Is $\langle\mathbb Q^+, *\rangle$ a monoid?

Question

Q: Given the set of positive rational numbers $\mathbb Q^+$, the operation is multiplication$~*$.
Is $\left<\mathbb Q^+, *\right>$ a monoid?

My answer is:

$ \forall x, y, z \in \mathbb Q^+$, then $ x*y \in \mathbb Q^+ $. So $\mathbb Q^+$ is a closed set.
And $ x*(y*z) = (x*y)*z $. So it is associative under operation multiplication, thus $\mathbb Q^+$ is a semigroup.

In addition, $$ 1*x= x*1 = x$$ so $\mathbb Q^+$ has an identity element $1$.

In summary, $\left<\mathbb Q^+, *\right>$ a monoid.

Is my answer correct?

Answer 1

Your answer is basically correct, but in a couple of places the terminology and notation are a little off.

• It’s not really correct to say that $\Bbb Q^+$ is a closed set: in this context closure is a joint property of set and an operation, so you should say that $\Bbb Q^+$ is closed under $*$.

• The statement it is associative under [the] operation multiplication isn’t quite right: associativity is a property of binary operations, so it’s the operation $*$ itself that is associative.

• A semigroup requires a binary operation as well as a set, so it’s $\langle\Bbb Q^+,*\rangle$ that’s a semigroup, not $\langle\Bbb Q^+\rangle$. Similarly, it’s the semigroup $\langle\Bbb Q^+,*\rangle$ that has an identity element.

Answer 2

You know that the set $\Bbb Q$ of rational numbers is a (commutative) monoid under multiplication $\ast$ (it would have been a nice group, but $0$ has no multiplicative inverse).

The set $\Bbb Q^+$ of strictly positive rational numbers is a subset of $\Bbb Q$, which is closed under $\ast$ and contains the identity element $1$. Hence $(\Bbb Q^+,\ast)$ is a submonoid of $(\Bbb Q,\ast)$, this is $(\Bbb Q^+,\ast)$ is a monoid itself.

Further, since every the element of $\Bbb Q^+$ have a unique inverse in $\Bbb Q^+$, you have that $(\Bbb Q^+,\ast)$ is also an (abelian) group.