**Properties of rings** in groups, let** (R, +, ×)** be a ring. Since **(R, +)** is a group, **all the group properties** for addition hold in R.** (real number)**

The **unique identity** of the additive group is called the **zero element of the ring**, and we denote it by 0. **Thus, a + 0 = 0 + a = a** for every a ∈ R (a. is a part of real number).

We denote the** unique** **additive inverse** of an element a by – a and write as a + (-**a) = a -a = 0 and a + (-b)= a – b.** We note that the properties of the **additive** **inverse** and the zero element give us the cancellation law.

If** a + b = a + c, b = c, a** unique solution **x = b – a,** of the equation a + x = b and the rule – (- a) = a, if n ∈ Z and n > 0.

Furthermore,

we define **n×a = a + a + a +……..a(n term).**

If n< 0 we define** n.a = (-a)+(-a)+(-a)…….(-a)(n terms).** And** if n = 0,**

** we define 0a = 0,** Where** 0 ∈ Z **on the left side of the equation and **0 ∈ R** on the right side.

The reader may note that n.a, is not to be considered As a product of n and a. in the ring, for the integer+ n may not be in the ring at all.

**The equation 0×a = 0 holds also 0 ∈ R** on both sides.

** Properties of rings in groups:**

**EXAMPLE BY THEOREM:**

**If (R, +, ×)** is a ring with additive identity 0, then, for all a, b ∈ R we have

**(1) a×0 = 0×a = 0**

**(2) a(-b) = (-a)b = -ab **

**(3) (-a)(-b)= ab **

**PROOF:**

**(1) a 0 = a(0 +0) = a 0 + a 0**

**Hence, 0 + a 0 =a 0 + a 0**

by the cancellation law for the** additive group (R, +)**,

we have

**0 = a 0**

likewise,

** 0 a = (0 + 0)a = 0 a + 0 a implies 0 a = 0**

**(2) a(-b) = (-a)b = -a b **

**proof:**

**By definition**, – (ab) is the element which added to **ab gives 0.** Thus, in order to show

**a(-b) = -(ab),** we must show **a(-b) +ab = 0** by the left distribution Law

**a (-b) +ab = a (-b+ b) = a 0 = 0**

similarly, (-a) b+ab = (-a + a) b = ob= 0

The result follows from the above equation.

**(3) proof:**

**Using operation adopted in 2**

(-a)(-b)= -(a(-b)) = -(-(ab)= ab

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