# Discontinuous function prove

I want to find an answer to this case If function f is discontinuous then √f is discontinuous. I am trying to disprove it and I put counter examples but it seems correct. Is this true? Can you help me please

Look at it this way: if $$\sqrt f$$ is continuous, then $$f$$ is continuous. This is clearly true, as the product of continuous functions is continuous (of course, you should start by saying that you are working over real numbers, that $$f$$ is positive, etc ...)

• you mean by contrapositive? Jan 27, 2021 at 10:09
• Yes. When looking at statement involving concepts such as "discontinuous", "infinite", "irreducible", defined as a negation of another property, it is often easier to look at the contrapositive because you can actually use those definitions directly. Jan 27, 2021 at 10:14
• yes i was thinking about contrapositive but i was not sure about it because i put if |f| is continuous then f is continuous but it is not true that if f is discontinuous then |f| is discontinuous Jan 27, 2021 at 10:22
• That is false: take the function $f \colon \mathbb R \to \mathbb R$, mapping $x$ to $-1$ if it is negative, and to $1$ otherwise. Then $|f|$ is constant (and so it is continuous) but $f$ is discontinuous at $0$. An implication and its contrapositive are completely equivalent logical statement. Jan 27, 2021 at 11:43
• yes i said it is not true that if f is discontinuous then |f| is discontinuous and i used the same counter example to prove it. I was talking about the contrapositive. Jan 27, 2021 at 11:51

If squeezing of the domain is allowed then this is your counterexample:

$$f(x)= 1$$ for $$x>0$$ and $$f(x)= -1$$ for $$x\le0$$ which is discontinuous but the square root of it i.e. $$\sqrt{f(x)}$$ is continuous on its domain.

• it is true why we put counter example? Jan 27, 2021 at 10:23
• As I said in answer , if squeezing of domain $f(x)$ is allowed. Jan 27, 2021 at 10:36