Thread: 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

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:



so squaring:



but , so:



and taking square roots gives:



when what has been shown is that:



The offending part of the argument is shown in the attachment

RonL

Moreover, the statement 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.

Originally Posted by JaneBennet

Moreover, the statement 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

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.

Please see the new solution and give your opinion about it.

This part

By equating rational and irrational parts, we get,

breaks down if 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 must be irrational. You must be sure is irrational in order for the proof to work.

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?

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,


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 !