Proof $x+a \in J$ if $J := \mathbb{R}\setminus\mathbb{Q}$ and $x \in J$, $a \in \mathbb{Q}$

Question

Let $J := \mathbb{R}\setminus\mathbb{Q}$.

How can one prove that for all $x\in J$ and $a \in \mathbb{Q}$ it holds that $x+a \in J$?

And for all $x \in J$, $a \in \mathbb{Q}$, $a \neq 0$ it holds that $ax \in J$?

If I take $x = 2$ and $a = \frac{1}{3}$ then $x+a$ would be $2\frac{1}{3}$, but why is it in $\mathbb{R}\setminus\mathbb{Q}$?

I think that the Archimedean property needs to be applied here, but I don't know how. It states that if $x$ is an element of $K$, then there exists a natural number n such that $n > x$.

If both are proven, does that mean that $J$ lies dense in $\mathbb{R}$?

Answer 1

If $x+a\in\mathbb{Q}$, then $$ x=(x+a)-a\in\mathbb{Q} $$ which is a contradiction to $x\in\mathbb{R}\setminus\mathbb{Q}$.

Similarly, if $ax\in\mathbb{Q}$, then $$ x=\frac{ax}{a}\in\mathbb{Q} $$

The key fact is that $\mathbb{Q}$ is closed under subtraction and division (by nonzero rationals).

Your proposed counterexample doesn't work because $2\in\mathbb{Q}$, so $2\mathrel{\color{red}{\notin}}J$.

Why is $J$ dense in $\mathbb{R}$? Prove that, given $a,b\in\mathbb{Q}$, with $a<b$, there exists $x\in J$ such that $a<r<b$.

First we prove this for $a=0$ and $b=1$: easy, because $r=\sqrt{2}/2$ satisfies the requirement.

Now $a<r<b$ is the same as $0<r-a<b-a$, which is the same as $$ 0<\frac{r-a}{b-a}<1 $$ Use the previous results to finish up.

Answer 2

one can assume that $a+x\in\mathbb{Q}$ , and therefore $x+\frac{m}{n}=\frac{p}{q}$, where $\frac{m}{n}$ is the representation of $a$ as a rational number, and $\frac{p}{q}$ is the representation of the result. then we get $x=\frac{p}{q}-\frac{m}{n}\in\mathbb{Q}$, which is a contradiction. the exact same argument can be made for multiplication.

Answer 3

Your set $J:= \mathbb{R}\setminus\mathbb{Q}$ means, that all contents of $J$ are irrational, so arbitary $x \in J$ cannot be represented as a fraction $\frac{p}{q}$, while $p, q\in \mathbb{Z}$.

1. $x+a\in J$ for $a\in \mathbb{Q}$ - let's take any $x \in J$ and add to it any rational number $a$, which can be represented as a fraction of two integers. Then the expression $x+a = x+\frac{a_{num}}{a_{den}} \in J$. You can visualise it by taking $x=\sqrt{2}$ and then try adding/substracting any number - the number you get will always have an irrational number in its nominator, which means that $x+a\notin \mathbb{Q}$.
2. $xa\in J$ for $a\in \mathbb{Q}$ - similarly to the first point.

When it comes to the example you gave, so $x=2$ and $a=\frac{1}{3}$, you are assuming wrong $x$, because your $x$ is not irrational number, thus $x+a=2\frac{1}{3}=\frac{7}{3}\in \mathbb{Q}$.