Least Upper Bound of a set.

Question

I'd like to confirm my answer for the following problem. I thought I had done it correctly, but a friend of mine got a different answer and would like to see where I (or he) went wrong.

Prove that $A=\{ x \in \Bbb R: x^2 < 1-x \}$ is nonempty and bounded above. Prove what its least upper bound is.

My attempt is as follows:

Fix x in A. Thus, $$x^2 <1-x \Rightarrow x^2+x<1$$ This suggests three cases.

Case 1: $(x>0)$ $$ x^2+x >0 \ \ \Rightarrow \ \ 0

Any help is greatly appreciated, thank you.

Answer 1

You showed 1 is the upper bound for $x^2 + x$ not $A$.

Answer 2

From what I see, case I is not finished, Case II is OK, Case III is wrong since $x^{2}+x$ can be negative, even if $x<0$.

However, you don't need to consider these cases. You just have $x^{2}+x-1<0$. It has solutions $x\in (\frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2})$, which is clearly bounded. Search 'solving quadratic inequalities' if you don't know how exactly you can get this answer.