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 $f(x) = \begin{cases}1 &x \in \mathbb{Q} \\0 &x \not \in \mathbb{Q} \end{cases}$ $g(x) = 1,~x \in \mathbb{R}$ $x \in \mathbb{Q} \Rightarrow f(x) = g(x) = 1$ $x \not \in \mathbb{Q} \Rightarrow f(x) = 0 \neq 1 = g(x)$ 
March 12th, 2018, 06:30 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 685 Thanks: 461 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 $\mathbb{R}$.


Tags 
question, sets 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Infinite sets question  shunya  Elementary Math  2  October 29th, 2015 09:47 AM 
Quick question on Sets?  maths4computing  Applied Math  3  August 25th, 2013 01:30 PM 
Help! Question regarding open sets  chappyform  Real Analysis  2  April 28th, 2013 06:19 PM 
I need help with these question on Sets...  Britmouth  Applied Math  6  August 31st, 2010 07:08 PM 
sets theroy question  sepehr  Applied Math  3  April 3rd, 2010 03:46 AM 