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 March 11th, 2018, 10:24 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 242 Thanks: 4 Question on Sets If $f(x) = g(x)$ $\forall x \epsilon Q$ then $f(x) = g(x)$ $\forall x \epsilon R$ Is this always true ? Since the set of Reals contains both Rational & Irrational numbers, I feel it is not always true. But someone kindly help me with this problem. Thank you! March 11th, 2018, 11:55 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476 It's rather trivial to construct a functions $f(x),~g(x)$ where it's not true $f(x) = \begin{cases}1 &x \in \mathbb{Q} \\0 &x \not \in \mathbb{Q} \end{cases}$ $g(x) = 1,~x \in \mathbb{R}$ $x \in \mathbb{Q} \Rightarrow f(x) = g(x) = 1$ $x \not \in \mathbb{Q} \Rightarrow f(x) = 0 \neq 1 = g(x)$ Thanks from Lalitha183 March 12th, 2018, 06:30 AM   #3
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 Originally Posted by Lalitha183 If $f(x) = g(x)$ $\forall x \epsilon Q$ then $f(x) = g(x)$ $\forall x \epsilon R$ Is this always true ? Since the set of Reals contains both Rational & Irrational numbers, I feel it is not always true. But someone kindly help me with this problem. Thank you!
It is definitely not true as was pointed out. However, it is interesting that the only hypothesis which needs to be added to make it true is continuity. If $f,g$ are continuous functions defined on $\mathbb{R}$ which agree on the rationals, then they must agree on all of $\mathbb{R}$. Tags question, sets 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 shunya Elementary Math 2 October 29th, 2015 09:47 AM maths4computing Applied Math 3 August 25th, 2013 01:30 PM chappyform Real Analysis 2 April 28th, 2013 06:19 PM Britmouth Applied Math 6 August 31st, 2010 07:08 PM sepehr Applied Math 3 April 3rd, 2010 03:46 AM

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