Prove that the binary operation a*b=a does not contradict the theorem 2.1.8

Question

Suppose we have the binary operation ∗ defined on ℝ by ∗ = . Then every ∗ = for every ∈ ℝ. For example, ∗ 2 = and ∗ 7 = . Why doesn’t this contradict Theorem 2.1.8

Theorem 2.1.8: Let be a set with a binary operation ∗. If there exists an identity element ∈ , then it is unique.

I am having a hard time showing why this operation doesn't violate theorem 2.1.8

Answer 1

The relation $a * 2 = a$ shows that $2$ is a right identity element, not an identity element. Similarly for $7$. There is no theorem saying that a right identity element is necessarily unique.

(However, Theorem 2.1.8 can be extended to the slightly more general statement that if a binary relation has a left identity element and a right identity element, then these two elements are equal. Thus, in the presence of a left identity element, it is indeed true that the right identity element is unique (if it exists), because any two right elements would have to both be equal to the left identity element. But your relation $*$ has no left identity element, so nothing forces the right identity element to be unique.)

Answer 2

Let $S$ be any set, equipped with the operation $xy = y$ for all $s, y \in S$. This operation is associative and hence $S$ is a semigroup. However, if $S$ contains at least two elements, then $S$ is not commutative and has no identity (an element $1$ of a semigroup $T$ is an identity if $1x = x = x1$ for all $x \in T$).

However, it is possible to add an identity to your semigroup $S$ by adding a new element $1$ to $S$ and by setting $1x = x = x1$ for all $x \in S \cup \{1\}$. I let you verify that $1$ is the (unique) identity of $S \cup \{1\}$. No contradiction with the given theorem.