(1) Prove that the cancellation law of multiplication is equivalent in a commutative ring to the assertion that the product of non-zero factors is not zero.
(2) 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.
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To prove that the cancellation law of multiplication is equivalent in a commutative ring to the assertion that the product of non-zero factors is not zero, we need to demonstrate two parts: (1) cancellation law implies non-zero product, and (2) non-zero product implies cancellation law.
(1) Cancellation law implies non-zero product:
Assume the cancellation law of multiplication holds in a commutative ring R, which states that for any elements a, b, and c in R, if a * b = a * c and a is not equal to zero, then b = c.
To prove that the product of non-zero factors is not zero, let's assume for contradiction that there exist non-zero elements b and c such that b * c = 0.
Since b and c are non-zero, we can conclude that neither b nor c is zero. Thus, using the cancellation law, we have b * c = 0 implies b = 0, which contradicts our assumption that b is non-zero.
Therefore, the assumption that there exist non-zero elements b and c such that b * c = 0 leads to a contradiction, and we can conclude that the product of non-zero factors is not zero.
(2) Non-zero product implies cancellation law:
Assume that in a commutative ring R, the product of non-zero factors is not zero.
We want to show that the cancellation law holds in R, which states that for any elements a, b, and c in R, if a * b = a * c and a is not equal to zero, then b = c.
Let's assume a * b = a * c, where a is not equal to zero.
If we multiply both sides by the inverse of a (denoted as a^(-1)), we have a^(-1) * (a * b) = a^(-1) * (a * c).
Using the commutative property of multiplication, we can rearrange the terms as (a^(-1) * a) * b = (a^(-1) * a) * c.
Since a^(-1) * a is equal to 1 (the multiplicative identity element), we have 1 * b = 1 * c.
By the definition of the multiplicative identity, we know that 1 * b is equal to b, and 1 * c is equal to c.
Therefore, we have b = c, which demonstrates the cancellation law.
In conclusion, we have shown that the cancellation law of multiplication is equivalent in a commutative ring to the assertion that the product of non-zero factors is not zero.
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For any given Integral Domain, the law of Trichotomy states that for any two elements a and b in the integral domain, exactly one of the following must be true: a = b, a < b, or b < a.
To show that in an integral domain D, the square of non-zero elements is positive, we need to demonstrate that for any non-zero element a in D, a^2 is greater than zero.
Let's assume that a is a non-zero element in the integral domain D.
Since D is a commutative ring, we have the cancellation law of multiplication, which implies that if a * b = a * c and a is not equal to zero, then b = c.
Now, consider the product a * a, which is equal to a^2.
If a^2 is equal to zero, then we have a * a = 0.
By applying the cancellation law, we can conclude that a = 0, which contradicts our assumption that a is non-zero.
Therefore, the assumption that a^2 is equal to zero leads to a contradiction.
Hence, we can conclude that in an integral domain D, the square of non-zero elements is positive.
(1) To prove that the cancellation law of multiplication is equivalent in a commutative ring to the assertion that the product of non-zero factors is not zero, we need to show both implications.
First, let's assume the cancellation law of multiplication holds in a commutative ring R. The cancellation law states that if a * b = a * c, and a is not zero, then b = c. We want to prove that if the product of two non-zero factors is zero, then the cancellation law holds.
So, suppose we have two non-zero factors, a and b, such that a * b = 0. We need to show that if a * b = a * c, then b = c. Since a is not zero, we can multiply both sides of the equation by a^-1 (the multiplicative inverse of a) without changing the equality:
a^-1 * (a * b) = a^-1 * (a * c)
Using associativity and commutativity of multiplication, we can rewrite the equation as:
(a^-1 * a) * b = (a^-1 * a) * c
Since a * a^-1 = a^-1 * a = 1 (the multiplicative identity), we get:
1 * b = 1 * c
And simplifying further:
b = c
This shows that the cancellation law holds when the product of non-zero factors is not zero.
Now, let's prove the other implication. We assume that the product of non-zero factors is not zero in a commutative ring R. We want to show that if a * b = a * c, then b = c.
Suppose a * b = a * c. We can subtract a * c from both sides to obtain:
a * b - a * c = 0
Factoring out a, we get:
a * (b - c) = 0
Since the product of non-zero factors is not zero, we conclude that b - c = 0, which implies b = c.
This completes the proof that the cancellation law of multiplication is equivalent in a commutative ring to the assertion that the product of non-zero factors is not zero.
(2) In an integral domain D, the law of trichotomy states that for any elements a and b in D, exactly one of the following is true: a = b, a < b, or b < a.
To show that in an integral domain D, the square of non-zero elements is positive, we need to prove that for any non-zero element a in D, a^2 > 0.
Let's assume that a^2 ≤ 0 for some non-zero element a in D. We want to show that this leads to a contradiction.
Since D is an integral domain, it's closed under multiplication, which means that a^2 is also in D. Based on the assumption, a^2 ≤ 0, we have two cases:
Case 1: a^2 = 0
If a^2 = 0, then applying the cancellation law of multiplication, we get a = 0. However, this contradicts the assumption that a is non-zero.
Case 2: a^2 < 0
If a^2 < 0, then considering the element -a, we have (-a)^2 = a^2 > 0. But this contradicts the assumption that the square of non-zero elements is not positive.
In both cases, we arrive at a contradiction. Therefore, our assumption that a^2 ≤ 0 for a non-zero element a in D is false. Hence, we can conclude that the square of non-zero elements in an integral domain D is positive.
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