Math4302 Modern Algebra (Lecture 1)

$+$ is a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.

---

$\cdot$ is a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.

---

division is not a binary operation on $\mathbb{Z}$ or $\mathbb{R}$. (Consider 0)

---

Generally, we can define a binary operation over sets whatever we want.

---

Let $X=\{a,b,c\}$ and we can define the (Cayley) table for binary operation as follows:

*	a	b	c
a	a	b	b
b	b	c	c
c	a	b	c

---

If we let $X$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$.

then $(f+g)(x)=f(x)+g(x)$,

$(f g)(x)=f(x)\circ g(x)$,

$(f\circ g)(x)=f(g(x))$, are also binary operations.

$(f\circ g)(x)=f(g(x))$, is not generally commutative, consider constant functions $f(x)=1$ and $g(x)=0$.

Suppose $X=\{a,b,c\}$

*	a	b	c
a	a	b	b
b	b	c	c
c	a	b	c

is not associative, take $a,b,c$ as examples.

$$a*(b*c)=a*c=b\neq (a*b)*c=b*c=c$$

Suppose $e_1$ and $e_2$ are identity elements of $X$. Then $e_1*e_2=e_2*e_1=e_1=e_2$.

$0$ is the identity element of $+$ on $\mathbb{Z}$ or $\mathbb{R}$.

$1$ is the identity element of $\cdot$ on $\mathbb{Z}$ or $\mathbb{R}$.

identity zero $f(x)=0$ is the identity element of $(f+g)(x)=f(x)+g(x)$.

identity one $f(x)=1$ is the identity element of $(f\circ g)(x)=f(g(x))$.

identity function $f(x)=x$ is the identity element of $(f\circ g)(x)=f(g(x))$.