We will proceed by induction, using the statement:If Max(A,B) = k, then A = BThis is true for k = 1, if Max(A,B) = 1 then A = 1 and B =1 (since A and B are positive integers) so A = BAssume it is true for all k < NIf Max(A,B) = N, then Max(A-1,B-1) = N - 1, since N - 1 < N we know A - 1 = B - 1 so A = BTherefore, for any k, if Max(A,B) = k, then A = BTherefore for any two positive integers, A = BTherefore, all positive integers are equal

The proof breaks at N=2. You assume implicitly that the function Max(A,B) is defined for couples of positive integers (A,B).Take A=1, B=2. Max(A,B)=2 and Max(A-1,B-1) is not defined (i.e. you cannot pull back from the initial argument for N=1).

Is that really what proofs look like? I don't get the joke.

I quite enjoyed this. I hadnt seen this one before.

QuoteOriginally posted by: DJAverageThe proof breaks at N=2. You assume implicitly that the function Max(A,B) is defined for couples of positive integers (A,B).Take A=1, B=2. Max(A,B)=2 and Max(A-1,B-1) is not defined (i.e. you cannot pull back from the initial argument for N=1).You've got it. You chose a formal way to express it, in the definition of Max(A,B). As an applied mathematician, I would single out the link from Max(A-1,B-1) = N-1 to A - 1 = B - 1. This statement is only assumed true if A - 1 and B - 1 are positive integers, therefore it fails if A or B = 1.

QuoteOriginally posted by: zarnywhoopI quite enjoyed this. I hadnt seen this one before.I'm not trying to be annoying, I am genuinely curious what the punchline is. I'm not a math person. The proof is obviously dumb, but maybe I just don't know what induction means.I'll be blunt: You tellin' me this thing is hard to see right through? Did you really not read a couple lines and say "whatever" after like three seconds?

I think I do qualify as a math person. Yes, I saw the flaw on sight - most 'paradox' proof involving induction rely on the inductive step failing on an early case. But I still enjoyed it. A joke being corny doesnt stop it being funny, though a lot depends on your frame of mind. If I'd been in a bad mood when I read it I might have had a different reaction. Also I'm glad to see Aaron posting again. A development which I think should be encouraged.

QuoteOriginally posted by: farmerThe proof is obviously dumb, but maybe I just don't know what induction means. I'll be blunt: You tellin' me this thing is hard to see right through? Did you really not read a couple lines and say "whatever" after like three seconds?To a non-math person, the "dumb" part of the argument is probably the introduction of the maximum function, which seems irrelevant at first, and the argument by induction. Induction is rarely used outside of mathematics because it relies on the inductive statement being true in a very strict sense. In non-mathematical argument, it usually fails. For example, I might say, "killing an innocent, healthy adult is murder"; and "if it's murder to kill someone tomorrow then it must be murder to kill them today"; and conclude that abortion is murder. But the argument relies too heavily on strict, either/or, definitions of "humanity" and "murder." Since it's hard to imagine something changing from lump of cells to human in a day, or from unobjectionable to murder overnight, the argument carries some rhetorical weight. But logically, all I've shown is if abortion is not murder, there must be gradations in the concept of either humanity or murder, or I there must be some special day (quickening, perhaps, or birth) in human development.To a math person, it's unsurprising to see the maximum function or the induction argument, both are common in this kind of proof. While the error is easy to find, if you didn't know the conclusion was absurd, it's also easy to miss. For example, suppose I was trying to prove for two positive integers A and B:A/B + B/A <= Max(A,B) + 1/Max(A,B)This is true if Max(A,B) = 1. Suppose I then proved that:(A - 1)/(B - 1) + (B - 1)/(A - 1) <= Max(A-1,B-1) + 1/Max(A-1,B-1) implies A/B + B/A <= Max(A,B) + 1/Max(A,B)I think most people would accept this proof at first glance, failing to notice that A-1 and B-1 need not be positive integers (that is, they can be zero).

very nice good for interviews!A

In the same vein, I quite like this fallacy which "shows" that Pi is rational:Let u_n = [10^n x Pi]/10^n, where [.] denotes integer partu_0 = 3 is rationalu_1 = 3.1 is rationalu_2 = 3.14 is rational...For all n, u_n is rationalSince u_n converges to Pi as n goes to infinity, we conclude that Pi is rationalAgain, it does take a mathematical mind to be amused...Oh and this one which is slightly more subtle (deemed the math version of the Y2K bug): 1.999... = 2.000...Indeed 1.999... = 1 + 9 x sum_k=1^infty{0.1^k} = 1 + 0.9 x sum_k=0^infty{0.1^k} = 1 + 0.9 x 1/(1 - 0.1) = 2 = 2.000...e.

But your second one is true, 1.999. . . does equal 2.The Greeks used something similar to your pi fallacy to argue against numerical computation in geometry (which does often lead to subtle errors). Consider a square of side 1. To go from one corner to the diagonally opposite corner is path of length 2. In Cartesian terms (which, of course the Greeks didn't know and were in fact using this argument to attack) you can go from (0,0) to (0,0.5) to (0.5,0.5) to (0.5,1) to (1,1), the total path length is still 2. For any n, you can go from (0,0) to (0,1/n) to (1/n,1/n) to (1/n,2/n) to . . . to (1,1) and the path length is still 2. So the limit argument shows the diagonal of a unit square is 2, but a simple geometric argument shows it should be the square root of 2.

QuoteOriginally posted by: farmerIs that really what proofs look like? I don't get the joke.I find in maths, as well as in other things in life, one should really learn to walk before they can run .

QuoteOriginally posted by: AaronBut your second one is true, 1.999. . . does equal 2.not if you require unicity of decimal representation (which is quite handy), in which case you need to bar any infinite sequence of 9's

QuoteOriginally posted by: eiriamjhFor all n, u_n is rationalSince u_n converges to Pi as n goes to infinity, we conclude that Pi is rationalAgain, it does take a mathematical mind to be amused...I believe this task will be solved by 99% of math students in less than 5 seconds after they hear the statement in bold.The denseness of rational numbers in the set of real numbers is so famous and is used in a lot of math proofs. It's a really nice tool, when you prove something only for rational numbers and then extend the proof to all real numbers by constructing a limiting sequence of rational numbers to that real number.