Why is the product of two negative numbers a
positive number?
Math 323, Section 3
Here’s the formulation we settled on for what we wanted to prove.
Theorem 0.1 Let n and m be integers.
1. If n > 0 and m < 0, then nm < 0
2. If n < 0 and m < 0, then nm > 0
3. If n = 0 or m = 0, then nm = 0.
In discussing how to prove this, we came up with a list of properties of numbers
that we decided to regard as axioms. They were
• Existence of the operations of addition and multiplication.
• Commutativity and associativity of addition and multiplication.
• The distributive law.
• Existence of a number 0 such that n + 0 = n for all integers n.
• Existence of a number 1 such that 1 · n = n for all integers n.
• Give an integer n, there exists a unique integer −n such that n + (−n) = 0.
Then we came up with the following arguments: for part (3) of the theorem, we
wrote
1+0
(1 + 0)n
1·n+0·n
n+0·n
(n + 0 · n) + (−n)
0·n
=
=
=
=
=
=
1
defining property of 0
1·n
1·n
distributive law
n defining property of 1
n + (−n)
0
defining property of −n (also properties of addition)
Then we managed to use this and the distributive law to show that
n(−m) = −(nm)
(1)
The argument went like this:
m + (−m) = 0
defining property of 0
n(m + (−m)) = 0
we just proved n · 0 = 0
nm + n(−m) = 0
distributive law
n(−m) = −(nm)
defining property of −(nm)
Two questions: Have we proved what we set out to prove? How do we know
our axioms are consistent?
23 August 2004
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