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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1323.0. "Rationalizing sqrt x + ... + sqrt z = 0." by TRACE::GILBERT (Ownership Obligates) Fri Nov 02 1990 15:35

Given the equation
 
sqrt x + sqrt y = 0
 
we can "rationalize it" by bringing sqrt y to the right side
and then squaring both sides.  You get x^2-y^2=0.
 
To rationalize
 
sqrt x + sqrt y + sqrt z = 0
 
subtract sqrt z from both sides, then square.
Isolate the remaining radical on one side and square again.
This gets rid of all radicals.  The result is
 
 4    4    4     2 2     2 2     2 2
x  + y  + z  - 2x y  - 2y z  - 2z x  = 0.
 
With 4 variables, you can bring two square roots to each side,
square both sides, bring the two radicals to the same side,
square again; and then finally you are left with an
equation with one radical which you can bring to one side
and square a final time.
 
The paper
 
P. R. Stein and C. Zemach, On the Rationalization of a Sum of Surds,
Advances in Applied Mathematics, 8(1987)393-404
 
says "The reader can easily convince himself that the
elementary method does not work for n>4"
and then goes on to give a complicated method of rationalizing
 
sqrt v + sqrt w + sqrt x + sqrt y + sqrt z = 0.
 
I am not so easily convinced.
 
Can anyone see an "elementary" method of rationalizing
this expression?

(Stan Rabinowitz sent this to me, asking that I post it)
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1323.1assuming real valued variables ...GUESS::DERAMODan D'EramoFri Nov 02 1990 16:398
	I would say, since sqrt v is >= 0 for whatever variable v
	you plug in, that

		sqrt v + sqrt w + sqrt x + sqrt y + sqrt z = 0.

	implies v = w = x = y = z = 0.

	Dan
1323.2An Elementary SolutionVAXRT::BRIDGEWATERBlasting out of the past.Fri Nov 02 1990 17:1938
    All variable substitutions below are to simplify expressions, not
    some magic transformation.  The tricks, if any, are deciding which
    radicals to bring together and which side gets the nonradical terms.
    In one step we produce sqrt(wz) on both sides of the equation and
    in another step we produce sqrt(wxyz) on both sides.


    sqrt(v) + sqrt(w) + sqrt(x) + sqrt(y) + sqrt(z) = 0

    Rearranging and squaring both sides:

    [sqrt(v) + sqrt(w) + sqrt(x)]^2 = [-sqrt(y) - sqrt(z)]^2

    v+w+x+ 2*[sqrt(vw) + sqrt(vx) + sqrt(wx)] = y+z+ 2*sqrt(yz)

    let a = [y + z - v - w - x]/2.  Then, rearranging and squaring
    both sides:

    [sqrt(vw) + sqrt(vx)]^2 = [a + sqrt(yz) - sqrt(wx)]^2

    vw+vx+ 2v*sqrt(wx) = a^2+yz+wx + 2a*sqrt(yz) - 2a*sqrt(wx) - 2*sqrt(wxyz)

    (2v+2a)*sqrt(wx) - 2a*sqrt(yz) = a^2+yz+wx-vw-vx - 2*sqrt(wxyz)

    let b = [a^2+yz+wx-vw-vx]/2.  let c = v+a.  Then, rearranging and
    squaring both sides:

    [c*sqrt(wx) - a*sqrt(yz)]^2 = [b - sqrt(wxyz)]^2

    (c^2)wx + (a^2)yz - 2ac*sqrt(wxyz) = b^2 - 2b*sqrt(wxyz) + wxyz

    Rearranging and squaring both sides:

    [2(b-ac)*sqrt(wxyz)]^2 = [b^2 + wxyz - (c^2)wx -(a^2)yz]^2

    and substituting back in for a,b, and c should complete the problem.
    
    - Don
1323.3Impossible with 6 variables?VAXRT::BRIDGEWATERBlasting out of the past.Fri Nov 02 1990 17:536
    sqrt(u) + sqrt(v) + sqrt(w) + sqrt(x) + sqrt(y) + sqrt(z) = 0

    looks impossible to rationalize.  I'm not sure how to attempt a
    proof, though.

    - Don
1323.4shouldn't assume real valued variables...?CHOVAX::YOUNGThe OOL's are not what they seem.Fri Nov 02 1990 20:026
    Re .1:
    
    Not necessarily Dan.  The variable may not be real numbers, they may be
    complex variables.  Or, even more likely, they might be polynomials.
    
    --  Barry
1323.5GUESS::DERAMODan D'EramoSat Nov 03 1990 05:3610
        Even as it is, it is wrong.  Take the simplest case,
        sqrt x + sqrt y = 0.  Follow the directions in .0 and
        bring sqrt y to the other side, sqrt x = - sqrt y.  Now
        square both sides:  x = y or x - y = 0, not x^2 - y^2 = 0
        as stated.  Likewise with three variables, bringing sqrt
        z to one side and squaring leaves a single term z on that
        side, and a second squaring yields a a quadratic term in
        z.  But the "solution" given contains z^4.
        
        Dan
1323.6ALLVAX::JROTHIt's a bush recording...Sat Nov 03 1990 19:0422
   It seems you have to be free to choose arbitrary branches on the roots
   to obtain any nontrivial solutions, but this is not stipulated clearly.

   Suppose we have a homogenous sum of n terms a + b + c + .. = 0.

   Then there will result 2^n/2 possible equations that are "true in the sense
   of rationalization" by bringing subsets of the terms to one side of the
   equal sign, squaring, and bringing the terms back.

   If you can find a linear superposition of the equations that cancels
   the cross terms without cancelling the other terms you can get a
   rationalization, which will always be a symmetric form in the variables
   (I think this has to be the case.)

   If not, then the equations can be treated the same way, yielding a
   larger set.  There will be a lot of symmetry, but there may be enough
   freedom to do the cases of 6 or more variables.

   This is just an idea, I don't know if it is the right way to think
   about the problem.

   - Jim
1323.7elementary?HERON::BUCHANANcombinatorial bomb squadSun Nov 04 1990 15:3224
1323.8should have used mapleALLVAX::JROTHIt's a bush recording...Sun Nov 04 1990 23:044
   Sigh... that's the first thing I tried, but I made an algebraic
   error which mislead me into thinking it couldn't work.

   - Jim
1323.9EAGLE1::BESTR D Best, sys arch, I/OMon Nov 05 1990 16:4821
re .0:

>Given the equation
> 
>sqrt x + sqrt y = 0
> 
>we can "rationalize it" by bringing sqrt y to the right side
>and then squaring both sides.  You get x^2-y^2=0.

Do you mean squaring both sides twice ?

Squaring once gives:

sqrt(x) = - sqrt(y)
x = y

When you raise equations to powers like this don't you sometimes introduce
extraneous roots ?

Also, you treating sqrt as a function (single valued with positive range)
or a relation ?