Question 1203441
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I'll work on part (i) only, and leave parts (ii) and (iii) for the student to do.


If the operator star was associative, then a*(b*c) = (a*b)*c must be true. 


Let's do a bit of scratch work
x*y=(x+y)-(x·y)
a*b=(a+b)-(a·b)
b*c=(b+c)-(b·c)



Then we can say:
a*(b*c)=a*((b+c)-(b·c))
a*(b*c)=(a+(b+c)-bc)-a((b+c)-(b·c))
a*(b*c)=(a+(b+c)-bc)+(-a(b+c)+abc)
a*(b*c)=(a+b+c-bc)+(-ab-ac+abc)
a*(b*c)=(a+b+c)+(-bc-ab-ac+abc)
Be careful not to mix up the star operator with the multiplication symbol.



And furthermore,
(a*b)*c = ((a+b)-(a·b))*c
(a*b)*c = ((a+b)-(a·b)+c)  - ((a+b)-(a·b))*c 
(a*b)*c = ((a+b)-(a·b)+c)  - (ac+bc)+(abc)
(a*b)*c = ((a+b)-(a·b)+c)  + (-ac-bc)+(abc)
(a*b)*c = (a+b+c-ab) + (-ac-bc+abc)
(a*b)*c = (a+b+c) + (-ab-ac-bc+abc)


In summary we found these equations
a*(b*c)=(a+b+c)+(-bc-ab-ac+abc)
(a*b)*c = (a+b+c) + (-ab-ac-bc+abc)


Compare the right hand sides of a*(b*c) and (a*b)*c 
Both have the same exact terms. Compare the terms carefully.


Therefore, we have proven that a*(b*c) = (a*b)*c
This shows the operator star is associative when we have defined the operator star to be x*y = (x+y) - xy
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