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Math problem in the first lesson of senior one. .
A∈S, it must be11-a ∈ s.

So 2∈S, then a=2 at this time.

Then1/(1-2) =-1∈ s

Then1/[1-(-1)] =1/2 ∈ s

Then1/(1-1/2) = 2 ∈ s.

So as to enter the cycle.

Because there are no identical elements in the collection.

So there are only three elements in s.

So there must be two other elements.

They are-1 and 1/2 respectively.

If there is only one

Then from a∈S, it must be11-a ∈ s.

There are no identical elements in the collection.

So only a= 1/( 1-a)

Ah-ah? = 1

Answer? -a+ 1=0

Equation has no solution

So there can't be just one.