Monoid proof

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January 25th 2010, 07:43 AM
pikminman

Monoid proof

I'm a little bit stuck on an assignment question.

The question is:

Quote:

Let c be a fixed positive integer, and let * denote the binary operation on the set Z of integers defined be the formula

$x * y = xy + c(x+y) + c^2 - c$

for all integers x, y, and z.

Is (Z, *) a monoid? If so, what is it's identity element?

I already proved that (Z, *) is a semigroup. I also showed that x*e = e*x . I just don't know how to find e. (That is, if e exists at all)

This is what I have so far:

$x*e = xe + c(x+e) + c^2 - c$

$<br /> e*x = ex + c(e+x) + c^2 - c<br /> <br />$

Any help would be much appreciated.

EDIT: I'm pretty sure it's NOT a monoid. Do you think I need to prove this though? Can I just say something along the lines of "clearly, there is no identity element" :P

$x * e = xe + c(x+e) + c^2 - c \neq x$
January 25th 2010, 12:35 PM
girdav

I guess you proved associativity.
The law $*$ is commutative, hence you only have to find $e$ such as $x*e= x$ .
You can isolate $e$ in this equation.
January 25th 2010, 12:41 PM
pikminman

Quote:

Originally Posted by girdav

I guess you proved associativity.
The law $*$ is commutative, hence you only have to find $e$ such as $x*e= x$ .
You can isolate $e$ in this equation.

Well how could I do that? I'm really confused.
January 25th 2010, 01:18 PM
girdav

We have $x*e =x$ so $xe+c\left(x+e\right) +c^2-c =x \Rightarrow e\left(x+c\right)+cx+c^2-c = x$ so we obtain
$e\left(x+c\right)= \left(1-c\right)x+c\left(1-c\right)$
January 25th 2010, 01:44 PM
Drexel28

Quote:

Originally Posted by girdav

We have $x*e =x$ so $xe+c\left(x+e\right) +c^2-c =x \Rightarrow e\left(x+c\right)+cx+c^2-c = x$ so we obtain
$e\left(x+c\right)= \left(1-c\right)x+c\left(1-c\right)$

You of course need to verify that when solved the above yields an identity element in the underlying set of the supposed monoid.