# Hard to prove problem

• Dec 20th 2007, 03:59 AM
spandan_shr
Hard to prove problem
I am a 10th standard student living in Nepal. I recently found a strange problem of Algebra. A teacher of mine solved the above problem in the some way (some steps jumped). The problem, as well as the solution is included in the PDF file attached herewith.

Now, I want to know whether or not the question and/or the solution are correct. Please give your opinion about the problem and the solution.

--
Spandan Shrestha
A.V.M. High School
Kathmandu, Nepal
• Dec 20th 2007, 04:21 AM
CaptainBlack
Quote:

Originally Posted by spandan_shr
I am a 10th standard student living in Nepal. I recently found a strange problem of Algebra. A teacher of mine solved the above problem in the some way (some steps jumped). The problem, as well as the solution is included in the PDF file attached herewith.

Now, I want to know whether or not the question and/or the solution are correct. Please give your opinion about the problem and the solution.

--
Spandan Shrestha
A.V.M. High School
Kathmandu, Nepal

The solution is invalid.

It uses the argument that:

$U-V = A-B$

so squaring:

$(U-V)^2=(A-B)^2$

but $(A-b)^2=(B-A)^2$, so:

$(U-V)^2=(B-A)^2$

and taking square roots gives:

$U-V = B-A$

when what has been shown is that:

$|U-V| = |B-A|$

The offending part of the argument is shown in the attachment

RonL
• Dec 20th 2007, 04:35 AM
JaneBennet
Moreover, the statement $\sqrt[3]{x+\sqrt{y}}=a+\sqrt{b}\ \Rightarrow\ \sqrt[3]{x-\sqrt{y}}=a-\sqrt{b}$ is false. As a counterexample, take x = 0, y = 64, a = 1, b = 1.

Your teacher probably set this question as a “spot the fallacy” exercise for his/her students. :)
• Dec 20th 2007, 06:38 AM
CaptainBlack
Quote:

Originally Posted by JaneBennet
Moreover, the statement $\sqrt[3]{x+\sqrt{y}}=a+\sqrt{b}\ \Rightarrow\ \sqrt[3]{x-\sqrt{y}}=a-\sqrt{b}$ is false. As a counterexample, take x = 0, y = 64, a = 1, b = 1.

Your teacher probably set this question as a “spot the fallacy” exercise for his/her students. :)

There may be some side condition that have been omitted, since this looks
like it is related to the Ferro-Tartaglia-Cardarno formula for the roots of a
cubic.

RonL
• Dec 21st 2007, 09:25 PM
spandan_shr
New solution: is THAT correct?
Thanks everyone for the reply. :)

When I told the teacher who solved this question about this, he admitted the flaw, and solved it in another way, which seems absolutely correct to me. Again, the PDF file has been attached herewith.

• Dec 22nd 2007, 06:05 AM
JaneBennet
This part

Quote:

$x+\sqrt{y}\ =\ (a^3+3ab)+\sqrt{(3a^2+b)^2b}$
By equating rational and irrational parts, we get,
$x=a^3+3ab\ \mbox{and }y=(3a^2+b)^2b$
breaks down if $\sqrt{y}$ is rational. (I suppose x, y, a and b are all assumed to be rational numbers.) Just because there’s a square-root sign doesn’t mean that $\sqrt{y}$ must be irrational. You must be sure $\sqrt{y}$ is irrational in order for the proof to work.
• Dec 25th 2007, 02:25 AM
spandan_shr
Quote:

when what has been shown is that:
|U-V| = |B-A|
Please prove that to me (or provide me with the link to the proof). I'm really confused about this one, whether I should take the absolute value, or both the positive and negative signs.

@JaneBennet:
Can you too give me some explanation?
• Dec 25th 2007, 06:37 AM
Isomorphism
Quote:

Originally Posted by spandan_shr
Please prove that to me (or provide me with the link to the proof). I'm really confused about this one, whether I should take the absolute value, or both the positive and negative signs.

@JaneBennet:
Can you too give me some explanation?

Hi spandan_shr,

$x^2 = y^2 \Rightarrow |x| = |y|$
The "or" here is dangerous. It means either of x = +y or x=-y is possible. Naturally both cannot hold simultaneously unless x=y=0 !