Solve Ineq. in Spivak Calculus 4(ii): 5-x2>8

[Saladsamurai]

Homework Statement

I am doing the HW in Spivak's calculus (problem 4 (ii) ) on inequalities. The problem statement is:

find all x for which

5-x² > 8

The Attempt at a Solution

I know this is a simple problem, but bear with me for a moment. I want someone who is familiar with Spivak to tell me what the right way to do this is. Pretend for a moment that I know virtually nothing (it shouldn't be too hard ) and that all that I know comes from the first chapter in Spivak's textbook. I start the solution like this

5-x²<8

5-x² + x² -8 < 8 + x² - 8

x² > -3Now, I am not sure how to 'rigorously' finish the solution. It is clear that this is true for all numbers x. Is it enough to say that

x² = x*x and since we already showed in a previous example that ab > 0 if a,b > 0 OR a,b < 0.

Is that the RIGHT way to do this? Thanks. I am still trying to get a feel for this text and to answer the problems without using any of my prior knowledge.

 

[jbunniii]

You've written two different inequalities that don't agree:

$$5-x^2 >-3$$
and
$$5 - x^2 < 8$$

Which one is correct?

In general, try rearranging the inequality so that you get all the terms on one side and zero on the other side.

For example, you might end up with something like

$$x^2 - 3x + 2 > 0$$

Then factor the polynomial into a product of monomials:

$$(x - 1)(x - 2) > 0$$

and apply what you know about what ab > 0 implies.

 

[jbunniii]

P.S. If you really do end up with

$$x^2 > -3$$

then as you say, it's obviously true for all x. I'm sure Spivak proves somewhere that

$$x^2 \geq 0$$

for all real x. And clearly 0 > -3, right? Do you have another theorem that you can use to chain together these two inequalities?

 

[Saladsamurai]

P.S. If you really do end up with

$$x^2 > -3$$

then as you say, it's obviously true for all x.

Yes. I edited the OP.

I'm sure Spivak proves somewhere that

$$x^2 \geq 0$$

for all real x. And clearly 0 > -3, right? Do you have another theorem that you can use to chain together these two inequalities?

Yes, we proved that x² >= 0 somewhere. Should I have another theorem? Probably huh ... Probably something to the affect of if a>b and b>c then a>c. Is that what you are going for? Thanks for the help jbunniii.

Also, out of curiosity, what is the 'proper' way to present the answer? Would it be to restate the inequality along with the interval on which it is valid? i.e.

$$5-x^2 > 8 \qquad x\in\mathbb{R}$$

 

[jbunniii]

You could write something like

$$\{ x : 5 - x^2 < 8\} = \mathbb{R}$$

which in words means "the set of numbers x satisfying the inequality is all of $\mathbb{R}$"

Probably something to the affect of if a>b and b>c then a>c. Is that what you are going for?

That's the one.

 

[Saladsamurai]

You could write something like

$$\{ x : 5 - x^2 < 8\} = \mathbb{R}$$

which in words means "the set of numbers x satisfying the inequality is all of $\mathbb{R}$"



That's the one.

Okie dokie! Thanks again