1. ## [SOLVED] Binary Operations

Suppose that two binary operations, denoted by $\oplus$ and $\odot$, are defined on a nonempty set S, and that the following conditions are satisfied $\forall x,y,z\in S$.

(1) $x\oplus y$ and $x\odot y$ are in S
(2) $x\oplus (y\oplus z)=(x\oplus y) \oplus z$ and $x\odot (y\odot z)=(x\odot y) \odot z$
(3) $x\oplus y=y\oplus x$

Also, $\forall x\in S$ and $\forall n\in \mathbb{Z}^+$, the elements $nx$ and $x^n$ are defined recursively as follows:

$1x=x^1=x$
if $kx$ and $x^k$ have been defined, then $(k+1)x=kx\oplus x$ and $x^{k+1}=x^k\odot x$

Which of the following must be true?

(i) $(x\odot y)^n=x^n\odot y^n$ $\forall x,y\in S$ and $\forall n\in\mathbb{Z}^+$

(ii) $n(x\oplus y)=nx\oplus ny$ $\forall x,y\in S$ and $\forall n\in\mathbb{Z}^+$

(ii) $x^m\odot x^n=x^{m+n}$ $\forall x\in S$ and $\forall m,n\in\mathbb{Z}^+$
This is one obviously true.

I am struggling with proving or disproving 1 and 2

2. Originally Posted by dwsmith
Suppose that two binary operations, denoted by $\oplus$ and $\odot$, are defined on a nonempty set S, and that the following conditions are satisfied $\forall x,y,z\in S$.

(1) $x\oplus y$ and $x\odot y$ are in S
(2) $x\oplus (y\oplus z)=(x\oplus y) \oplus z$ and $x\odot (y\odot z)=(x\odot y) \odot z$
(3) $x\oplus y=y\oplus x$

Also, $\forall x\in S$ and $\forall n\in \mathbb{Z}^+$, the elements $nx$ and $x^n$ are defined recursively as follows:

$1x=x^1=x$
if $kx$ and $x^k$ have been defined, then $(k+1)x=kx\oplus x$ and $x^{k+1}=x^k\odot x$

Which of the following must be true?

(i) $(x\odot y)^n=x^n\odot y^n$ $\forall x,y\in S$ and $\forall n\in\mathbb{Z}^+$

(ii) $n(x\oplus y)=nx\oplus ny$ $\forall x,y\in S$ and $\forall n\in\mathbb{Z}^+$

(ii) $x^m\odot x^n=x^{m+n}$ $\forall x\in S$ and $\forall m,n\in\mathbb{Z}^+$
This is one obviously true.

I am struggling with proving or disproving 1 and 2
I think (i) is false, but I don't think my demonstration constitutes a rigorous proof.

We can rewrite

$(x\odot y)^n=\underbrace{(x\odot y)\odot (x\odot y) \odot ... \odot (x\odot y)}_{n \text{ instances of }(x\odot y)}$

Since $\odot$ is not commutative, we cannot rearrange terms.

I think (ii) is true and can be proven from the definition of multiplication and conditions (2) and (3).

We can rewrite

$n(x\oplus y)=\underbrace{(x\oplus y) \oplus (x\oplus y) \oplus \cdots \oplus (x\oplus y)}_{n \text{ instances of }(x\oplus y)}$

Now with commutativity and associativity we can simply rearrange terms to get

$n(x\oplus y)=\underbrace{x \oplus x \oplus \cdots \oplus x}_{n \text{ instances of } x} \oplus \underbrace{y \oplus y \oplus \cdots \oplus y}_{n \text{ instances of } y}$

$= nx\oplus ny$

3. Why does $n(x\oplus y)$ mean $(x\oplus y)\oplus (x\oplus y)....$? I don't see how you deciphered that from what was giving.

4. Originally Posted by dwsmith
Why does $n(x\oplus y)$ mean $(x\oplus y)\oplus (x\oplus y)....$? I don't see how you deciphered that from what was giving.
Well $1x = x$

$2x = 1x \oplus x = x \oplus x$

$3x = 2x \oplus x = (x \oplus x) \oplus x = x \oplus x \oplus x$

...

5. $nx=(n-1+1)x=(n-1)x\oplus x=(n-2+1)x\oplus x=(n-2)x\oplus x \oplus x.....$

6. Originally Posted by dwsmith
$nx=(n-1+1)x=(n-1)x\oplus x=(n-2+1)x\oplus x=(n-2)x\oplus x \oplus x.....$
This relies on

$(n_1+n_2)x = n_1x \oplus n_2x$

which I suppose is obvious since it's just the $\oplus$ version of (iii), nevertheless it still requires proof.

The way I wrote it down, weak induction could be used for a formal proof.

I'm curious as to your obvious proof of (iii), since I would have used

$x^n = \underbrace{x \odot x \odot \cdots \odot x}_{n \text{ instances of } x}$

from which (iii) follows trivially.

7. $x^{k+1}=x^k\odot x$

This is why 3 is obvious.