# Prove that `a_ n in (0,1)` `a_(n+1)=(1 - sqrt a_n)` `a_1 in (0,1)`

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You need to use mathematical induction to test if `a_n in (0,1), ` hence, you need to perform the two steps, starting with the base case, such that:

`a_1 in (0,1)` is given by the problem

Performing the inductive step you need to test the following statement, such that:

`a_k in (0,1), then a_(k+1) in (0,1)`

Considering that the statement `a_k in (0,1)` holds, you need to test if `a_(k+1) in (0,1)` , such that:

`a_(k+1) = (1 - sqrt a_k)`

Since `a_k in (0,1) => sqrt(a_k) in (0,1),` such that:

`0 < sqrt(a_k) < 1 => 1 - 0 > 1 - sqrt(a_k) > 1 - 1 => 1 - sqrt(a_k) in (0,1) => a_(k+1) in (0,1)` .

**Hence, performing mathematical induction yields that both steps holds, hence `P(n) = a_n in (0,1)` holds for n natural.**