Prove this variation: If a+b ∝ a-b, prove that a^2+b^2 ∝ ab

[RChristenk]

Homework Statement

If $a+b \varpropto a-b$, prove that $a^2+b^2 \varpropto ab$

Relevant Equations

Basic fractions and algebra

$a+b \varpropto a-b$ means $a+b = k_1(a-b)$

$(a+b)^2=k_1^2(a-b)^2$

$a^2+b^2+2ab=a^2k_1^2+b^2k_1^2-2abk_1^2$

$2ab(1+k_1^2)=a^2(k_1^2-1)+b^2(k_1^2-1)$

$2ab(k_1^2+1)=(a^2+b^2)(k_1^2-1)$

$a^2+b^2=ab(\dfrac{2k_1^2+2}{k_1^2-1})$

Since $a^2+b^2 \varpropto ab$ means $a^2+b^2=abk_2$, compared to what I got, I can say $k_2 = \dfrac{2k_1^2+2}{k_1^2-1}$. Therefore $a^2+b^2 \varpropto ab$ has been proven. Is this a correct? Thanks.

 

[PeroK]

$a = 3, b = 1 \ \Rightarrow \ (a+b) = 2(a -b)$

$a^2 + b^2 = 10, \ ab = 3$

 

[FactChecker]

Have you stated the problem exactly and completely? Maybe I'm missing something but it seems false to me.

 

[PeroK]

Have you stated the problem exactly and completely? Maybe I'm missing something but it seems false to me.

It's definitely false, given I provided a counterexample!

 

[Mark44]

Homework Statement: If $a+b \varpropto a-b$, prove that $a^2+b^2 \varpropto ab$
Relevant Equations: Basic fractions and algebra

$a+b \varpropto a-b$ means $a+b = k_1(a-b)$

What seems to be missing in the problem statement here is the idea that $a^2 + b^2$ must be shown to be in the same proportion to ab as a + b is to a - b.

IOW, if a + b = k(a - b), then $a^2 + b^2 = kab$.

 

[PeroK]

What seems to be missing in the problem statement here is the idea that $a^2 + b^2$ must be shown to be in the same proportion to ab as a + b is to a - b.

IOW, if a + b = k(a - b), then $a^2 + b^2 = kab$.

That doesn't seem possible.

 

[Mark44]

That doesn't seem possible.

I agree. My comment was about I believe what the intent of the problem was, taking into account your counterexample. As a very simple example of what I am talking about, show that if $x \varpropto y$ then $2x \varpropto 2y$. In the two resulting equations, the same constant applies to both.

What the OP is doing is something like this:
Show that $x \varpropto y \Rightarrow x^2 \varpropto y^2$.
$x \varpropto y \Rightarrow x = ky$ for some nonzero k
$x = ky \Rightarrow x^2 = k^2y^2 \Rightarrow x^2 = K y^2 \Rightarrow x^2 \varpropto y^2$

But if x = 4 and y = 2, then x = 2y, but $x^2 = 16 \ne 8 = 2y^2$.
x and y aren't in the same proportion as $x^2$ and $y^2$.

 

[RChristenk]

The question just states if $a+b \varpropto a-b$, prove that $a^2+b^2 \varpropto ab$.

I think the constants are different. So $a+b = k_1(a-b)$ and $a^2+b^2 = k_2ab$.

$\Rightarrow (a+b)^2=k_1^2(a-b)^2$

$\Rightarrow a^2+2ab+b^2=a^2k_1^2-2abk_1^2+b^2k_1^2$

Now using $a^2+b^2 = k_2ab$ and replacing the left side of the above equation, get:

$\Rightarrow k_2ab+2ab=a^2k_1^2-2abk_1^2+b^2k_1^2$

$\Rightarrow k_2ab+2ab+2abk_1^2=(a^2+b^2)k_1^2$

$\Rightarrow ab(k_2+2+2k_1^2)=(a^2+b^2)k_1^2$

$\Rightarrow a^2+b^2 = ab\dfrac{(2k_1^2+k_2+2)}{k_1^2}$

So since $\dfrac{(2k_1^2+k_2+2)}{k_1^2}$ is a new constant of variation, $a^2+b^2 \varpropto ab$ is proven. That's just what I'm thinking.

 

[renormalize]

The question just states if $a+b \varpropto a-b$, prove that $a^2+b^2 \varpropto ab$.

I think the constants are different. So $a+b = k_1(a-b)$ and $a^2+b^2 = k_2ab$.

$\Rightarrow (a+b)^2=k_1^2(a-b)^2$

$\Rightarrow a^2+2ab+b^2=a^2k_1^2-2abk_1^2+b^2k_1^2$

Now using $a^2+b^2 = k_2ab$ and replacing the left side of the above equation, get:

$\Rightarrow k_2ab+2ab=a^2k_1^2-2abk_1^2+b^2k_1^2$

$\Rightarrow k_2ab+2ab+2abk_1^2=(a^2+b^2)k_1^2$

$\Rightarrow ab(k_2+2+2k_1^2)=(a^2+b^2)k_1^2$

$\Rightarrow a^2+b^2 = ab\dfrac{(2k_1^2+k_2+2)}{k_1^2}$

So since $\dfrac{(2k_1^2+k_2+2)}{k_1^2}$ is a new constant of variation, $a^2+b^2 \varpropto ab$ is proven. That's just what I'm thinking.

But this definition of "proportionality" is tautological. Any two quantities $x$ and $y$ are "proportional" ($y=kx$) if one is free to set $k=y/x$. A proper proportionality requires that $k$ be independent of $x,y$.

 

[PeroK]

The question just states if $a+b \varpropto a-b$, prove that $a^2+b^2 \varpropto ab$.

I think the constants are different. So $a+b = k_1(a-b)$ and $a^2+b^2 = k_2ab$.

$\Rightarrow (a+b)^2=k_1^2(a-b)^2$

$\Rightarrow a^2+2ab+b^2=a^2k_1^2-2abk_1^2+b^2k_1^2$

Now using $a^2+b^2 = k_2ab$ and replacing the left side of the above equation, get:

$\Rightarrow k_2ab+2ab=a^2k_1^2-2abk_1^2+b^2k_1^2$

$\Rightarrow k_2ab+2ab+2abk_1^2=(a^2+b^2)k_1^2$

$\Rightarrow ab(k_2+2+2k_1^2)=(a^2+b^2)k_1^2$

$\Rightarrow a^2+b^2 = ab\dfrac{(2k_1^2+k_2+2)}{k_1^2}$

So since $\dfrac{(2k_1^2+k_2+2)}{k_1^2}$ is a new constant of variation, $a^2+b^2 \varpropto ab$ is proven. That's just what I'm thinking.

Apart from when zero is involved, all real numbers are proportional to each other. This question only makes sense if the numbers involved are integers. In this case, if $a, b \ne zero$, then:
$$a^a + b^2 = (\frac{a^2 + b^2}{ab})ab$$And there is no need to do any calculations. And nothing to prove.

Moreover, $a + b = k(a - b)$ is true for all real numbers $a, b$. So, there is no "if" involved. It's only when $a, b$ and $k$ are integers that this condition makes sense.

 

[PeroK]

Homework Statement: If $a+b \varpropto a-b$, prove that $a^2+b^2 \varpropto ab$

Perhaps $a$ and $b$ are supposed to be variables or parameters? Then $a+b \varpropto a-b$ implies $a \varpropto b$ and the result is trivial.

I originally assumed it was an integer equation. The question needs some context.

 

[FactChecker]

Perhaps $a$ and $b$ are supposed to be variables or parameters? Then $a+b \varpropto a-b$ implies $a \varpropto b$ and the result is trivial.

IMO, this is the answer that the problem had in mind.