Hi, I’ve been stuck on this problem for a week and can’t seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.

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- May 14th 2006, 07:41 AMcoopsterdudeStuck on radicals...please help!
Hi, I’ve been stuck on this problem for a week and can’t seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.

- May 14th 2006, 06:48 PMThePerfectHacker
I developed a theorem (imcomplete) sometime ago, that if you have an

**integral***linear combination*of**irrational***distinct*radicals equal to each other then there corresponding parts are equal.

For example, if

where are integers then,

.

Now if you have,

where are integers is just impossible,

If you have that,

where are integers then,

Thus,

Then,

is just impossible in integers.

This is my reasoning that why I would not expect a simpler solution with the integers. - May 15th 2006, 09:21 AMcoopsterdude
Thanks for all your help and I'm am so sorry about breaking all those rules.

- May 15th 2006, 09:50 AMCaptainBlackQuote:

Originally Posted by**coopsterdude**

(using the convention that denotes the positive

root.)

Just goes to show that it never does to be too dogmatic PH :D

RonL - May 15th 2006, 10:23 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

Note I made a mistake in doing

the wrong problem.

If you used my method then you will see that it works ;)

-------------------------

I base it on the fact that,

is always irrational whenver all are integers and the b_i are all non-squares and a_i are not zero.

Then given

=

You can write,

where

Which is rational an impossibility unless a_i=c_i giving zeros. - May 16th 2006, 05:48 PMRanger SVOQuote:

Originally Posted by**CaptainBlack**

- May 16th 2006, 06:38 PMThePerfectHackerQuote:

Originally Posted by**Ranger SVO**

You attempt to find integers such as,

Thus,

Thus,

Then, by my previous post,

Look at the second equation,

This, is equivalent to

Since are integers the only possibilities are,

Now we check which one of these satisfies the second equation,

we see that,

Thus,

Both of these are true, indeed,

But we need to chose,

Because square roots are non-negative and,

- May 17th 2006, 05:49 AMRanger SVO
Thank you, I understand it now but I wonder if a High School Math Teacher would have intended a simpler solution.

http://i7.photobucket.com/albums/y284/wrf01a/eqa.jpg

I am finished with finals and had time to look at the problem.