- 28 Jun '12 10:40Choose a positive irrational number r and form two sequences;

n x (1 + r); n = 1, 2, 3, ...

m x (1 + 1/r); m = 1, 2, 3, ...

That gives two sequences of irrational numbers. Round them all down to the nearest integer. Say, with r = pi;

r = 3.141

1/r = 0.318

4.14 8.28 12.43, ... rounds down to 4, 8, 12, 16, 20, 24, 28, 33, ...

1.32 2.64 3.95, ... round down to 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, ... 31, 32, 34, ...

It seems all positive integers feature in one sequence or the other, but never both.

Is that true for pi?

If so, is that true for all other positive irrational numbers too? - 08 Jul '12 19:27

Hmm*Originally posted by talzamir***Choose a positive irrational number r and form two sequences;**

n x (1 + r); n = 1, 2, 3, ...

m x (1 + 1/r); m = 1, 2, 3, ...

That gives two sequences of irrational numbers. Round them all down to the nearest integer. Say, with r = pi;

r = 3.141

1/r = 0.318

4.14 8.28 12.43, ... rounds down to 4, 8, 12, 16, 20, 24, 28, 33, ...

1.32 2. ...[text shortened]... oth.

Is that true for pi?

If so, is that true for all other positive irrational numbers too?

We can certainly prove any rational number does collide

Assume the rational number is X/Y, where X and Y are integers

at a potential collision point:

n(1+X/Y) = m(1+Y/X)

n/m = (1+Y/X)/(1+X/Y)

n/m = (X + YX)/X * Y/(Y + YX)

n/m = (XY + XY^2)/(XY + YX^2)

as the top and bottom of the RHS are integers, there will be a collision here, ie:

(XY+XY^2)(1+r) = (XY + YX^2)(1+1/r)