ONE IS EQUAL TO ZERO … HOW COME?

Some people may think it is a weird question and say “Nonsense!” But if you ask the question to someone who has studied abstract algebra, you’ll possibly get another answer. This article introduces the concept of ring in mathematics.

Definition

A non-empty set R is called a ring if two binary operations, denoted by + and ⋅, are defined in R such that all the properties below are satisfied:

1.  (a + b) ∈ R for every a, b ∈ R
2.  a + b = b + a for every a, b ∈ R
3. (a + b) + c = a + (b + c) for every a, b, c ∈ R
4. There exists 0 ∈ R such that a + 0 = a for every a ∈ R.
5. For every a ∈ R there exists an element -a ∈ R such that a + (-a) = 0
6. a⋅b ∈ R for every a, b ∈ R
7. a⋅(b⋅c) = (a⋅b)⋅c for every a, b, c ∈ R
8. For every a, b, c ∈ R both a⋅(b+c) = a⋅b + a⋅c and (b+c)⋅a = b⋅a + c⋅a apply.

The operation + is usually called the addition operation (simply, addition), while the ⋅ operation is usually called the multiplication operation (simply, multiplication). 0 in Property 4 is often referred to as additive identity element or zero element. A ring R with the operations + and ⋅ is usually denoted by (R,+,-). Also note that Properties 1 to 5 above are the properties of abelian groups. As a consequence, we have an equivalent definition of a ring as follows.

A non-empty set R is called a ring if two binary operations, denoted by + and ⋅, are defined in R such that all the properties below are satisfied:

1.  (R,+) is an abelian group
2. a⋅b ∈ R for every a, b ∈ R
3. a⋅(b⋅c) = (a⋅b)⋅c for every a, b, c ∈ R
4. For every a, b, c ∈ R both a⋅(b+c) = a⋅b + a⋅c and (b+c)⋅a = b⋅a + c⋅a hold.

Is the set of all natural numbers, , with the conventional addition + and multiplication ⋅ operations a ring? In there is no zero element, violating Property 4 above. Hence, the set of all natural numbers with these two operations fails to be a ring.

On the other hand, the set of all integers, , with the usual addition + and multiplication ⋅ operations is a ring. fulfills all the properties (1 to 8) above, so is a ring. Similarly, and with the usual addition + and multiplication ⋅ are rings.

In there is a multiplicative identity element, denoted by 1. For every a it holds that a⋅1 = 1⋅a = a. However, not all rings have the multiplicative identity element. For example, consider the set . The set 2ℤ satisfies all the eight rings’ properties, but 1 . If R is a ring with addition and multiplication operations + and ⋅ and there is an element 1 ∈ R such that 1⋅a = a⋅1 = a for every a ∈ R, then (R,+,∙) is called a ring with unit element or ring with unity.

In a ring, the addition has to be commutative (Property 2), but the multiplication does not. Consider the set M₂ which is defined as follows.

M₂ is the set of all real-valued square matrices of order 2. It can be shown that (M₂,+,∙) is a ring. If A, B ∈ M₂, in general A∙B B∙A, so the multiplication is not commutative.  A ring (R,+,∙) is said to be commutative if a∙b = b∙a for every a, b ∈ R. Such a ring is called a commutative ring. Therefore, (M₂,+,∙) is not a commutative ring.

The simplest ring is the trivial ring (R,+,∙) with R = {e}, which is a set containing one and only one member denoted by e. The addition in R is defined as e + e = e and the multiplication is defined as e∙e = e. With this definition, it is easy to prove that R is a ring with unity. Since e + e = e, the only possibility is e = 0 (as a consequence of Property 4) and because e.e = e, the only possibility is e = 1. So we have proved that 0 = 1!

0 = 1 only holds in the trivial ring. If (R,+,∙) is a ring containing more than one member, the ring is called a nontrivial ring. In a nontrivial ring, is it possible that 0 = 1? In a nontrivial ring, 0 ≠ 1. As a proof, assume that (R,+,∙) a nontrivial ring. Let x be any member of R. Suppose that 0 = 1. This implies x = x∙1 = x∙0 = 0. [The theorem x⋅0 = 0 is  proved below.] Since x ∈ R is arbitrary and x = 0, it follows that all elements of R are 0. In other words, R is the trivial ring. This contradicts the assumption that (R,+,∙) is a nontrivial ring. So, in a nontrivial ring, 0 ≠ 1.

Proof that x⋅0 = 0 for every x ∈ (R,+,⋅)

Let x be any member of (R,+,⋅). By Property 8, x⋅0 = x⋅(0 + 0) = x⋅0 + x⋅0. Since R is a group under addition, this equation implies that x⋅0 = 0. So, for every x ∈ (R,+,⋅) x⋅0 = 0 (q.e.d.)