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 May 15th, 2017, 06:41 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 217 Thanks: 2 Number of points of continuity The number of points of continuity of the function f = $| x^2 - 1|$ if $x$ is irrational, $0$ if $x$ is rational A) $0$. B) $1$. C) $2$. D) Infinite I hope if you substitute irrational numbers in the function we get a rational answer because of the "square". So the function is continuous at infinite points. If I'm wrong correct me please, thank you
 May 15th, 2017, 07:09 PM #2 Senior Member   Joined: Aug 2012 Posts: 1,772 Thanks: 481 This depends on $f$. In any event $\pi^2$ is irrational. Or is the expression the function? Not clear.
May 15th, 2017, 07:21 PM   #3
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Quote:
 Originally Posted by Maschke This depends on $f$. In any event $\pi^2$ is irrational. Or is the expression the function? Not clear.
$f$ is a function and it is defined as $|x^2-1|$ if $x$ is rational
$0$ if $x$ is irrational

 May 15th, 2017, 08:34 PM #4 Senior Member   Joined: Aug 2012 Posts: 1,772 Thanks: 481 Oh right. For some reason I read that as "the number of points of continuity" of the function was that many. So we have this: $$f(x) = \left\{ \begin{array}{ll} 0 & x \in \mathbb Q \\ | x^2 - 1| & x \notin \mathbb Q \end{array} \right.$$ Let's see. The first thing is that it's zero on the rationals and we know that a continuous function is determined by its values on the rationals. For example if it's continuous at $\pi$ then since the function is $0$ on each of $3, 3.1, 3.14, \dots$ it must be $0$ at $\pi$. But $\pi^2 - 1 \neq 0$ so $f$ can't be continuous at any irrationals. The same argument applies to rationals in fact. Every rational is also the limit of a sequence of rationals so if $f$ is continuous at a rational it must be $0$ there too. So the only values where the function can be $0$ and also $x^2 - 1 = 0$ are $\pm 1$. I don't know why they are confusing the issue with the absolute value. So I would go with C) 2.
May 15th, 2017, 10:07 PM   #5
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Quote:
 Originally Posted by Maschke Oh right. For some reason I read that as "the number of points of continuity" of the function was that many. So we have this: $$f(x) = \left\{ \begin{array}{ll} 0 & x \in \mathbb Q \\ | x^2 - 1| & x \notin \mathbb Q \end{array} \right.$$ Let's see. The first thing is that it's zero on the rationals and we know that a continuous function is determined by its values on the rationals. For example if it's continuous at $\pi$ then since the function is $0$ on each of $3, 3.1, 3.14, \dots$ it must be $0$ at $\pi$. But $\pi^2 - 1 \neq 0$ so $f$ can't be continuous at any irrationals. The same argument applies to rationals in fact. Every rational is also the limit of a sequence of rationals so if $f$ is continuous at a rational it must be $0$ there too. So the only values where the function can be $0$ and also $x^2 - 1 = 0$ are $\pm 1$. I don't know why they are confusing the issue with the absolute value. So I would go with C) 2.

I understood it.
Thank you so much for the explanation

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