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a # b = a + b – ab

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a # b = a + b – ab

by Ahmed MS » Sun Jun 12, 2011 11:34 pm
If # is defined by a # b = a + b - ab, then which is true?

a # b = b # a
a # 0 = a
(a # b) # c = a # (b # c)
a. I
b. II
c. I & II
d. I & III
e. I, II & III

The answer of the above problem is e. I am looking for a short and smart way to solve this problem.

Cheers!
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by Anurag@Gurome » Mon Jun 13, 2011 12:17 am
Ahmed MS wrote:If # is defined by a # b = a + b - ab, then which is true?

a # b = b # a
a # 0 = a
(a # b) # c = a # (b # c)
  • I. a # b (a + b - ab) and b # a = b + a - ab ----> TRUE
    II. a # 0 = a + 0 - a*0 = a ----> TRUE
    III. (a # b) # c = (a # b) + c - (a # b)c = (a + b - ab) + c - (a + b - ab)c = (a + b + c - ab -bc - ca + abc) = a + (b + c - bc) - a(b + c - bc) = a + (b # c) - a(b # c) = a # (b # c) ----> TRUE
The correct answer is E.
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by sunilramu » Mon Jun 13, 2011 4:12 pm
a # b = b # a
a # 0 = a
(a # b) # c = a # (b # c)


a=1, b=2, c=3

1+2-2 = 2+1-2 +> 1=1 true
1+0-0 = 1 =>1=1 true
(1+2-2)#3 = 1 # (2+3-6) => 1#3 = 1#-1 => 1+3-3=1-1+1 => 1=1 true
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by MBA.Aspirant » Mon Jun 13, 2011 4:37 pm
Ahmed MS wrote:If # is defined by a # b = a + b - ab, then which is true?

a # b = b # a
a # 0 = a
(a # b) # c = a # (b # c)
a. I
b. II
c. I & II
d. I & III
e. I, II & III

The answer of the above problem is e. I am looking for a short and smart way to solve this problem.

Cheers!
I) a#b = b#a

a+b-ab = b+a-ba

II)a#0 = a

a+0-a*0= a-0 = a

III) (a # b) # c = a # (b # c)

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

a+b-ab+c - ac-bc+abc = a+b+c-bc - ab-ac+abc
-ab = - ab

or you can use numbers

a=1 b=2 c=3
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