Proof by contradiction that irrational numbers conform to $f(x)=x^2$

Question

I am having some difficulty with this proof for my Real Variables class.

I know that $f(x)$ is a continuous function defined on $R^1$, and $f(x)=x^2$ for any rational $x$. I also have the definition of $e$ to work with, where $$e=\sum \frac1{j!}$$ that I believe can be used in the proof.

I need to prove that $f(x)=x^2$ for irrational $x$ as well. I know I need to use some sort of contradiction as well. This is what I have so far:

We know that $f$ is continuous, so it is continuous for irrational numbers. So for every $\varepsilon >0$, there exists a $\delta>0$ such that $$|x^2-f(p)|<\varepsilon$$ for all points where $$|x-p|<\delta$$ (I used the distance formula for the real numbers, because that is what $f(x)$ is defined on).

Now I assume that $f(x)\ne x^2$ for an irrational $x$. If this is true, it also holds for $e$.

How do I proceed from here? I can't just say $$x^2 - (NOT p^2)<\varepsilon$$, can I? That seems much too vague to me. The follow-up questions for this problem involve series, if that affects any of the answers.

Thank you for any assistance or hints!

Answer 1

I think the cleanest way to use the following two facts:

1. A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous $\iff$ for all convergent sequences $x_n \rightarrow L$, $f(x_n) \rightarrow f(L)$.

(In the event that you haven't seen this before, a proof can be found in $\S 10.3$ of these notes.)

2. For every real number $L$, there is a sequence of rational numbers $\{x_n\}$ such that $x_n \rightarrow L$.

to deduce:

3. If $f,g: \mathbb{R} \rightarrow \mathbb{R}$ are both continuous functions such that $f(x) = g(x)$ for all $x \in \mathbb{Q}$, then $f = g$.