Same functions over the set of rational numbers
Same functions over the set of rational numbers
If two functions, both defined from R to R, take same values for each rational argument, they are equal. Need a proof of it.
"Je le vois, mais je ne le crois pas!"  Georg Ferdinand Ludwig Philipp Cantor
Re: Same functions over the set of rational numbers
It's false. Take
\[f(x)=\begin{cases}1\quad\text{if }x\in\mathbb Q\\
0\quad\text{otherwise}\end{cases}\]
and
\[g(x)=\begin{cases}1\quad\text{if }x\in\mathbb Q\\
2\quad\text{otherwise}\end{cases}\]
\[f(x)=\begin{cases}1\quad\text{if }x\in\mathbb Q\\
0\quad\text{otherwise}\end{cases}\]
and
\[g(x)=\begin{cases}1\quad\text{if }x\in\mathbb Q\\
2\quad\text{otherwise}\end{cases}\]
"Everything should be made as simple as possible, but not simpler."  Albert Einstein
Re: Same functions over the set of rational numbers
Ooops, I forgot to mention. Its for continuous functions
"Je le vois, mais je ne le crois pas!"  Georg Ferdinand Ludwig Philipp Cantor
Re: Same functions over the set of rational numbers
It's a definition. The statement is the sum of the definition of continuous functions and the definition of real numbers.

 Posts: 1
 Joined: Sun Jul 15, 2012 11:56 pm
Re: Same functions over the set of rational numbers
On the contrary, assume that the functions $f$ and $g$ are different at some point $\alpha \in \mathbb{R}$. Take $\epsilon = f(\alpha)g(\alpha) >0$.
Since $f$ and $g$ are continuous, there exists $\delta > 0$, such that $f(x)f(\alpha)<\frac{\epsilon}{3}$ and $g(x)g(\alpha)< \frac{\epsilon}{3}$, for all $x\alpha<\delta$
Since the set of rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$, $\exists q \in \mathbb{Q}$ s.t. $q\alpha<\delta$. Thus we note that
\begin{eqnarray}
&&\epsilon = f(\alpha)g(\alpha) = f(\alpha)f(q)+g(q)g(\alpha) \leq f(\alpha)f(q)+ g(\alpha)g(q) \\
&<& 2\frac{\epsilon}{3}
\end{eqnarray}
Which is a contradiction !
Since $f$ and $g$ are continuous, there exists $\delta > 0$, such that $f(x)f(\alpha)<\frac{\epsilon}{3}$ and $g(x)g(\alpha)< \frac{\epsilon}{3}$, for all $x\alpha<\delta$
Since the set of rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$, $\exists q \in \mathbb{Q}$ s.t. $q\alpha<\delta$. Thus we note that
\begin{eqnarray}
&&\epsilon = f(\alpha)g(\alpha) = f(\alpha)f(q)+g(q)g(\alpha) \leq f(\alpha)f(q)+ g(\alpha)g(q) \\
&<& 2\frac{\epsilon}{3}
\end{eqnarray}
Which is a contradiction !
 Fm Jakaria
 Posts: 79
 Joined: Thu Feb 28, 2013 11:49 pm
Re: Same functions over the set of rational numbers
Suppose f and g are two such functions. Write L(A(n)), for some function A, whose domain is not bounded upside by some constant; to be the limit of A when n tends to infinity, where n belongs to the domain of A.
Define a function h from R to R, so that h(x) = f(x)  g(x). Then for all rational x, h(x) = 0.
Choose an arbitrary irrational E.
Again, define a function D as we want that satisfy the following conditions:
1) It's domain is the set of natural numbers, and range is the set of rationals.
2) D(n) < D(n+1) for all natural n.
3) L(D(n)) = E.
As both f and g are continuous, so is h. Now
h(E) = h(L(D(n))) = L(h(D(n))) = L(0) = 0.
Then f(E) = g(E). This concludes that f(x) = g(x) for all x, as desired.
Define a function h from R to R, so that h(x) = f(x)  g(x). Then for all rational x, h(x) = 0.
Choose an arbitrary irrational E.
Again, define a function D as we want that satisfy the following conditions:
1) It's domain is the set of natural numbers, and range is the set of rationals.
2) D(n) < D(n+1) for all natural n.
3) L(D(n)) = E.
As both f and g are continuous, so is h. Now
h(E) = h(L(D(n))) = L(h(D(n))) = L(0) = 0.
Then f(E) = g(E). This concludes that f(x) = g(x) for all x, as desired.
You cannot say if I fail to recite
the umpteenth digit of PI,
Whether I'll live  or
whether I may, drown in tub and die.
the umpteenth digit of PI,
Whether I'll live  or
whether I may, drown in tub and die.

 Posts: 2
 Joined: Tue May 08, 2018 7:38 pm
Re: Same functions over the set of rational numbers
After all, I can safely say that now I know all the basics functions it’s just down to the random details