4 ms·
Fair enough. Take 3/2 in that case.
by rhn_mk1 4y ago
Fair enough. Take 3/2 in that case.
- hgsgm 4y agoDoesn't work, because in k * (3/2 - 1)(3/2) = k'(9/4) - k(3/2) (3/2)^2 = 9/4 is not an integer, you don't get k' < K so there is no contradiction. You'd only get an integer if 4 divides k', which doesn't happen, because k = 2 and k' = k(3/2 -1) = k/2 = 1
- rhn_mk1 4y agoThanks, you're right. I missed that there was supposed to be a √2*√2 in the expansion of the last line.