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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

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2 Answers 2

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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 ...)

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  • $\begingroup$ you mean by contrapositive? $\endgroup$
    – Lama Bitar
    Jan 27, 2021 at 10:09
  • $\begingroup$ 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. $\endgroup$
    – Numbra
    Jan 27, 2021 at 10:14
  • $\begingroup$ 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 $\endgroup$
    – Lama Bitar
    Jan 27, 2021 at 10:22
  • $\begingroup$ 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. $\endgroup$
    – Numbra
    Jan 27, 2021 at 11:43
  • $\begingroup$ 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. $\endgroup$
    – Lama Bitar
    Jan 27, 2021 at 11:51
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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.

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

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