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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
by 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 agoCould you explain why the left hand side must be zero?
- yk 10y agoI 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 agoThe 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 agoYou are right, see my edit above.