sequence {0,1} adding and multiplication

Question

Define addition of a set of symbols {0, 1} by:

$$0 + 0 = 0, \ 0 + 1 = 1, \ 1 + 0 = 1, \ 1 + 1 = 0$$

Define multiplication by n where n is a non-negative integer by:

$$0·0=0, \ 0·1=0, \ n·0=0, \ n·1=1+1+···+1 \ for \ n>0$$

How can you prove that if $n$ is odd then $n·1=1$ and that if $n$ is even then $n·1=0$.

I am having a really hard time trying to understand this.

I was thinking that I should sub. $n=2k+1$ because it is odd and try working from there. I was wondering if anyone could give me a hint or start me off.

Answer 1

Consider $n=2k+1, \ k\in\mathbb{Z}$. We have:

$2k+1=1+1+1....+1$ (we add $1$ $2k+1$ times). Since $1+1=0$, then $2k+1=0+1+1....+1$ (we add $1$ $2k-1$ times). But $0+1=1$, so now $2k+1=1+...+1$ (we add $1$ $2k-3$ times). Following this algorithm we'll get $2k+1=0+1=1$.

The proof for $2k=0$ follows the exact same algorithm.

Note (may be of interest to you):

The given operation $+$ is, in fact, the logical $exclusive \ or$. See https://en.wikipedia.org/wiki/Exclusive_or.

Answer 2

I'm not quite sure where you're stuck, but since you ask for a hint, let me try to give one.

Your multiplication operation is defined by iterating addition. $2 \cdot 1 = 1 + 1$. Note there are exactly $2$ ones. Similarly, $5 \cdot 1 = 1 + 1 + 1 + 1 +1$. Exactly $5$ copies.

Now by your table above, we can compute these. You say above $1 + 1 = 0$, so $2 \cdot 1 = 0$. Then (assuming associativity of $+$, which you can check), we can write the second example as $1 + 1 + 1 + 1 + 1 = (1 + 1) + (1+1) + 1) 2 \cdot 1 + 2 \cdot 1 + 1 = 0 + 0 + 1 = 0 + 1 = 1$

Can you see how to generalize this?