As I read through the statement it says the player making the next to last mistake wins. In my example black makes the next to last mistake and loses. So I consider this a counterexample to the statement. As far as I know every statement allowing a counterexample is falsified and hence not true. Quod erat demonstrandum.
Your argument is logically compelling. In order to be intellectually honest, I must admit my error.
Well done, herdwars! 
God, how I love wisdom!
Its not your error its Tartakower s, and usually there is a lot of wisdom in what he says.
As a mathematician I am of course trained in logic...
Nonsense. You are wise far beyond your years, young man. In the short time we've corresponded, you have taught me much — about you, about life, about myself.
Its not your error its Tartakower s, and usually there is a lot of wisdom in what he says.
As a mathematician I am of course trained in logic...
Ah! That explains it, then. I admire mathematicians, and hold a great respect for mathematics on the whole. It is, after all, a branch of philosophy (under epistemology).
Yes Philosophy and Mathematics are very entangled. I like philiosophic questions and aguments a lot, but sometimes they are to deep for me. I read your statements on why Kant is wrong but could not totally understand them. Probably I have to rethink them.
Yes Philosophy and Mathematics are very entangled. I like philiosophic questions and aguments a lot, but sometimes they are to deep for me. I read your statements on why Kant is wrong but could not totally understand them. Probably I have to rethink them.
Recall your studies of general relativity. You'll get it. I have every confidence!
It's definitely not true in all cases. Imagine if white makes the last mistake and loses. But the mistake before that (the penultimate mistake) was also made by white. Then white has made the mistake before the last one but still did not win.
I agree with the statement in spirit and in the spirit it was intended. But we can't justify it by logic.
I think we've all come to agreement on that. Good thread.
As I read through the statement it says the player making the next to last mistake wins. In my example black makes the next to last mistake and loses. So I consider this a counterexample to the statement. As far as I know every statement allowing a counterexample is falsified and hence not true. Quod erat demonstrandum.
You know there's a mathematician in the house when someone says Q.E.D.
At any rate, I'm a bit of a mathematician myself, albeit currently without a degree, and what I've always enjoyed most is simply solving a good, tough problem. This one, which I came across about two weeks ago, will probably be easy for someone of @herdwars' stature, but was particularly memorable and I think illustrates nicely what I'm trying to get across.
Find all natural numbers below 1,000,000 with exactly 77 divisors.
(No calculators or Internet searches allowed 
)
This chatbox is not fit for mathematical reasoning but the solution I find would be
4096 x 3 x 25
4096 x 3 x 49
4096 x 7 x 25
4096 x 9 x 5
4096 x 9 x 7
4096 x 9 x 11
4096 x 9 x 13
4096 x 9 x 17
4096 x 9 x 19
4096 x 9 x 23
The reasoning goes as follows
Every number can be written as p1^a1 x ..... pk^ak the pi being different primes, a1,..,ak at least 1.
The numberof divisors for such a number is (a1+1) x.....x(ak+1) - 1.
In order this to be 77 the product should be 78 = 2x3x13.
This limits the choices for k (being at most 3) and a1,a2 and a3.
I leave it as an excersise to see why k=1 or k=2 are impossible... the numbers would become tobig even if one picks the smallest two primes.
So k=3 a1=1, a2=2 and a3 = 12.
ITis easy to see that the listed possibilities are the only ones satisfying all conditions....
Strangely, I got a different answer. It went like this:
First of all, if a natural number is to have an odd number of divisors, it must be a perfect square.
Why? For a non-square natural number, its divisors can be written in the form of a number of pairs a * b where a < sqrt n < b.
Secondly, any natural number can be written in the form (p1^a1)(p2^a2)(p3^a3).... in which case the number of divisors is equal to (a1+1)(a2+1)(a3+1)... (I assume the difference between our answers is because I counted both 1 and n as divisors of n whereas you chose not to count one of those and got different results.) 77 = 11*7, so k=2, a1=10 and a2=6.
That leaves us with two possible answers:
3^10 x 2^6 = 3,779,136 > 1,000,000
2^10 x 3^6 = 746,496
Oh yes i was thinking to complicated again, yes 1 should be counted as a divisor...... and i didn t do that....
Still you have to consider the case k=1... but the numbers would be to big.....
Whatever. It's enough that we can debate something respectfully on the Internet That's yet another great thing about math. Nobody cares about it enough to start a flame war
All of them, isn't it? I mean, it's kind of like the "which months have 28 days riddle."
So actually I solved two problems
1) Determine the numbers less then 1,000,000 having exactly 77 divisors bigger then 1
2) Determine the numbers less then 1,000,000 having exactly 77 real divisors (i.e. divisors not being the number itself)
As I read through the statement it says the player making the next to last mistake wins. In my example black makes the next to last mistake and loses. So I consider this a counterexample to the statement. As far as I know every statement allowing a counterexample is falsified and hence not true. Quod erat demonstrandum.