[math-fun] Puzzle about non-empty sets
The sum of the elements of the empty set is 0; the product of the elements of the empty set is 1. What is the smallest non-empty set of real numbers that shares this property with the empty set? (What if we require all the numbers to be rational? I haven't figured this part out yet.) Jim Propp
(What if we require all the numbers to be rational? I haven't figured this part out yet.)
It can be done with 4 numbers: {-2, -1/2, 1/2, 2} and cannot be done with fewer numbers. In particular, any solution with 3 rationals must necessarily correspond to a solution of the Diophantine equation: a b (a + b) = n^3 where a and b are positive integers. We can wlog assume a and b are coprime, so {a, b, a + b} are pairwise coprime. Consequently, for their product to be a cube, they must each be cubes. In particular, we obtain a positive integer solution to: x^3 + y^3 = z^3 which was proved by Gauss to have no solutions. Contradiction. Best wishes, Adam P. Goucher
Nice argument. I had gotten that I needed to find integers a and b such that: a*b*(a+b) was a perfect cube, but I wasn't sure if a solution existed. I had missed that you could require them to be coprime. Tom Adam P. Goucher writes:
(What if we require all the numbers to be rational? I haven't figured this part out yet.)
It can be done with 4 numbers:
{-2, -1/2, 1/2, 2}
and cannot be done with fewer numbers. In particular, any solution with 3 rationals must necessarily correspond to a solution of the Diophantine equation:
a b (a + b) = n^3
where a and b are positive integers. We can wlog assume a and b are coprime, so {a, b, a + b} are pairwise coprime. Consequently, for their product to be a cube, they must each be cubes. In particular, we obtain a positive integer solution to:
x^3 + y^3 = z^3
which was proved by Gauss to have no solutions. Contradiction.
Best wishes,
Adam P. Goucher
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Very very nice! Jim Propp On Friday, July 1, 2016, Adam P. Goucher <apgoucher@gmx.com> wrote:
(What if we require all the numbers to be rational? I haven't figured this part out yet.)
It can be done with 4 numbers:
{-2, -1/2, 1/2, 2}
and cannot be done with fewer numbers. In particular, any solution with 3 rationals must necessarily correspond to a solution of the Diophantine equation:
a b (a + b) = n^3
where a and b are positive integers. We can wlog assume a and b are coprime, so {a, b, a + b} are pairwise coprime. Consequently, for their product to be a cube, they must each be cubes. In particular, we obtain a positive integer solution to:
x^3 + y^3 = z^3
which was proved by Gauss to have no solutions. Contradiction.
Best wishes,
Adam P. Goucher
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
*SPOILER ALERT* For reals the smallest number is 3. Here's one set of values that works: -1, -1/phi, phi where phi = (sqrt(5)+1)/2. With rationals it can be done with 4: -2, -1/2, 1/2, 2 I'm not sure if it can be done with 3 rationals. Tom James Propp writes:
The sum of the elements of the empty set is 0; the product of the elements of the empty set is 1.
What is the smallest non-empty set of real numbers that shares this property with the empty set?
(What if we require all the numbers to be rational? I haven't figured this part out yet.)
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
-
Adam P. Goucher -
James Propp -
Tom Karzes