# Thread: Spivak's Calculus - Proof

1. ## Spivak's Calculus - Proof

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

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

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

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

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

2. ## Re: Spivak's Calculus - Proof

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

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

so

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

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

3. ## Re: Spivak's Calculus - Proof

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

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

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

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 $a = -2.1$ (like you said) then what you want is

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

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

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 $a = -2.1$ (like you said) then what you want is

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

$1 - 2.1 + 1 = -.1 \ngtr 0$
Yes, the statement must be true for all $x$and $y$ not $0$, as the problem requests. But the problem mainly asks for which numbers $a$ is the statement true. Of course, the statement is true if $-2 < a < 2$, but clearly the statement isn't $only$ 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 $a$ 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 $1$, 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
$\Rightarrow a^2 < 4$,
$\Rightarrow a < \pm2$.
ehm...

7. ## Re: Spivak's Calculus - Proof

Originally Posted by RogueDemon
Yes, the statement must be true for all $x$and $y$ not $0$, as the problem requests. But the problem mainly asks for which numbers $a$ is the statement true. Of course, the statement is true if $-2 < a < 2$, but clearly the statement isn't $only$ 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 $a$ 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 $1$, 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?

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