visram04 wrote:

If operation # for all integers is defined as a#b=a+b-ab, which of the following is must be ture for all integers a,b and c?

i) a#b=b#a

ii) a#0=a

iii) (a#b)#c=a#(b#c)

Answer is all 3. Can someone help me understand i) and iii)?

Given: a#b=a+b-ab(i) a#b=b#a

a#b = a+b-ab

b#a = b+a-ba --> we can rewrite this as a+b-ab

which is same as a#b. Thus it is true

(ii) a#0=a

a#0 = a+0-(a*0) = a. Thus true

(iii) (a#b)#c=a#(b#c)

(a#b)#c

Consider a#b = x. Thus (a#b)#c ==> x#c => x+c-(x*c)

Now a#b = x => a+b-ab =x. Sub in the above equation

a+b-ab+c-[(a+b-ab)*c) = a+b+c-ab -[ac+bc-abc] = a+b+c-ab-ac-bc+abc--(1)

a#(b#c)

Consider b#c =y. Thus a#(b#c) --> a#y -> a+y-ay

Now b#c=y --> b+c-bc =y. Sub in the above equation

a+b+c-bc-[a*(b+c-bc)] --> a+b+c-bc-[ab+ac-abc] --> a+b+c-bc-ab-ac+abc --(2)

(1) =(2)

Thus (a#b)#c=a#(b#c) is true.

Hence all are true

_________________

I'm happy, if I make math for you slightly clearer

And yes, I like kudos ¯\_(ツ)_/¯