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

3 Answers
Apr 17, 2017

# 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#
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However #f^2# is continuous:

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

Apr 17, 2017

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#.

Apr 17, 2017

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.