first place.

A penny saved is worth two in the bush.

Fuck mathematics! I'm balancin' my checkbook!

I'd be very interested to see people take a crack at 5. It's purely a reasoning question. I basically did it straight off; it will be particularly interesting to see (though to go into details why would give it away at the moment) if that's down to being trained to think like a mathematician, or simply because it's quite an easy question.

And somebody else please post some other cool puzzles.

4 and 5 seem to prove themselves for me, you start empty at one out of multiple gas stations that have enough fuel to take you around the mountain, so you're able to go around the mountain and you never said there had to be any trains, so there could just happen to be 3 towns next to each other with car transportation in between. The proof is in the pudding. The other ones are a bit beyond me.

Sorry, number 4 is kind of hard to express clearly. I had the same misinterpretation when I first read it... of course, that's not what it means, the answer is totally trivial. No, the total amount of fuel in all of the fuel stops combined is enough to travel around the mountain once, not in each and every fuel stop. Hopefully it makes much more sense as a puzzle now. It is not immediately obvious that you can do it, because some stops might not contain enough fuel to make it to the next stop, even if combined there is enough fuel to make the trip once.

As to number 5, it's true that I didn't say there had to be any trains, and so it is possible that each town could just have a bus service. But that is not a proof that this is necessarily true for the kind of situation I described; although it is possible that there aren't any trains, it is also possible that it isn't the case that there aren't any trains.

You could definitely do number 1 if you took a crack at it, just replace it with,

1. I draw 6 circles on a piece of paper. Prove that they divide the paper into no more than 32 regions.

The answer is very pretty.

Originally Posted by Xei

1. I draw N circles on a piece of paper. Prove that they divide the paper into no more than N(N - 1) + 2 regions.

I don't know how to prove this without just doing it, which is kind've pointless I guess.

Originally Posted by Xei

2. x is any number on the number line such that x + 1/x is a whole number. Prove x^n + 1/x^n is also a whole number, for any positive whole number n.

Hmmmm, yep..... ok, yep
Yes, I know some of these words....

Originally Posted by Xei

3. We have the set of consecutive numbers from 1 to some even number, for instance, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Now we take one more than half of the numbers from this set, for instance, {2, 3, 4, 7, 9, 10}.

a) Prove that in this new set, there will be two numbers with no common factors (for instance, 4 and 9).

b) Prove that in this new set, one number will divide another (for instance, 3 and 9).

Oh god....
See, again, I could just do this a few times, but it's not really proving anything, one needs to know maths to prove it.

Originally Posted by Xei

4. You wish to drive around a mountain. There are fuel stops at various places on the road, and the total amount of fuel in them is exactly enough to make the journey once. Prove that it's possible to make the journey in a car with an empty fuel tank from some fuel stop.

Even if they all have unequal amounts of fuel?

I don't think it's always possible. Say 100km is the full distance. The first fuel stop is only full enough to go 1km and the next fuel stop is 2km away. Fail straight away.
Or was this worded incorrectly?

Originally Posted by Xei

5. There are six towns. Between each pair of towns is either a bus service or a train service, but never both. Prove that you can travel through three towns in a loop (without visiting any others) with only one type of transport.

Does this have to be true for five towns?

I don't know what "pairs of towns" means, but I'm going to assume they're laid out like this. (Oh wait I get it now, but this suffices anyway)

Actually fuck I don't even need a diagram.
Why is this a question? Just take a bus between each town. Or do you not know whether it will have a train or a bus before you get there?
If so, it's impossible. It may work one time, if there is a bus to each town but not another time, if there's a train fucking up your whole plan.

I know how you're thinking about it, but it doesn't work IRL.

For example in number 4, thinking like a mathematician you would say "Okay well if we start at the stop (which could be in the middle or whatever) which has the most fuel, we can suddenly make it the rest of the way".
But thinking in IRL terms, you just start out at the start of the mountain.