# [SOLVED] Binary Operations

#### dwsmith

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

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

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

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

Which of the following must be true?

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

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

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

I am struggling with proving or disproving 1 and 2

#### undefined

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

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

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

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

Which of the following must be true?

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

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

(ii) $$\displaystyle x^m\odot x^n=x^{m+n}$$ $$\displaystyle \forall x\in S$$ and $$\displaystyle \forall m,n\in\mathbb{Z}^+$$
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

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

Since $$\displaystyle \odot$$ 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

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

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

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

$$\displaystyle = nx\oplus ny$$

• dwsmith

#### dwsmith

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

#### undefined

MHF Hall of Honor
Why does $$\displaystyle n(x\oplus y)$$ mean $$\displaystyle (x\oplus y)\oplus (x\oplus y)....$$? I don't see how you deciphered that from what was giving.
Well $$\displaystyle 1x = x$$

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

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

...

#### dwsmith

MHF Hall of Honor
$$\displaystyle nx=(n-1+1)x=(n-1)x\oplus x=(n-2+1)x\oplus x=(n-2)x\oplus x \oplus x.....$$

#### undefined

MHF Hall of Honor
$$\displaystyle nx=(n-1+1)x=(n-1)x\oplus x=(n-2+1)x\oplus x=(n-2)x\oplus x \oplus x.....$$
This relies on

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

which I suppose is obvious since it's just the $$\displaystyle \oplus$$ 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

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

from which (iii) follows trivially.

#### dwsmith

MHF Hall of Honor
$$\displaystyle x^{k+1}=x^k\odot x$$

This is why 3 is obvious.