How to prove that if $-1<x<0$ then $x^2 + x < 0$?

Question

I am trying to prove an equivalence.

I have already proved that:

$$x^2 + x < 0 \implies -1 < x < 0 $$

using a sub-proof by cases, in which I used the fact that when $xy < 0$, $x$ and $y$ have opposite signs. So I just factored the expression and went from there.

However, I can't seem to find an appropriate way of proving the second implication?

$$-1 < x < 0 \implies x^2 + x < 0$$

Answer:

I solved it by determining that x + 1 > 0, then since x < 0, (x)( x + 1 ) < 0 , then $x^2$ + x < 0

Answer 1

Hint: $(x+1)>0$, and multiplying an inequality by a positive number doesn't change the inequality arrows.

Answer 2

To show $-1 < x < 0 \Rightarrow x^2 + x < 0$, observe that, since $0 < \vert x \vert < 1$, we have $0< \vert x \vert^2 < \vert x \vert$; to see this, just multiply the inequality $0 < \vert x \vert < 1$ through by $\vert x \vert$. Next, note that for $x$ in the range $-1 < x < 0$, we have $\vert x \vert^2 = x^2$ and $\vert x \vert = -x$. Thus $x^2 + x = \vert x \vert^2 - \vert x \vert < 0$ since $\vert x \vert^2 < \vert x \vert$. QED

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!