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

$\displaystyle 1+\sqrt{2}+\sqrt{3}=x+y\sqrt{2}+z\sqrt{3}$

where $\displaystyle x,y,z$ are integers then,

$\displaystyle x=1,y=1,z=1$.

Now if you have,

$\displaystyle 1+\sqrt{2}+\sqrt{3}=x+y\sqrt{5}$

where $\displaystyle x,y$ are integers is just impossible,

If you have that,

$\displaystyle \sqrt{16-6\sqrt{3}}=x+y\sqrt{3}$

where $\displaystyle x,y$ are integers then,

$\displaystyle 16-6\sqrt{3}=(x+y\sqrt{3})^2$

Thus,

$\displaystyle 16-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$

Then,

$\displaystyle \left\{ \begin{array}{c}x^2+3y^2=16\\ 2xy=6$ 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**

$\displaystyle

\sqrt{12-6\sqrt{3}}=3-\sqrt{3}

$

(using the convention that $\displaystyle \sqrt{}(\cdot)$ 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

$\displaystyle \sqrt{16-6\sqrt{3}}$ the wrong problem.

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

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

I base it on the fact that,

$\displaystyle a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$

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

Then given

$\displaystyle a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$=$\displaystyle c_0+c_1\sqrt{b_1}+...+c_n\sqrt{b_n}$

You can write,

$\displaystyle d_0+d_1\sqrt{b_1}+...+d_n\sqrt{b_n}=0$ where $\displaystyle d_i=a_i-c_i$

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 $\displaystyle x,y$ such as,

$\displaystyle \sqrt{12-6\sqrt{3}}=x+y\sqrt{3}$

Thus,

$\displaystyle 12-6\sqrt{3}=(x+y\sqrt{3})^2$

Thus,

$\displaystyle 12-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$

Then, by my previous post,

$\displaystyle \left\{ \begin{array}{c}12=x^2+3y^2\\ -6=2xy$

Look at the second equation,

$\displaystyle -6=2xy$

This, is equivalent to

$\displaystyle xy=-3$

Since $\displaystyle x,y$ are integers the only possibilities are,

$\displaystyle (x,y)=(-1,3),(3,-1),(-3,1),(1,-3)$

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

we see that,

$\displaystyle \begin{array}{c} (3)^2+3(-1)^2=12\\ (-3)^2+3(1)^2=12$

Thus,

$\displaystyle 3-\sqrt{3}$

$\displaystyle \sqrt{3}-3$

Both of these are true, indeed,

$\displaystyle (\sqrt{3}-3)^2=(3-\sqrt{3})^2=12-6\sqrt{3}$

But we need to chose,

$\displaystyle 3-\sqrt{3}$

Because square roots are non-negative and,

$\displaystyle \sqrt{3}-3<0$ - 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.