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If $f$ and $g$ are continuous on $\mathbb R$ and $f\left(\frac{p}{q}\right)=g\left(\frac{p}{q}\right)$ for all non-zero integers, $p$ and $q$, then is it true that $f(x)=g(x),\forall x \in \mathbb R$?

Can someone please explain how to go about this? Any help would be appreciated thanks.

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  • $\begingroup$ Define $h(x)=f(x)-g(x)$ for all $x\in \mathbb{R}$. We Know that for every nonzero rational $r\in \mathbb{Q} / \{0\}$, there is nonzero intgers $p$ and $q$ such that $r=\frac{p}{q}$, and so $f(r)=g(r)=0$ i.e., $$h(x)=0$$ for all $x\in \mathbb{Q} / \{0\}$. Now $\frac{1}{n}\rightarrow 0$ and so $0=h(\frac{1}{n})\rightarrow h(0)$ i.e., $h(0)=0$. Therefore $h$ is continuous function and $h(x)=0$ on $\mathbb{Q}$. Now you can get the result from the answer that posted. $\endgroup$ – Deliasaghi Sep 20 '15 at 19:12
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So you mean $f(x)=g(x)$ for all $x\in\mathbb Q$? Then we have $f=g$ since $\mathbb Q$ is dense in $\mathbb R$: Each $x\in \mathbb R$ is the limit of a sequence of rational numbers. Continuity of $f$ and $g$ gives you $f(x)=g(x)$.

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Consider a function h= f-g, h is also continuous on R. now h(x) =0 for all rationals x You need to show that h(x)=0 fol all irrationals too. Take a sequence of(xn) rationals converging to say x(irrational), you can do so as rationals are dense in R. Now as h is continuos at x , therefore by sequential criterion (h(xn))converges to h(x) , but (h(xn)) is a zero sequence , therefore by uniqueness of limit h(x)=0. thus h(x)=0 for all real no.'s thus f=g.

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