# Prove Existence of Real Number

1. May 18, 2014

### knowLittle

1. The problem statement, all variables and given/known data
a.) Prove $\exists x, x \in \mathbb{R} | x^3 -x^2 = 5$
I know that x = 2.1163, but how do I find this without a calculator?

b.) Prove that $\not \exists x, x \in \mathbb{R} | x^4 - 2x^2 +2 =0$

3. The attempt at a solution
x^2 ( x-1 )=5 for part a)?

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b.)
I find it easier. If x =0, then it's false.
$x^2 ( x^2 -2 )+2 = 0 \\ x^2 \geq 0 \\ x =1 \vee -1 \\ 1(-1) +2 =1 \neq 0$
Then since x is only greater than 1, it is false that x exists that satisfies the equation.

What do you think?

Last edited: May 18, 2014
2. May 18, 2014

### haruspex

You're not asked to find its value, only to prove such a number exists.
Have you typed the question correctly? Should it say $\nexists$?
I cannot even begin to follow your reasoning. Why must x be 1 or -1?
Try applying the usual formula for solving quadratics.

3. May 18, 2014

### knowLittle

I said 1 or -1 because I was trying the base case after 0, but yeah let me try the quadratic formula.

4. May 18, 2014

### knowLittle

Can I do this to the quadratic equation? :
$x^4 - 2x^2 +2 = 0 \\ (x^2)^2 - 2(x^2) +2 =0 \\$
And, I solve for $x^2 = -1 +- \sqrt{-1}$

5. May 18, 2014

### haruspex

So is there a real x satisfying that? If not, why not?

6. May 18, 2014

### knowLittle

$\not\exists x | x \in \mathbb{R},$ since $x = \sqrt{x^2} = \sqrt{ -1 +- \sqrt{-1} }, x \in \mathbb{C}$

7. May 18, 2014

### haruspex

$x \in \mathbb{C}$ does not prove it is not in $\mathbb{R}$.
Work backwards. Suppose x is real. What does that tell you about x2?

8. May 18, 2014

### LCKurtz

Try using the intermediate value theorem.

Try completing the square with $x^2$ as the variable.

9. May 19, 2014

### knowLittle

Thanks, Kurtz!

$(x^2 - 1)^2 = -1$, since the square of any number is always non-negative, it is impossible that x is real.

10. May 19, 2014

### Ray Vickson

What tools are you allowed to use? Calculus makes these easy.

In (a), let $f(x) = x^3-x^2$. We have $f(0) = 0 < 5$ but $f(x) \to \infty$ for $x \to \infty$, so $f(x_0) > 5$ for some large-enough $x_0$. By continuity there is an $x \in(0,x_0)$ giving $f(x) = 5$.

For (b), look at $f(x) = x^4 - 2 x^2 = x^2(x^2-2)$. This a quadratic in $y = x^2$, whose minimum is at $y = 1$, giving $f_{\min} = f(1) = -1>-2$. Therefore, it is impossible to find any $x$ giving $f(x) = -2$.

11. May 19, 2014

### Curious3141

Are you familiar with Descartes' Rule of Signs? It gives a very quick solution here.

Treat that as a quadratic equation in $x^2$ and consider the discriminant.