For any given integral domain, state the law of trichotomy, and show that in an integral domain D, the square of non-zero elements are positive
Thanks.
The law of trichotomy states that for any two elements a and b in an integral domain, one and only one of the following three conditions hold: a = b, a < b, or a > b.
To show that in an integral domain D, the square of non-zero elements are positive, we can use the properties of integral domains.
First, let's consider a non-zero element a in D. We want to show that a^2 > 0, i.e., the square of a is positive.
Since a ≠ 0, we know that a is either positive or negative in an integral domain by the law of trichotomy. Let's consider the two cases:
1. If a > 0, then a can be expressed as a product of itself: a = a * a. Since a and a are both positive, their product is also positive. Therefore, a^2 > 0.
2. If a < 0, then we can write -a > 0. Using the same reasoning as in the first case, (-a)^2 = (-a)(-a) > 0. Since (-1)(-1) = 1, we have a^2 = (-a)^2 = (-1)(-1)(a)(a) = 1 * a * a = a * a = a^2 > 0.
In both cases, we have shown that a^2 > 0 for any non-zero element a in an integral domain D.
Note that this result holds specifically for integral domains, which are commutative rings with unity where no zero divisors exist. It may not hold in other types of rings or number systems.