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March 21st, 2018, 05:49 AM   #1
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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.
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March 21st, 2018, 08:36 AM   #2
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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.
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March 22nd, 2018, 03:42 AM   #3
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Quote:
Originally Posted by romsek View Post
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 $x=1$ & $x=-1$
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