Proof

[Thales12345]

GIVEN:
a, b and c are rational numbers, different from 0
a + b + c = 0

TO PROVE:
1/a^2 + 1/b^2 +1/c^2 is the square of a rational number

Does anyone have an idea how I should start on this? I have no clue...

 

[Cubist]

GIVEN:
a, b and c are rational numbers, different from 0
a + b + c = 0

Then [MATH]c = -\left( \frac{a_1}{a_2} + \frac{b_1}{b_2} \right)[/MATH] where \(a_1,a_2,b_1,b_2\) are integers

And continue...

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[Thales12345]

I don’t understand why that could be useful.
I’ve come to this:
(a2/a1)^2 + (b2/b1)^2 + (c2/c1)^2

maybe I can prove it by contradiction?

 

[Cubist]

I’ve come to this:
(a2/a1)^2 + (b2/b1)^2 + (c2/c1)^2

I was thinking that instead of (c2/c1)^2, you'd use the expression for c that is in post#2 (but first - you'd need to put it over a common denominator).

Then, the goal is to express 1/a^2 + 1/b^2 +1/c^2 as (list of factors#1) / (list of factors#2), and hopefully both of the lists of factors will contain each factor twice (therefore it will be something squared). I must admit that I have not done the whole question myself, but this is the strategy that I would attempt.

 

[Dr.Peterson]

I don’t understand why that could be useful.
I’ve come to this:
(a2/a1)^2 + (b2/b1)^2 + (c2/c1)^2

maybe I can prove it by contradiction?

You wouldn't prove something is rational by contradiction; I'd use contradiction to prove something irrational.

But Cubist's method will work, if you have the patience. The numerator and denominator will factor as squares without too much trouble.

 

[Cubist]

But Cubist's method will work, if you have the patience. The numerator and denominator will factor as squares without too much trouble.

Yes, now I've done it too. I found the numerator to be quite a challenge. I made a couple of substitutions (u=? & v=?) that helped me.

But, I wonder if there's an easier method to prove this

 

[Thales12345]

Yes, now I've done it too. I found the numerator to be quite a challenge. I made a couple of substitutions (u=? & v=?) that helped me.

Can you show me how you did that? When I put everything over a common denominator, the denominator is a square but the numerator isn’t.

 

[Dr.Peterson]

Can you show me how you did that? When I put everything over a common denominator, the denominator is a square but the numerator isn’t.

Please do as we ask and show us YOUR work as far as you get. That's how it works here. Once we see where you stopped or went wrong, we'll have something to discuss.

 

[Thales12345]

So, I‘ve typed everything out. That way it’s much easier for you to follow/read. My denominator is a perfect square, but my numerator isn’t. How can begin with the substitution? What can I replace by other letters? The proof is done when the numerator is a perfect square, right?

 

[Dr.Peterson]

View attachment 28108
So, I‘ve typed everything out. That way it’s much easier for you to follow/read. My denominator is a perfect square, but my numerator isn’t. How can begin with the substitution? What can I replace by other letters? The proof is done when the numerator is a perfect square, right?

I can tell you that I never fully expanded my numerator, because I saw some common pieces that would help in factoring it. Those pieces included \((a_1^2b_2^2+a_2^2b_1^2)\) and \(a_1^2a_2^2b_1^2b_2^2\). This is a very useful technique: Use what you're given whole, if you can, before mashing it up.

(Actually, I avoided subscripts and used a/b, c/d, e/f to keep things cleaner, so what I saw were \((a^2d^2+b^2c^2)\) and \(a^2b^2c^2d^2\). Similarly, my denominator was \(a^2c^2(a^2d^2+b^2c^2)\). Visibility is important.)

 

[Cubist]

@Thales12345 I too never fully expanded the numerator (although your expanded numerator seems correct - but go back to your pre-expanded version)

For visibility I made the following two substitutions in the numerator (which I simplified separately to the denominator) \( u=a_1 b_2 \) and \( v=a_2 b_1 \) and this left the numerator with only two variables. You might want to use x and y if your written u and v look similar!

 

[Thales12345]

For visibility I made the following two substitutions in the numerator (which I simplified separately to the denominator) \( u=a_1 b_2 \) and \( v=a_2 b_1 \) and this left the numerator with only two variables. You might want to use x and y if your written u and v look similar!

I worked it out your way (your u is my e and your v is my f). This is my numerator now.

I’m stuck. I don’t see how I can put everything in 1 term under 1 (even) exponent.

 

[Dr.Peterson]

First, factor out (e+f)^2 from the first two terms.

Then expand (e+f)^2.

Then look for something that looks like a perfect square (a^2 + 2ab + b^2).

 

[Thales12345]

is ef my a when I use (a^2 + b^2 + 2ab)?
do I have to add and subtract things to make that formula?
is it useful to work out the first term?

 

[Dr.Peterson]

Basically, you just have to try things and see what happens, guided by a sense of what things need to look like.

Do you see that 2? You'd like that to be a factor of the term 2ab, so split your first term into two, one being 2ef(e^2+f^2). Then what might a and b be? Do you see the same a and b elsewhere being squared?

 

[lex]

Spoiler

[MATH]\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{(a+b)^2}\\ =\dfrac{a^2(a+b)^2+b^2(a+b)^2+b^2a^2}{(ab(a+b))^2}\\ =\dfrac{a^2(a+b)^2+b^2(2a^2+2ab+b^2)}{(ab(a+b))^2}\\ =\dfrac{a^2(a+b)^2+2b^2(a(a+b))+b^4}{(ab(a+b))^2}\\ =\dfrac{(a(a+b)+b^2)^2}{(ab(a+b))^2}[/MATH]which is the square of the rational number

[MATH]\dfrac{a(a+b)+b^2}{ab(a+b)} [/MATH]
Note, bottom line is not 0, since a, b, c ≠ 0, and (a+b)≠0 since (a+b)=0 implies c=0

 

[lex]

Or let [MATH]2u=a+b, 2v=a-b[/MATH]then the expression is:
[MATH]\dfrac{1}{4u^2}+\dfrac{1}{(u+v)^2}+\dfrac{1}{(u-v)^2}\\ =\dfrac{(u^2-v^2)^2+4u^2(u-v)^2+4u^2(u+v)^2}{(2u(u^2-v^2))^2}\\ =\dfrac{9u^4+6u^2v^2+v^4}{(2u(u^2-v^2))^2}\\ =\left(\dfrac{3u^2+v^2}{2u(u^2-v^2)}\right)^2[/MATH]

 

[Cubist]

I like it @lex , we don't need to show the result can be written as [MATH] \left( \frac{\text{integer}}{\text{another integer}} \right)^2 [/MATH]. We just need to show it as [MATH] \left( \frac{\text{rational}}{\text{another rational}} \right)^2 [/MATH].

 

[Cubist]

Another option, which helps to eliminate more variables quickly, let \(b=xa\) therefore x is rational and the expression becomes...

[MATH]\frac{1}{a^2}+\frac{1}{x^2a^2}+\frac{1}{a^2(1+x)^2}[/MATH]
[MATH] =\frac{ x^2\cancel{a^4}(x + 1)^2 +\cancel{a^4}(x+1)^2 + x^2\cancel{a^4} }{ x^2a^{\xcancel{6}2}(x+1)^2} [/MATH]
[MATH] =\frac{ x^2(x + 1)^2 +( x^2+ 2x + 1) + x^2 }{ x^2a^2(x+1)^2 } [/MATH]
[MATH] =\frac{ \left(x^2 + x\right)^2 +2(x^2 + x) + 1 }{ x^2a^2(x+1)^2 }[/MATH]
[MATH] =\left( \frac{ x^2 + x + 1}{ xa(x+1) }\right)^2 [/MATH]

 

[Cubist]

Or let [MATH]2u=a+b, 2v=a-b[/MATH]...

Or, mixing Lex's idea with post#19, let [MATH]2u=a+b, 2uv=a-b[/MATH]
[MATH]\dfrac{1}{4u^2}+\dfrac{1}{u^2(1+v)^2}+\dfrac{1}{u^2(1-v)^2}\\ =\dfrac{\cancel{u^4}(1-v^2)^2+4\cancel{u^4}(1-v)^2+4\cancel{u^4}(1+v)^2}{4u^{\xcancel{6}2} (1-v^2)^2}\\ =\dfrac{v^4+6v^2+9}{(2u(1-v^2))^2}\\ =\left(\dfrac{v^2+3}{2u(1-v^2)}\right)^2[/MATH]