Prove or disprove: For every number $s \in \mathbb{R} \setminus \mathbb{Q}$ there is a $t \in \mathbb{R}$ with $s+t \in \mathbb{Q}$

Question

Prove or disprove: For every number $s \in \mathbb{R} \setminus \mathbb{Q}$ there is a $t \in \mathbb{R}$ with $s+t \in \mathbb{Q}$.

First of all, I thought what exactly means $s\in\mathbb{R}\setminus\mathbb{Q}$. I think it means all numbers $\in \mathbb{R} \setminus \mathbb{N}, \mathbb{Z}, \mathbb{Q}$.

So what will remain to use then? The irrational numbers.

Let $s=\sqrt{2}$ and let $t=0$. The result will be $\sqrt{2}$ which is $\notin \mathbb{Q}$.

Thus the statement is wrong.

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Did I do everything correctly?

Answer 1

The statement is correct. One (of many) very simple way of doing this is taking $t=-s$.

Answer 2

You have not disproved the statement.

The statement is: "For every irrational number $s$, there is some number $t$ such that $s+t$ is rational," not ". . . for every number $t$, $s+t$ is rational." If you want to disprove the statement, you need to find some irrational $s$ such that no matter what $t$ you pick, $s+t$ is always irrational.

This will be difficult to do, since the statement is in fact true. HINT: given $s$, can you think of a number which when added to $s$ gives $0$? How about $2$? Or ${1\over 3}$? Or . . .

Answer 3

Let s be irrational number.

Take t=-s .

Then s+t=0.