Let ${\mathbb{F}_2} = \left\{ {0,1} \right\}$, with addition defined by $0 + 0 = 1 + 1 = 0$ ; $0 + 1 = 1 + 0 = 1$, and multiplication by $0 \cdot 0 = 0 \cdot 1 = 1 \cdot 0 = 0$; $1 \cdot 1 = 1.$
I need help with the following:
- Show that $+$ and $\cdot$ are associative and commutative.
- Show that $0+u=0=1\cdot u$ whenever $u \in {\mathbb{F}_2}$.
For question 1. I have:
Addition:
Commutativity: $0+1=1+0=1$, and;
Associativity: $0+\left(1+0 \right)=\left(0+1 \right)+0=1$ .
Multiplication:
Commutativity: $0 \cdot 1=1 \cdot 0=0$, and;
Associativity: $0\left(1\cdot0 \right)=\left(0 \cdot 1 \right)0=0$ .
Is this the right way of showing addition and multiplication are commutative and associative?
For question 2.:
I'm having a hard time understanding this, because if $u=0$, then $0+0=0=1\cdot 0$, which is fine, but if $u=1$, then $0+1 \not = 0 \not =1\cdot 1$. Or am I missing something?
"Is this the right way" Yes, but you need to check all other possible combinations.
For commutativity, it's quite obvious that $1+1 = 1+1$ and so on, but for assoviativity you need to check $1+(0+1) = (1+0) + 1$, and all other conceivable combinations. There are $8$ of them in total. Some of them follow immediately from commutativity (like $1+(1+1) = (1+1)+1$), but just checking one of the combinations isn't quite enough.
Also, the checking needs to be done slowly and carefully. Preferably, you should have something like $$ (0+1) +0 = 1+0 = 0\\ 0+(1+0) = 0+1 = 0 $$ spelled out explicitly for at least one, but possibly all of the combinations you check.
Number 2. likely has a typo in it, as expressed in the comments above, and when that's fixed up it should be much easier.