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 <ted.ko...@gmail.com> 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/
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