# $A_\sqrt{2}$ isn't a Dedekind cut?

In a problem we consider a cut of $\mathbb{Q}$ a subset $A\subset\mathbb{Q}$ that fulfills:

• $A\neq\emptyset$

• $\forall (q,q')\in A\times\mathbb{Q},\,q'<q\Rightarrow q'\in A$

• $\forall q\in A,\,\exists q'\in A,\,q<q'$

In a question of an problem, I'm asked to prove that $A_\sqrt{2}=\left\{p\in\mathbb{Q},\,p^2<2\right\}$ is a cut of $\mathbb{Q}$.

Well since $1\in A_\sqrt{2}$ we have $A_\sqrt{2}\neq\emptyset$.

If we take the numbers $1$ and $-2$, we have $1\in A_\sqrt{2}$ and $(-2)\in\mathbb{Q}$ and $-2w1$, but $(-2)\not\in A_\sqrt{2}$ because $(-2)^2=4\ge 2$ and so:$$\exists(q,q')\in A_\sqrt{2}\times\mathbb{Q},\,q'<q\vee q'\not\in A_\sqrt{2}$$

Then $A_\sqrt{2}$ isn't a cut of $\mathbb{Q}$. Please I'm I wrong? Where's the mistake?

Thank you.

By the way, in this article of Wikipedia about the Construction of the real numbers, it's said:

"A real number $r$ is any subset of the set $\mathbb{Q}$ of rational numbers that fulfills the following conditions:

1. $r$ is not empty

2. $r\neq\mathbb{Q}$

3. $r$ is closed downwards. In other words, for all $x,y^\in\mathbb{Q}$ such that $x<y$, if $y\in r$ then $x\in r$

4. $r$ contains no greatest element. In other words, there is no $x\in r$ such that for all $y\in r$, $y\le x$"

This definition isn't different from the one in the problem, except that $r\neq\mathbb{Q}$ wasn't said, a condition that I think important.

Anyway, after that it's said:

"As an example of a Dedekind cut representing an irrational number, we may take the positive square root of $2$. This can be defined by the set $A=\left\{x\in\mathbb{Q},\,x<0\vee x\times x<2\right\}$.

Here again, if we take $-1$ instead of $1$ and we take $-3$ we find that $A$ isn't a real number.

Note that $-3\notin\{p\in\Bbb Q\mid p^2<2\}$, since $(-3)^2=9>2$. So this set is not downwards closed (doesn't satisfy the second condition in your three conditions). This is a common mistake that one makes when writing an exercise without paying full attention to the details (and it happens to everyone). You should notify whoever assigned this exercise that there is a mistake.
Note that if we take $\{p\in\Bbb Q\mid p^2<2\lor p<0\}$, then this is indeed the cut which defines $\sqrt2$, namely this is the set $A_{\sqrt2}$. It is not empty, it is downwards closed, and it does not have a maximal element.