If domain of f is a and f(f(x))=x^2, show that if A=C (complex), then f(f(x)) is not x^2.

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

Since the problem provides the information that `f(f(x)) = x^2` , hence `f(f(f(x))) = f(x^2)`  such that `f^2(x) = f(x^2).`

Assuming that `x=0 =gt f^2(0) = f(0) =gt f(0) in {0;1}`

Assuming that `x=1 =gt f^2(1) = f(1) =gt f(1) in {0;1}`

Assuming that `x=-1 =gt f^2(-1) = f(1) =gt f(-1) in {-1;0;1}`

But if `f(-1) =-1 =gtf(f(-1)) = f(-1) = -1 =gt -1=1,`  that is absurd `=gtf(-1) in {0;1} `

Assuming that `x=i =gtf^2(i) = f(i^2) = f(-1)in {0;1} =gt f(i) in {-1;0;1}`

Using the relation provided by the problem yields `f(f(i)) = i^2 = -1` , but `f(f(i)) in {0,1}`  and `-1 !in {0;1}, ` hence, the property `f(f(x)) = x^2`  is not true for A=C.

Sources:

We’ve answered 318,989 questions. We can answer yours, too.

Ask a question