# Thread: Prove or disprove help

1. ## Prove or disprove help

I can't firgure out how to show that this prove or disprove problem is false:

Prove or disprove that there exists a real number x such that x^2 + x + 1 ≤ 0.

As i said i know the problem is false for every number. I just don't know how to show it. I thought maybe by cases with x ≤ 0 and x ≥ 0.

2. Originally Posted by jconfer
I can't firgure out how to show that this prove or disprove problem is false:

Prove or disprove that there exists a real number x such that x^2 + x + 1 ≤ 0.

As i said i know the problem is false for every number. I just don't know how to show it. I thought maybe by cases with x ≤ 0 and x ≥ 0.
Assume there exists a $y \in \mathbb{R}$ such that

$(y^2+y) \leq -1$

Then either:
(i) $y\geq 0$
(ii) $-1
(ii) $y \leq -1$

Consider (i), i.e. $y \geq 0$
Then since y is positive, y+1 is certainly positive $\Rightarrow y(y+1) > 0$ Contradiction. Hence y<0

Consider (ii), i.e. $-1
Then $y^2 + y \leq -1$
$y(y+1) \leq -1$
But since y > -1, we certainly have that y+1 is positive and $0 $\Rightarrow y(y+1) > -1$ Contradiction. Hence y must be less than or equal to 1.

Consider (iii), i.e. $y \leq -1$
Then $y \leq -1 \Rightarrow y+1 <0 \Rightarrow y(y+1) > 0$. Contradiction.

Hence, there exists no such $y \in \mathbb{R}$.

EDIT: I don't know why I used y instead of x. Sorry.

3. ## Does this work?

There are three cases that need to be considered: $|x| \leq 1$ and $x \geq 1$ and $x \leq -1$

Case 1:
$|x| < 1$
$\rightarrow x+1 > 0$
since $x^2 \geq 0$ for all real numbers, we can conclude that $x^2+x+1 > 0$

Case 2:
$x \geq 1$
$\rightarrow x^2 +x \geq 1^2+1 > 0$
$\rightarrow x^2+x+1>0$

Case 3:
$x \leq -1$
$\rightarrow x^2 \geq 1$
$\rightarrow x^2 +x \geq 1 + (-1) \geq 0$
$\rightarrow x^2+x+1>0$

Our three cases cover all possible real values of x, and in each case $x^2+x+1>0$. So we have exhaustively demonstrated that there is no real number x such that $x^2+x+1 \leq 0$

4. thank you both for your help.