Prove or disprove help

• Mar 27th 2008, 06:29 PM
jconfer
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.
• Mar 27th 2008, 07:00 PM
WWTL@WHL
Quote:

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 $\displaystyle y \in \mathbb{R}$ such that

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

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

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

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

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

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

EDIT: I don't know why I used y instead of x. Sorry.
• Mar 27th 2008, 07:16 PM
teuthid
Does this work?
There are three cases that need to be considered: $\displaystyle |x| \leq 1$ and $\displaystyle x \geq 1$ and $\displaystyle x \leq -1$

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

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

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

Our three cases cover all possible real values of x, and in each case $\displaystyle x^2+x+1>0$. So we have exhaustively demonstrated that there is no real number x such that $\displaystyle x^2+x+1 \leq 0$
• Mar 27th 2008, 08:06 PM
jconfer
thank you both for your help.