This one wants a proof. Work it on paper, read the official solution, then mark
yourself honestly — the ladder only means something if the record is true.
Official solution
25.41. Suppose that e=m/n, where m and n are natural numbers. Then according to problem 25.39
0<nm−(2+2!1+…+n!1)<n!n1
After multiplying by n! we get that 0<a<1/n, where a is an integer. This cannot be
Source: NuminaMath-1.5,
licensed Apache-2.0.
Statement and solution reproduced as published; topic, difficulty and ordering added
by this site.