Good start! The idea is there. To make the proof precise, I'd do two things:
- Very skeptically home in on phrases like, "Choose $n, p$ such that..." and "Now introduce $p_1$ such that...". Why can you do that? What exactly are these numbers you're choosing and introducing?
- Make the phrase "better estimate" more precise. For example, how do you know your better estimate gets you arbitrarily close? The sequence $1.1 + 2^{-n}$ is successively better approximations of $1$, but it never actually gets arbitrarily close.
To start you off, I'll restate the proposition in question:
Let $R$ be an arbitrary real number. For every $\epsilon > 0$, there exist $n, p$ such that $$\bigg| R - \frac{p}{2^{n/2}}\bigg| < \epsilon. $$
Now your job is to address the challenge, "Here is $\epsilon$. What $p, n$ have you got for me?"
You are trying to proof by contrapositive that for all $x,y\in\mathbb{R},$ if $x$ is rational and $y$ is irrational then $x+y$ is irrational.
The contrapositive of this statement is
For all $x, y \in \mathbb{R},$ if $x+y$ is rational, then $x$ irrational or $y$ is rational.
Using logic notation, let $P,Q,R$ be statements, note that
$$P \to (Q \vee R) \iff (P \wedge \neg Q) \to R.$$
Hence to prove this statement, you can suppose $P$ and $\neg Q$, and derive $R.$ And you can do it using the proof by contradiction, you assume that $P$ and $\neg Q$ are true and $R$ is false and then derive a contradiction. This proves that $(P \wedge \neg Q) \to R$ is true, which is equal to the contrapositive we were ask to prove. Then we are done.
Proof: Let $x, y \in \mathbb{R},$ such that $x+y$ and $x$ are rational. Suppose that $y$ is irrational. Let’s derive a contradiction. Since $x+y$ and $x$ are rational, then there are integers $a,b,c,d$ with $b \neq 0 \neq d$ such that $x+y=\frac{a}{b}$ and $x=\frac{c}{d}.$ Substituting $x$ in $x + y,$ we get
$$\frac{c}{d}+y=\frac{a}{b}.$$
Simplifying this expression, we get that $y=\frac{ad-bc}{bd},$ where $ad-bc$ and $bd$ are integers, with $bd \neq 0.$ Therefore $y$ is rational, which is a contradiction. Hence we prove that, if $x+y$ is rational, then $x$ irrational or $y$ is rational. By contrapositive, we conclude that if $x$ is rational and $y$ is irrational then $x+y$ is irrational. $\square$
Best Answer
The statement you're trying to prove is $\forall a,b\, (a+b\notin \Bbb{Q} \implies a\notin \Bbb{Q} \text{ or } b \notin \Bbb{Q})$. This is simply the symbolic translation of the statement "for every $a,b$, if $a+b$ is irrational then atleast one of $a$ or $b$ is irrational".
Here, the statement $X$ is "$a+b\notin \Bbb{Q}$", and the statement $Y$ is "$a\notin \Bbb{Q} \text{ or } b \notin \Bbb{Q}$". So, the contrapositive of "for every $a,b$ ($X \implies Y$)" is "for every $a,b$ $(\neg Y \implies \neg X)$", which in this case is:
and this is what you argued.