Solving Homework Statement: Proving a/b=c/d=e/f to ace/bdf

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The discussion centers on proving the equation (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) = ace/bdf under the assumption that a/b = c/d = e/f. Participants clarify that the specific form of the numerator and denominator is a given and not essential for understanding the proof's requirements. The focus should be on manipulating one side of the equation to demonstrate the equality, rather than deriving the expression from the ratios. It is emphasized that the proof relies on the assumption of the ratios rather than specific values. Ultimately, the goal is to show that the equality holds based on the provided relationships.
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Homework Statement



if a/b = c/d = e/f

then show that...

(a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) = ace/bdf

Homework Equations




1.I understand how to prove it but I don't understand algebraically how they got from
a/b = c/d = e/f to... (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3)
( or is this irrelevant and what I should focus on is the proving the theorem?)

2. second I don't understnad how (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) = ace/bdf?
the numerator does not simplify into the denominator UNLESS...
because a/b = c/d = e/f

(a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) could also equal (k^4 + 2k^3 - 3k^4)/(k^4 + 2k^3 - 3k^4) = 1

and ace/bdf could equal k^3/k^3 = 1

therefore 1=1 is true?
 
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Hi Miike012! :smile:

Start again …

"if a/b = c/d = e/f = p … "
 
then
a=bp, c= dp, and e=fp


I understand that part... is there something that you are hinting at?
 
Miike012 said:
1.I understand how to prove it but I don't understand algebraically how they got from
a/b = c/d = e/f to... (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3)
( or is this irrelevant and what I should focus on is the proving the theorem?)
It's irrelevant. The expression is a given. You don't care where it comes from.
2. second I don't understnad how (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) = ace/bdf?
the numerator does not simplify into the denominator UNLESS...
because a/b = c/d = e/f
The point of the problem is to show that if you assume a/b = c/d = e/f, then (a3b+2c2e-3ae2f)/(b4+2d2f-3bf3) = ace/bdf. All proofs are like this. You're given a set of assumptions, and you want to show that it then follows that some statement is true.
(a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) could also equal (k^4 + 2k^3 - 3k^4)/(k^4 + 2k^3 - 3k^4) = 1

and ace/bdf could equal k^3/k^3 = 1

therefore 1=1 is true?
It's certainly true if a, b, c, d, e, and f all equaled each other (and aren't equal to 0), the equality would hold, but that fact doesn't really help in your proof because you're using an assumption that wasn't a given in the problem. All you can assume is a/b = c/d = e/f.
 
wow, thank you ... that makes a lot more sense now.
 
A good way to write a proof like this is to start with one side of the equation and manipulate it until you end up with the other side:

\begin{align*}<br /> \frac{a^3b + 2c^2e - 3ae^2f}{b^4 + 2d^2f - 3bf^3} &amp; = \cdots \\<br /> &amp; = \cdots \\<br /> &amp; \vdots \\<br /> &amp; = \frac{abc}{def}<br /> \end{align*}<br />
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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