5 ms·

He writes 'but the left side is an integer'.... IMO the left side of the last equation is equal 0.... Any comments?

• widdma 10y ago
  Zero is an integer. The point is the right side is not. Since it's strictly greater than zero and less than 1.
• SidiousL 10y ago
  No, it's not zero. The sum is up to `a`, not up to infinity.
• chx 10y ago
  Erm, nope, it would be if the sum went to inf but it doesn't.
• yk 10y ago
  It is clear that the left hand side has to be zero, however I do not see a straightforward way to prove that without using the contradiction itself and he does not need it to be zero to choose that there is a contradiction, so the claim the left side is an integer sidesteps an entirely optional technical difficulty. [Edit:] The argument I had in mind is bad, my point is that the value of the left hand side does not matter for the proof, so it is not calculated. (I leave the comment up for context of the answers below.)
  • andars 10y ago
    Could you explain why the left hand side must be zero?
    • yk 10y ago
      I had something in mind like, assume that the last line is not a contradiction, then we have c( e- a) on the left and something smaller than one on the right. However, assume the last line is not a contradiction is false, so I can't build the argument atop of it.
  • widdma 10y ago
    The left side does not need to be zero, and it does not follow from the assumptions. The proof only uses the fact that the left side is an integer, since all terms are integers (a!/n! ∈ Z for a > n)
    • yk 10y ago
      You are right, see my edit above.