so many Hungarian names in one comic. I like it XD
Even with the correct Hungarian umlaut!
The font I use doesn't have the Hungarian umlaut so I have to add it manually after :-(
What! You can't insert TeX-code? :-)
My experience is that Sudoku solvers are also people who don't 'always' appreciate the beauty of proof by contradiction.
which is odd, because certain Sudoku can only be solved by assuming an answer at random, coming to a contradiction, and thereby learning the correct answer...
True. The problem they have with "proof by contradiction" is that it is too close to guessing: try a digit in a given cell, and check if you can fill out the puzzle. If not, try another alternative. Their preferred techniques do depend on an underlying "proof by contradiction", so there is no logical difference. Even more so, because in practice all the solvers strive to find the shortest possible "proof" that resolves the sudoku.
lololol so true
Actually the proof, as usually presented, is flawed. It uses the unstated, albeit valid, premise that "If the square of a (whole) number is even, then the number is itself even." To show the importance of this premise, try replacing 2 with an arbitrary large number -- that is a number that may or may not be a perfect square.
In hope of staving off undue criticism note that I am neither denying the result nor the process, merely the presentation.
The issue of unstated premises can of course lead to Charlie dodgson's version of "Achilles and the Tortoise".
Yes, the proof assumes and uses the primality of 2...but is that really such a problem?
Not really a problem, of course. But the proof also depends on our ability to partition the integers into odd/even camps. That in turn follows easily from the unique factorization of integers, of course, and ultimately from Euclid's algorithm.
But that is trivial. if an even number is squared, the result is even, since (2k)^2 = 4k^2 = 2 (2k^2). If an odd number is squared, the result is always odd (similarly, 2k+1 squared = 2 (2k^2 + 2k) + 1.) Since every number is one and only one of odd or even, any even squares will have an even root, and any odd squares will have and odd root.
For my money the catch in your logic is in "...every number is one and only one of odd or even..." Does that not depend on unique factorization of integers? Ok, not the full UFD-property, but limited to prime factor 2. Substitute this with the Euclidean property, if you prefer that.
I think the discussion demonstrates how non-trivial the failure to include this premis is.
Correct me if I'm wrong, but doesn't this proof rely on the statement that a/b is reduced, and show that it can never be reduced if sqrt(2) is rational?
Also, the paper is presented as a visual aid to a verbal statement; I think it's safe to say that the unwritten premises were verbally stated.
I guess there is a lesson there: before going down the path of proof-by-contradiction with the uninitiated, it'd probably be good to tell them what the plan is at the start. I.e.: patiently explain that the goal is to prove that X is *not* true, by assuming that X *is* true, and then showing step by step that, that will lead to a contradiction.
Furry cows moo and decompress.
Beauty does come with age.. but sometimes plastic surgery can help.
That girl is irrational.