4 ms·

The cheat is absolutely brilliant reasoning.

by chton 12y ago

jerf 12y ago
Ehh, sort of. It's logically flawed as phrased, in that the first bit is false: "If the problem is being posed, it must have a constant solution" is false; it could just as well be a niftily-simple symbolic solution too. (I say "as phrased" because as other commenters observe, there are mathematically valid ways to arrive at the result.) This reminds me of one of my favorite joke proofs from when I was in school. (It's so simple I'm sure it has other originations but AFAIK I independently recreated it.) "The problem begins with the phrase 'Show that...' or similar. All previous problems that began with that phrase have been provable. Therefore, by induction, the claim I am being asked to prove must be correct. QED." Unfortunately, this proof line hit a snag about halfway through my graph theory class in which we were assigned a problem out of the book that turned out to ask you to prove a false statement. (Well, a snag above and beyond the fact that that is an invalid inductive proof in general, ahem.) It was a typo and clearly accidental, but it was enough to break my proof forevermore. May you have better luck with it.
- thedufer 12y ago
  Its flawed as a "proof", but it might contextually justify the answer. That exact reasoning accounts for pretty much all of my limited success at middle school/high school math competitions. You can usually skip long, difficult steps by simply assuming there is a single constant answer (as the rules of the contests usually make clear must be the case).
- colanderman 12y ago
  My favorite "cheat" was on an exam in an algorithm-analysis class (lots of discrete math, infinite sums, and proofs). We were asked to prove or find a counterexample to some conjecture or other. Looking at the conjecture, and looking at the clock, I decided the only thing I could complete in time was to find a counterexample. Indeed within a few minutes I found one and wrote it down. I was the last to finish my exam and handed in my booklet; the prof looked it over and noted I was the only one (out of 5) in the class to get that particular question correct. I explained to him my "reasoning". He said, "yes, that is how you're supposed to do it!"
  - millstone 12y ago
    In a linear algebra class, we were asked to prove that a theorem is true if and only if X. (I think X may have been that a matrix had a nonzero determinant.) I had no idea how to do it, but I noticed the theorem was trivially true in the case of a 1x1 zero matrix. Since this matrix has a zero determinant, it spoiled the "only if" part. I wrote it as a counter-example, showing that "if AND only-if" was false, and didn't prove the "if." Full credit!
- chton 12y ago
  I took it to mean that the problem would not be given in this way unless there was an exact solution. In the stated version, there are no degrees of freedom specified, so there can be no degrees of freedom in the solution. In that way, the proof wouldn't work if the problem stated "A six inch high cylindrical hole is drilled through the center of a sphere with radius R. How much volume is left in the sphere?" In that case, the solution might be dependent on R, and we'd have to go through the long version. Without that mention of R, unless the problem is wrongly stated, it must be constant. So it's still an invalid inductive proof, but it's a lot stronger than just assuming that because there is a question there is a constant answer.
phkahler 12y ago
Agreed. Given that it is a puzzle and not a request for proof. Once you realize that the hole and sphere diameter are related and the situation has one degree of freedom not specified, it is perfectly valid to conclude that the answer must be independent of that and choose a case you can work out in your head.
nilkn 12y ago
It's much more impressive if you can show the answer is independent of the variables and then use the limiting case to find the answer. As another poster wrote, all you have to do is write down the integrals for the volumes and notice that in the difference the R^2 terms will cancel. Unfortunately, in this case the integrals are trivial to evaluate. I think a much more interesting problem would be one where the same approach I described works, except it's not tractable (or at least not easy) to do the integrals in your head.

The cheat is absolutely brilliant reasoning.

by chton 12y ago

The cheat is absolutely brilliant reasoning.

jerf 12y ago

Ehh, sort of. It's logically flawed as phrased, in that the first bit is false: "If the problem is being posed, it must have a constant solution" is false; it could just as well be a niftily-simple symbolic solution too. (I say "as phrased" because as other commenters observe, there are mathematically valid ways to arrive at the result.) This reminds me of one of my favorite joke proofs from when I was in school. (It's so simple I'm sure it has other originations but AFAIK I independently recreated it.) "The problem begins with the phrase 'Show that...' or similar. All previous problems that began with that phrase have been provable. Therefore, by induction, the claim I am being asked to prove must be correct. QED." Unfortunately, this proof line hit a snag about halfway through my graph theory class in which we were assigned a problem out of the book that turned out to ask you to prove a false statement. (Well, a snag above and beyond the fact that that is an invalid inductive proof in general, ahem.) It was a typo and clearly accidental, but it was enough to break my proof forevermore. May you have better luck with it.

thedufer 12y ago
Its flawed as a "proof", but it might contextually justify the answer. That exact reasoning accounts for pretty much all of my limited success at middle school/high school math competitions. You can usually skip long, difficult steps by simply assuming there is a single constant answer (as the rules of the contests usually make clear must be the case).
colanderman 12y ago
My favorite "cheat" was on an exam in an algorithm-analysis class (lots of discrete math, infinite sums, and proofs). We were asked to prove or find a counterexample to some conjecture or other. Looking at the conjecture, and looking at the clock, I decided the only thing I could complete in time was to find a counterexample. Indeed within a few minutes I found one and wrote it down. I was the last to finish my exam and handed in my booklet; the prof looked it over and noted I was the only one (out of 5) in the class to get that particular question correct. I explained to him my "reasoning". He said, "yes, that is how you're supposed to do it!"
- millstone 12y ago
  In a linear algebra class, we were asked to prove that a theorem is true if and only if X. (I think X may have been that a matrix had a nonzero determinant.) I had no idea how to do it, but I noticed the theorem was trivially true in the case of a 1x1 zero matrix. Since this matrix has a zero determinant, it spoiled the "only if" part. I wrote it as a counter-example, showing that "if AND only-if" was false, and didn't prove the "if." Full credit!
chton 12y ago
I took it to mean that the problem would not be given in this way unless there was an exact solution. In the stated version, there are no degrees of freedom specified, so there can be no degrees of freedom in the solution. In that way, the proof wouldn't work if the problem stated "A six inch high cylindrical hole is drilled through the center of a sphere with radius R. How much volume is left in the sphere?" In that case, the solution might be dependent on R, and we'd have to go through the long version. Without that mention of R, unless the problem is wrongly stated, it must be constant. So it's still an invalid inductive proof, but it's a lot stronger than just assuming that because there is a question there is a constant answer.

phkahler 12y ago

Agreed. Given that it is a puzzle and not a request for proof. Once you realize that the hole and sphere diameter are related and the situation has one degree of freedom not specified, it is perfectly valid to conclude that the answer must be independent of that and choose a case you can work out in your head.

nilkn 12y ago

It's much more impressive if you can show the answer is independent of the variables and then use the limiting case to find the answer. As another poster wrote, all you have to do is write down the integrals for the volumes and notice that in the difference the R^2 terms will cancel. Unfortunately, in this case the integrals are trivial to evaluate. I think a much more interesting problem would be one where the same approach I described works, except it's not tractable (or at least not easy) to do the integrals in your head.