Find exact upper and lower bounds of set

Question

Heloo, i have a set

A={x-x^2>0}

x belongs to rationals

Need to find exact lower and upper bounds of this set. Tried to draw a graphic but it didnt worked. Any hints or solutions will be apreciated.

Answer 1

We can write the condition as

$x (1 - x) > 0$.

When the inequality holds, the factors $x$ and $1 - x$ cannot both be negative, so they must both be positive, that is, both $x > 0$ and $1 - x > 0$. Rearranging gives that the latter is equivalent to $x < 1$, so $A$ is just $(0, 1) \cap \mathbb{Q}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there are points in the $A$ arbitrarily close to $0$ and $1$, which are thus respectively the greatest lower bound and least upper bound for $A$.

Answer 2

$0$ and $1$.

You know that $x(1-x) > 0 \Leftrightarrow ( x > 0 \wedge 1-x > 0 ) \vee ( x < 0 \wedge 1- x < 0 ) \ \Leftrightarrow 0 < x < 1 \vee 0 > x > 1 \Leftrightarrow x \in ( 0, 1 ) \cap \mathbb{Q}$ which is a dense set in $(0, 1)$

So supremum and infimum of this set is $1$ and $0$