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

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!

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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 $f = g$.

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About the third fact: So can I say that because $f$ is continuous, there exists a sequence $x_n -> e$ and therefore $x_n^2 -> e^2$? It feels like there is something missing, and I think it is because I don't quite understand it. – JohnDvorak Feb 20 '13 at 23:07
@Sylar: If by $e$ you mean the irrational number that is approximately $2.71828...$ then I don't understand what it has to do with the rest of your question. Rather you should take an arbitrary irrational number $L$, $g: \mathbb{R} \rightarrow \mathbb{R}$ by $g(x) = x^2$, and try to use the above facts to show that $f(L) = L^2$. – Pete L. Clark Feb 20 '13 at 23:11
Okay, I think I'm beginning to get it. I let L be an arbitrary irrational number, and from 2) I know that there is a convergent sequence of rationals approaching it, and then use 1) to show that because f is continuous, $f(x_n)$ approaches $f(L)$, so $x_n^2$ approaches $L^2$. After that, I need to use 3) to show that the function $f(x)$ I used in 1) is the same function. Where did you get 3) from? I've never heard anything like it. – JohnDvorak Feb 20 '13 at 23:15
@Sylar: 3) is just the general form of what this argument establishes: you don't need to prove it first. What you have at the moment is $x_n \rightarrow L$ and $x_n^2 = f(x_n) \rightarrow f(L)$. But now, since the function $g(x) = x^2$ is continuous, you can say what $x_n^2$ converges to, and that tells you what $f(L)$ is. – Pete L. Clark Feb 20 '13 at 23:23
By the way, I believe you that you've never heard of 3), and if you're a relative beginner in this area there's nothing wrong with that, but I promise you it's an entirely standard result. It is often phrased more generally, in terms of "dense subsets" of $\mathbb{R}$ (or a topological space...). – Pete L. Clark Feb 20 '13 at 23:25