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Seeking at least mathematically.
A is not empty, there is at least one element A, and according to the meaning of the question, there is also1(1-a) ∈ a, if1(1-a) = a.

Then a 2-a+ 1 = 0 has no real number solution, so there are at least two elements;

Because1(1-a) ∈ a, then1[1-(1-a))] = [(a-1)]

If (a- 1)/a=a, we also get a 2-a+ 1 = 0, which has no real number solution and at least three elements;

Because [(a- 1)/a]∈A, then1[1-(a-1)/a] = a ∈ a is not a new element;

Therefore, if set A is not an empty set, it has at least three elements.