1. Spivak's Calculus - Proof

(d) For which numbers $\displaystyle a$ is it true that $\displaystyle x^2 + axy + y^2 > 0$ whenever $\displaystyle x$ and $\displaystyle y$ are not both $\displaystyle 0$?

$\displaystyle (x + \frac{ay}{2})^2 + (\frac{\pm\sqrt{-((ay)^2 - 4y^2)}}{2})^2 > 0$ for $\displaystyle (ay)^2 - 4y^2 < 0$,
$\displaystyle \Rightarrow (x + \frac{ay}{2})^2 + (\frac{\pm\sqrt{-(ay)^2 + 4y^2)}}{2})^2 > 0$,
$\displaystyle \Rightarrow (x + \frac{ay}{2})^2 + \frac{-(ay)^2 + 4y^2}{4} > 0$,
$\displaystyle \Rightarrow (x + \frac{ay}{2})^2 + \frac{-(ay)^2}{4} + \frac{4y^2}{4} > 0$,
$\displaystyle \Rightarrow (x + \frac{ay}{2})^2 + \frac{-(ay)^2}{4} + y^2 > 0$,
$\displaystyle \Rightarrow x^2 + axy + \frac{(ay)^2}{4} + \frac{-(ay)^2}{4} + y^2 > 0$,
$\displaystyle \Rightarrow x^2 + axy + y^2 > 0$ for $\displaystyle (ay)^2 - 4y^2 < 0$.

$\displaystyle (ay)^2 - 4y^2 < 0$,
$\displaystyle \Rightarrow (ay)^2 < 4y^2$,
$\displaystyle \Rightarrow a^2y^2 < 4y^2$,
$\displaystyle \Rightarrow a^2 < 4$,
$\displaystyle \Rightarrow a < \pm2$.
Therefore, $\displaystyle x^2 + axy + y^2 > 0$ for $\displaystyle a < \pm2$.

Is this proof valid? When I sub in a value less than $\displaystyle -2$ for $\displaystyle a$, I end up with a result contrary to the conclusion of my proof. For instance:

$\displaystyle x^2 + axy + y^2 = (-5)^2 + (-100)(-5)(-10) + (-10)^2 = -4,875 > 0$.

2. Re: Spivak's Calculus - Proof

This is what I might do. Divide by $\displaystyle x^2$ so

$\displaystyle \dfrac{y^2}{x^2} + a \dfrac{y}{x} + 1 > 0$

so

$\displaystyle \left( \dfrac{y}{x} + \dfrac{a}{2}\right)^2 > \dfrac{a^2}{4} - 1$

To guarantee this for all x and y not zero, then we would require that $\displaystyle \dfrac{a^2}{4} - 1 < 0$ so $\displaystyle -2 < a < 2$.

3. Re: Spivak's Calculus - Proof

Your proof seems fine, but when I sub in a number for $\displaystyle a$ that is less than $\displaystyle -2$, such as $\displaystyle -2.1$, I get the following:

$\displaystyle x^2 + axy + y^2 = (-5)^2 + (-2.1)(-5)(-10) + (-10)^2 = 20 > 0$.

So although the statement above is true, it can't be true that the statement is only true for $\displaystyle -2 < a < 2$.

4. Re: Spivak's Calculus - Proof

But don't you want your statement to be true for all x and y not zero. For example, if you choose $\displaystyle a = -2.1$ (like you said) then what you want is

$\displaystyle x^2 - 2.1xy + y^2 > 0$ but if you choose $\displaystyle x = y = 1$ then

$\displaystyle 1 - 2.1 + 1 = -.1 \ngtr 0$

5. Re: Spivak's Calculus - Proof

Originally Posted by Danny
But don't you want your statement to be true for all x and y not zero. For example, if you choose $\displaystyle a = -2.1$ (like you said) then what you want is

$\displaystyle x^2 - 2.1xy + y^2 > 0$ but if you choose $\displaystyle x = y = 1$ then

$\displaystyle 1 - 2.1 + 1 = -.1 \ngtr 0$
Yes, the statement must be true for all $\displaystyle x$and $\displaystyle y$ not $\displaystyle 0$, as the problem requests. But the problem mainly asks for which numbers $\displaystyle a$ is the statement true. Of course, the statement is true if $\displaystyle -2 < a < 2$, but clearly the statement isn't $\displaystyle only$ true for those numbers. And, if I interpreted the question correctly (please correct me if I'm wrong), what is required is that I find all $\displaystyle a$ for which the statement is true. If I could just state a portion of that, I could easily say that the statement is true for one arbitrary number, such as $\displaystyle 1$, and that alone would qualify as a complete solution.

I'll be logging off now, but I'll be back later.

I've been trying to solve this problem for quite some time now. :/

6. Re: Spivak's Calculus - Proof

Originally Posted by RogueDemon
$\displaystyle \Rightarrow a^2 < 4$,
$\displaystyle \Rightarrow a < \pm2$.
ehm...

7. Re: Spivak's Calculus - Proof

Originally Posted by RogueDemon
Yes, the statement must be true for all $\displaystyle x$and $\displaystyle y$ not $\displaystyle 0$, as the problem requests. But the problem mainly asks for which numbers $\displaystyle a$ is the statement true. Of course, the statement is true if $\displaystyle -2 < a < 2$, but clearly the statement isn't $\displaystyle only$ true for those numbers. And, if I interpreted the question correctly (please correct me if I'm wrong), what is required is that I find all $\displaystyle a$ for which the statement is true. If I could just state a portion of that, I could easily say that the statement is true for one arbitrary number, such as $\displaystyle 1$, and that alone would qualify as a complete solution.

I'll be logging off now, but I'll be back later.

I've been trying to solve this problem for quite some time now. :/
Are you saying that the following statement is true?

$\displaystyle x^2-2.1xy+y^2>0$ whenever x and y are not both zero.