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

In summary, the problem states that if two non-zero integers, a and b, are in proportion to each other, then their squares are also in proportion to their product. However, the given proof is flawed as it is not specified that the constant of proportionality, k, must be the same for both equations. Additionally, the definition of proportionality used is tautological and only makes sense when the numbers involved are integers. Therefore, the statement cannot be proven and is not true for all real numbers.
  • #1
RChristenk
64
9
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.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
##a = 3, b = 1 \ \Rightarrow \ (a+b) = 2(a -b)##

##a^2 + b^2 = 10, \ ab = 3##
 
  • Like
Likes MatinSAR and FactChecker
  • #3
Have you stated the problem exactly and completely? Maybe I'm missing something but it seems false to me.
 
  • Like
Likes MatinSAR
  • #4
FactChecker said:
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!
 
  • Like
Likes MatinSAR and FactChecker
  • #5
RChristenk said:
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##.
 
  • #6
Mark44 said:
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.
 
  • #7
PeroK said:
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##.
 
  • #8
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.
 
  • #9
RChristenk said:
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##.
 
  • Like
Likes FactChecker and Mark44
  • #10
RChristenk said:
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.
 
  • Like
Likes FactChecker
  • #11
RChristenk said:
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.
 
  • Like
Likes FactChecker
  • #12
PeroK said:
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.
 

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

1. What does the symbol "∝" mean in this context?

The symbol "∝" denotes proportionality. When we say that \( a + b \propto a - b \), it means that \( a + b \) is proportional to \( a - b \), or mathematically, there exists a constant \( k \) such that \( a + b = k(a - b) \).

2. How do you start proving the given variation?

To start proving the variation, we use the given proportionality \( a + b \propto a - b \). This implies that \( a + b = k(a - b) \) for some constant \( k \). Then we manipulate this equation to find a relationship between \( a^2 + b^2 \) and \( ab \).

3. What steps are involved in deriving the relationship between \( a^2 + b^2 \) and \( ab \)?

First, express the given proportionality as \( a + b = k(a - b) \). Then, square both sides to find an expression involving \( a^2 \), \( b^2 \), and \( ab \). Simplify this expression to show that \( a^2 + b^2 \) is proportional to \( ab \).

4. Can you show the detailed mathematical steps to prove the variation?

Sure. Start with \( a + b = k(a - b) \). Squaring both sides gives:\[ (a + b)^2 = k^2(a - b)^2 \]Expanding both sides, we get:\[ a^2 + 2ab + b^2 = k^2(a^2 - 2ab + b^2) \]Rearrange this to:\[ a^2 + 2ab + b^2 = k^2a^2 - 2k^2ab + k^2b^2 \]Collecting like terms, we have:\[ a^2 + b^2 - k^2a^2 - k^2b^2 = -2k^2ab - 2ab \]Simplifying, we get:\[ (1 - k^2)(a^2 + b^2) = -2ab(1 + k^2) \]Therefore, \( a^2 + b^2 \propto ab \).

5. What does the result imply about the relationship between \( a \) and \( b \)?

The result \( a^2 + b^2 \propto ab \) implies that the squares of \( a \) and \( b \), when summed, are proportional to their product. This indicates

Back
Top