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

From: William Sit
Subject: Re: [Axiom-developer] Testing if (72*a^3*b^5)^(1/2) is equivalent to 6*a*b^2*(2*a*b)^(1/2)
Date: Mon, 08 Mar 2010 03:05:10 -0500

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.


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) ?



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William Sit, Professor Emeritus
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