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Let $p,q\in\mathbb{R}$ \ $\mathbb{Q}$, then $p-q$ is irrational.

This means that: $p,q\in\mathbb{R}$ \ $\mathbb{Q}$ $\Rightarrow p,q\in\mathbb{R}$ such that $p,q\notin \mathbb{Q}$

Thus, $p,q$ are irrational.

Proof by contradiction: Let's assume $p-q$ is rational, where $p=\frac { a }{ b }$ and $q=\frac { m }{ n } $ such that $a,b,m,n\in \mathbb{Z}$ and $b\neq 0$, $n\neq 0$

$$\frac { a }{ b } -\frac { m }{ n } \quad =\frac { an-mb }{ bn } $$

At this point, I get stuck because I don't know how to properly prove this from here. I feel like I had a good idea, but I quickly realized that it won't actually disprove the claim. Proofs are difficult for me because I have an idea of what I want to prove, but it's hard to organize my thoughts. What I do know is that $p-q$ being irrational is not always the case. The simplest counter example is if $p=q$ and $p=\sqrt { 2 } $. This would lead to $\sqrt { 2 } -\sqrt { 2 } =0$ which is rational.

How do I organize my thoughts and proceed?

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  • $\begingroup$ You need to start out by having a statement that is true before you prove it. It's a good sign that you got stuck and couldn't complete the proof. Maybe you meant to prove that if $p\in\mathbb Q$ and $q\in\mathbb R\setminus\mathbb Q$ then $p-q\in\mathbb R\setminus\mathbb Q$ or something similar? $\endgroup$ – skyking Oct 16 '15 at 11:45
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The statement is false; $\sqrt{2} - \sqrt{2} = 0$, which is rational.

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  • $\begingroup$ I understand that it is false. I think the best way to proceed is to disprove the claim by using a counter example. I just can't seem to organize my thoughts in any way to do so with a complete and formal mathematical proof. $\endgroup$ – Cherry_Developer Oct 16 '15 at 11:48
  • $\begingroup$ I have presented a counterexample. To prove that it is false that $x + 3 = 5$ for all $x \in \mathbb{R}$, it suffices to say that $x := 1$ is a counterexample to the statement. :) $\endgroup$ – Megadeth Oct 16 '15 at 11:52
  • $\begingroup$ One example is good enough? I don't need to prove it using algebraic manipulations and variables? $\endgroup$ – Cherry_Developer Oct 16 '15 at 11:53
  • $\begingroup$ For some propositions it may be difficult to construct a counterexample, but for simple statements often times a counterexample is obvious. Ask yourself this question: is $1 + 3 = 5$? If not, then $1$ is a counterexample to the statement. $\endgroup$ – Megadeth Oct 16 '15 at 11:55
  • $\begingroup$ Try not to be kidnapped by formal logic. $\endgroup$ – Megadeth Oct 16 '15 at 11:55

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