My Math Forum Question on Sets

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 March 11th, 2018, 10:24 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, 11:55 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,324 Thanks: 1233 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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Quote:
 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 $\mathbb{R}$.

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