Re: [Axiom-developer] Testing if (72*a^3*b^5)^(1/2) is equivalent to 6*a*b^2*(2*a*b)^(1/2)

[William Sit]

Every "number" has two square roots. The expression may be 
zero and may not be zero, depending on which of the four 
possible interpretations you put on the square root.

An expression like the one given can be interpreted at 
various levels in Axiom. Each "square root" can be 
interpreted as an algebraic element in an extension of the 
field Q(a,b) , where a, b are two algebraically 
independent element over the field of rational numbers Q 
(defined by a minimal polynomial) and the difference would 
be another algebraic element. Or, one can view a,b as 
standing for two elements in some ordered ring R, and the 
square root is always taken to be the positive square 
root, in which case, (a^2)^(1/2) = a in R only when a > 0 
in R.

There is also a difference (no pun intended) between 
testing x = y and testing x-y=0. If x = sqrt(72a^3b^5) and 
y = 6ab^2sqrt(2ab), then the minimal polynomials of x and 
of y over say Q(a,b) are the same and as algebraic 
elements over Q(a,b), x and y may be considered "equal". 
However, if I am not mistaken, x-y will be algebraic of 
degree 4 and not equal to 0.

William

On Mon, 8 Mar 2010 01:48:08 -0500
 Ted Kosan <address@hidden> wrote:
> I have been experimenting with Axiom to see how it 
> compares to other
> computer algebra systems.
>
> One of the things I tried testing was if Axiom could 
> determine if
> (72*a^3*b^5)^(1/2) was equivalent to 
> 6*a*b^2*(2*a*b)^(1/2):
>
> (2) -> (72*a^3*b^5)^(1/2) - 6*a*b^2*(2*a*b)^(1/2)
>
>          +------+
>          |   3 5        2 +----+
>  (2)  \|72a b   - 6a b \|2a b
>
>
> When I entered this expression into Wolfram Alpha, it 
> returned 0 as a result.
>
> Is Axiom capable of determining if (72*a^3*b^5)^(1/2) is 
> equivalent to
> 6*a*b^2*(2*a*b)^(1/2) ?
>
> Thanks,
>
> Ted
>
>
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William Sit, Professor Emeritus
Mathematics, City College of New York                  
Office: R6/202C Tel: 212-650-5179
Home Page: http://scisun.sci.ccny.cuny.edu/~wyscc/