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:

- (a + b) ∈ R for every a, b ∈ R
- a + b = b + a for every a, b ∈ R
- (a + b) + c = a + (b + c) for every a, b, c ∈ R
- There exists 0 ∈ R such that a + 0 = a for every a ∈ R.
- For every a ∈ R there exists an element -a ∈ R such that a + (-a) = 0
- a⋅b ∈ R for every a, b ∈ R
- a⋅(b⋅c) = (a⋅b)⋅c for every a, b, c ∈ R
- 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:

- (R,+) is an abelian group
- a⋅b ∈ R for every a, b ∈ R
- a⋅(b⋅c) = (a⋅b)⋅c for every a, b, c ∈ R
- 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 M_{2} which is defined as follows.

M_{2} is the set of all real-valued square matrices of order 2. It can be shown that (M_{2},+,∙) is a ring. If A, B ∈ M_{2}, 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

*. Therefore, (M*

**commutative ring**_{2},+,∙) 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.)