Question

is there a function f from the reals to the reals which is not continuous,but has a continous square?

Answer 1

`f(x) = { (-1, x lt 0), (1, x ge 0) :}`

Explanation:

Consider the following function

`f(x) = { (-1, x lt 0), (1, x ge 0) :}`

This has a discontinuity when `x=0`

However `f^2` is continuous:

`f^2(x) = { ((-1)^2, x lt 0), (1^2, x ge 0) :} = 1`

Answer 2

Yes, because `sqrt(y^2) = absy` which is not always `y`.

Explanation:

`f(x) = {(x,"if",x != 5),(-5,"if",x=5):}`

`f` is not continuous at `5`, but

`(f(x))^2 = x^2` is continuous for all `x`.

Answer 3

There's more!

Explanation:

In fact, there is a family of functions `f` which are never continuous in any point, and their square is continuous: they are called the Dirichlet functions: fix `n>0` an integer, then:

`f(x)=x^n` if `x` is rational, `-x^n` if `x` is irrational.

Are the required functions. Their square is always `x^{2n}`, which is continuous, but `f` is never continuous.