PROVE the Properties of Subtraction and Negatives For any el

(PROVE the Properties of Subtraction and Negatives). For any elements a and b of a ring R,

(1) a-a = 0.

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

(3) -(-a) = a.

(4) -(a + b) = (-a) + (-b). (5) -(a-b) = -a + b.

(6) (-a)(-b) = ab.

(7) a(b-c) = ab-ac and (b-c)a = ba-ca.

(8) if R has an identity 1, then (-1)a = -a.

Solution

Answar:-

The Properties of Subtraction and Negatives, For any elements a and b of a ring R is as follows.

Part(1):- a-a = 0.

a +(-a)= 0 additive property.

1*(a) -1(a) =0   multiplicative property

a(1-1) = 0   

a(0) =0 identity .            hence proved.

Part(2):- By denition -(ab) is the unique solution of ab +x =0

so any other solution of this equation must be equal to -(ab).

But x=a(-b) is also the solution,

since by distributive law and a,

ab + a(-b)

= a{b + (-b) }

= a(0)

Therefore     -(ab) = a(-b).

or (a)b = a(b) = (ab) [0=0b = (a+(a))b = ab+(a)b,so (a)b = (ab);

similarly,
0 = a0 = a(b + (b)) = ab + a(b), so a(b) = (ab)]

a(-b) = -(ab) = (-a)b.    hence proved

Part(3):- By definition,

-(-a) is a unique solution of (-a) +x =0

but x=0 is also the solution

so,   a = -(-a)

hence proved

Part(4)-by definition,

-(a+b) is the unique solution of (a+b) +x =0

But (-a) +(-b) is also the solution

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

so,

bu uniqueness

a+b = (-a) + (-b)

hence proved.

Part(5):- -(a-b) = -a + b.

by the definition od subtraction,

-(a+b) = - (a -(-b) )

-(a+b) = (-a) + (-(-b))           ( by rule # 3 and 5)

-(a+b)   = -a +b

hence proved.

Part(6):- (-a)(-b) = ab.

(-a)(-b) = -(a(-b) )            multiplicative property.

(-a)(-b) = -(-(ab) )                   (by ruke #4)

(-a)(-b) = (ab)                         (by rule #3)

hence proved.

Part(7):- a(b-c) = ab-ac and (b-c)a = ba-ca.

this is called distributive property.

a(b-c) = ab-ac     called left distributive property.

a(b-c)

= a(b) + a(-c)

= a(b) + (-a)(c)

so   = a(b) + (-a)(c)         from rule #2

then

= ab-ac     hence proved.

and

(b-c)a = ba-ca.   called right distributive property.

(b-c)a

=(b) (a) + (-c) (a)        multiplicative property.

=(b) (a) + (c) (-a)    by rule#2

so,

(b-c)a = (b) (a) - (c) (a)

(b-c)a = ba - ca   hence proved.

Part(8):- if R has an identity 1, then (-1)a = -a.

(-1)a

= -(1 a)               by rule#2

= -(a)

(-1)a =   -a     hence proved.