March 11th, 2018, 09:24 PM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  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, 10:55 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,011 Thanks: 1044 
It's rather trivial to construct a functions $f(x),~g(x)$ where it's not true $f(x) = \begin{cases}1 &x \in \mathbb{Q} \\0 &x \not \in \mathbb{Q} \end{cases}$ $g(x) = 1,~x \in \mathbb{R}$ $x \in \mathbb{Q} \Rightarrow f(x) = g(x) = 1$ $x \not \in \mathbb{Q} \Rightarrow f(x) = 0 \neq 1 = g(x)$ 
March 12th, 2018, 05:30 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 398 Thanks: 212 Math Focus: Dynamical systems, analytic function theory, numerics  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 $\mathbb{R}$.


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