Question 1203441: Which of the ff are associative binary operations? Show your solutions?
i. (ℤ,*), where x*y=(x+y)-(x·y) for all x,y,∈ℤ.
ii. (ℝ,*), where x*y=max (x,y) for all x,y,∈ℝ.
iii. (ℝ,*), where x*y= |x+y| for all x,y,∈ℝ.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
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
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! Which of the ff are associative binary operations? Show your solutions?
i. (ℤ,*), where x*y=(x+y)-(x·y) for all x,y,∈ℤ.
ii. (ℝ,*), where x*y=max (x,y) for all x,y,∈ℝ.
iii. (ℝ,*), where x*y= |x+y| for all x,y,∈ℝ.
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I will solve (ii) and (iii).
(ii) ((x*y)*z) = (max(x,y)*z) = obviously = max(x,y,z).
(x*(y,z)) = (x*max(y,z)) = obviously = max(x,y,z).
The expressions generate the same output for all x, y, z, so this operation (*) is associative.
(iii) Take x= 1, y= 2, z= -3. Then
((x*y)*z) = ||x+y| + z| = ||1+2| + (-3)| = |3 + (-3)| = |0| = 0.
(x*(y*z)) = |x + |y+z|| = |1 + |2+(-3)|| = |1 + |-1|| = |1 + 1| = |2| = 2.
The expressions generate different results for given x, y, z, so this operation (*) is NOT associative.
Solved.
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The lesson to learn is this:
To prove that the operation is associative, the proof should be done for all x, y, z.
To prove that the operation is not associative, it is enough to show it using special counter-examples.
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