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 March 21st, 2018, 05:49 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 242 Thanks: 4 Continuity of a Function The points of continuity of the function $f:R -> R$ defined by $f(x) =${ $|x^2-1|,$ if $x$ is irrational $0$, if $x$ is rational are a) $x= -1, x=0, x=1$ b) $x=-1, x=1$ c) $x=-1, x=0$ d) $x=0, x=1$ How to solve this type of question when rationals & irrationals are given as the condition. Thank you March 21st, 2018, 08:36 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 1) use the usual epsilon-delta proof 2) note that due to denseness of the reals in the rationals (and vice versa), rational and irrational numbers can be found arbitrarily close together. 3) Thus the only points where $f$ will be continuous is where the value of $f$ is equal for both rationals and irrationals 4) There are only two points where this will be true, and they are rather trivial to find. See if you can figure it out. Thanks from Lalitha183 March 22nd, 2018, 03:42 AM   #3
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 Originally Posted by romsek 1) use the usual epsilon-delta proof 2) note that due to denseness of the reals in the rationals (and vice versa), rational and irrational numbers can be found arbitrarily close together. 3) Thus the only points where $f$ will be continuous is where the value of $f$ is equal for both rationals and irrationals 4) There are only two points where this will be true, and they are rather trivial to find. See if you can figure it out.
3rd point helped me alot Thank you.
The points of continuity are $x=1$ & $x=-1$ Tags continuity, function Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MathAboveMeth Calculus 1 January 7th, 2017 03:55 AM condemath2 Calculus 6 June 6th, 2014 12:19 PM frankpupu Calculus 4 May 12th, 2012 08:21 AM jollysa87 Calculus 1 July 15th, 2009 06:58 PM Tear_Grant Calculus 2 April 19th, 2009 04:43 AM

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