Checking operations over a set

[IKonquer]

I am trying to the following operation * is closed under R (the real numbers).

x * y = x + 2y + 4

Check commutability - NOT commutative
x * y = x + 2y + 4
y * x = y + 2x + 4

Check associativity - NOT associative
x * (y * z) = x * (y + 2z + 4) = x + 2(y + 2z + 4) + 4 = x + 2y + 4z + 12
(x * y) * z = (x + 2y +4) * z = x + 2y + 4 + 2z + 4 = x + 2y + 2z + 8

Check for identity - NO identity which also means there is no inverse
x * e = x

x + 2e + 4 = x
e = -2

x * (-2) = x + (2)(-2) + 4 = x
(-2) * x = (-2) + 2x + 4 = 2x - 2

Is the work above correct?

Thanks in advance

 

[HallsofIvy]

Yes, it is. You should note in your last that an identity, e, must satisfy both e*x= x and x*e= x. What you have shown, of course, is that x*(-2)= x but (-2)*x\ne x.

 

[Wingeer]

Exactly what are you proving? That the operation forms a group structure over the set, or that the set is closed under the operation? If the last one it is enough to see whether two elements in the set under the operation is still in the set. If the first one you would have to check associativity, existence of an identity, existence of inverses for all elements in the set and at last that the set is closed under the operation.

 

[IKonquer]

Wingeer I am trying to show that the operation forms a group structure over the set.

When finding the identity, why should it be x * e = x and not e * x = x? I know I have to check both, but when finding the identity, my book tells me to use x * e = x to solve for the identity.

x * y = x + 2y + 4

x * e -> x + 2e + 4 = x -> e = -2
e * x -> e + 2x + 4 = x -> e = -x - 4

If there is an "x" term when solving for the identity, can I assume that solution is wrong?

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Also in another problem when solving for the identity via x * e = x, I found that 2 = x and from e * x = x, I found that e = (-x)/(1-x).

Does that mean there is no identity?

 

[Wingeer]

Yes. First use x * e = x to solve for the identity, then you check whether this identity is valid in e * x = x. If it is not, then you have shoved that the operation does not form a group structure on the set, and you're done.

Regarding the other problem, will you write it down so I can see the operation?
However, it does not mean that there is no identity. If the elements in the set are of a form such that every x in that equation gives 2, then you have an identity.

 

[IKonquer]

"Yes. First use x * e = x to solve for the identity, then you check whether this identity is valid in e * x = x. If it is not, then you have shoved that the operation does not form a group structure on the set, and you're done."

Right this is what my book tells me to do. But my question is why can't I first use e * x = x instead of x * e = x?

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x * y = x + 2y - xy

Check for identity

When solving x * e = x:

x + 2e - xe = x
2e = xe

e = 0 or if e is not equal to 0, then 2 = x.

I'm not sure what to make of x = 2, but e = 0 does not satisfy both x * e = x and e * x = x. So I'm assuming there would be no identity element.