# Given the law of composition x*y=xy+2ax+2y determine a and b if the law is communicative and associative.

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x*y = xy + 2ax + 2y

Since the law is communicative :

==> x*y = y*x

==> x*y = xy + 2ax + 2y.........(1)

==> y*x = yx + 2ay + 2x ........(2)

we know that (1) = (2) :

==> xy + 2ax + 2y = yx + 2ay + 2x

Combin terms with a on the left side:

==> 2ax - 2ay = 2x - 2y

Now factor 2a on left and 2 on right.

==> 2a(x-y) = 2(x-y)

Divide by (x-y) :

==> 2a = 2

==> a = 2/2 = 1

**==> a= 1 **

An operator o is commutative if X o Y = Y o X

Here x*y = xy + 2ax + 2y

=> y*x = yx + 2ay + 2x

now x*y = y*x

=> xy + 2ax + 2y = yx + 2ay + 2x

=> 2ax + 2y = 2ay + 2x

=> Equating the coefficients of x and y, we get 2a = 2

=> a = 1

**Therefore a is 1.**

( b has not been in the expression)

If the law is commutative then:

x*y = y*x

We'll write the law of composition for x*y:

x*y = xy+2ax+2y (1)

y*x = yx + 2ay + 2x (2)

We'll put (1) = (2):

xy+2ax+2y = yx + 2ay + 2x

We'll eliminate like terms:

2ax+2y = 2ay + 2x

The coefficients of x from both sides have to be equal:

2a = 2

We'll divide by 2:

**a = 1**

If the law is associative then:

(x*y)*z = x*(y*z) xy+2ax+2y

(xy+2ax+2y)*z = x*(yz+2ay+2z)

(xy+2ax+2y)z + 2a(xy+2ax+2y) + 2z = x(yz+2ay+2z) + 2ax + 2(yz+2ay+2z)

We'll remove the brackets:

xyz + 2axz + 2yz + 2axy + 4a^2x + 4ay + 2z = xyz + 2axy + 2xz + 2ax + 2yz + 4ay + 4z

We'll eliminate like terms (the bolded one):

**xyz** + 2axz + **2yz** + **2axy** + 4a^2x + **4ay** + 2z = **xyz** + **2axy** + 2xz + 2ax + **2yz** + **4ay** + 4z

We'll factorize by 2ax to the left side and by 2z to the right side:

2ax(z+2a) + 2z = 2z(x+2) + 2ax

There is no b in the expression!

The law is commutative and not "communicative"!