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Re: negative * negative
Posted:
Apr 5, 1996 4:15 PM
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At 10:17 AM 4/2/96, Steve Cottrell wrote:
>One of the standard theorems from Foundations of Algebra courses and
>which follows in a few steps from the field axioms is
>
> (-a)(-b) = ab. The proof is not difficult and it's readily
>
>accessible in any algebra foundations text. The reason that the
>product of two negatives is positive is a consequence of the field
>axioms.
Careful now. This proves that the product of the opposites of two elements
is equal to the product of those two elements. That is, -a is not
necessarily a negative number! Which goes back to the point raised earlier
on the overloading of the "-" sign.
Why is this such a big deal? Not necessarily because we need to
systematically teach students the different meanings, BUT... if there are
important mathematical issues that are being obscured, a muddled situation
for students is likely to result. Even if students need not be explicitly
instructed about all this, curriculum writers and teachers had best be VERY
aware.
A very common example of this muddling occurs when dealing with functions
like f(x) = -x ... more than a few students believe f(x) always has to be
negative. Thus, calling -x to be "negative x" is not at all a good idea...
Peace, Gary
PS: To finish up Steve's argument, we need to bring in ordering.
(i) If a>0 and b>0, then ab>0 ... BY DEFINITION (closure)
(ii) a>0 iff -a<0 ... BY DEFINITON (from trichotomy)
SO... If a<0 and b<0, then -a>0 and -b>0 ... by (ii)
Which means (-a)(-b)>0 ... by (i)
Now use the theorem Steve mentions above:
ab = (-a)(-b) > 0
Thus, the product of two negatives is positive.
W. Gary Martin
Curriculum Research and Development Group
University of Hawaii
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