# More on Binary Operations

• August 4th 2010, 12:23 PM
novice
More on Binary Operations
We know that addition, subtraction, multiplication, and divisision are binary operations.

Question:
Is raising a number to a power also a binary operation?
• August 4th 2010, 12:27 PM
Ackbeet
It is, because you must have a base and a power. You can think of binary operations as a predicate, or a function: a + b = sum(a,b). Similarly, exponentiation would have to look like this: x^y = power(x,y). A binary operation is like a function with two arguments. Make sense?
• August 4th 2010, 12:36 PM
novice
Quote:

Originally Posted by Ackbeet
It is, because you must have a base and a power. You can think of binary operations as a predicate, or a function: a + b = sum(a,b). Similarly, exponentiation would have to look like this: x^y = power(x,y). A binary operation is like a function with two arguments. Make sense?

Yes sir, it does make sense since binary operation is a function $\mathbb{R} \times\mathbb{R} \rightarrow \mathbb{R}$.

Now, since $min(\delta, 1)= \delta \vee 1$, it seems to me that it follows the function $\mathbb{R} \times\mathbb{R} \rightarrow \mathbb{R}$.

Does it mean that $min(\delta, 1)=\delta$ too is a binary operation?
• August 4th 2010, 12:41 PM
Ackbeet
Here you have to be careful. Are you talking about the function $\min(x,y)$? Or the function $f(\delta)=\min(\delta,1)$? The first is binary, the second unary.
• August 4th 2010, 12:59 PM
novice
Oh, yes, I was indeed very careless.
• August 4th 2010, 04:48 PM
novice
Quote:

Originally Posted by Ackbeet
Here you have to be careful. Are you talking about the function $\min(x,y)$? Or the function $f(\delta)=\min(\delta,1)$? The first is binary, the second unary.

Now, instead of $\min(x,y)$, I reverse it to $x*y= \max(x,y)$ in $\mathbb{Z}$. For this, I think 0 is an identity of $(\mathbb{Z},*)$since $\max(x,0)=x$ and $\max(0,y)=y$.

Again, if I change it to $x*y= \max(x,y)$ in $\mathbb{N}$. Will it be correct to say the 1 is an identity of $(\mathbb{N},*)$ since there is no integer less that 1?

My thinking is that if [tex]x>1, then $\max(x,1)=x$ and if $x=1$, nothing has changed $\max(x,1)=x$.

Yah?
• August 4th 2010, 05:24 PM
Ackbeet
If by $\mathbb{Z}$ you mean all the integers, then $0$ is not the identity, because $\max(-1,0)=0$, not $-1.$ The identity would have to be $-\infty$, which is not in $\mathbb{Z}$ unless you consider the "extended integers". I've never seen that, though I have seen the extended reals.

Now, if you consider $\mathbb{N},$ then you definitely do have an identity with the max function, as you say. You should be careful, incidentally, in your words. You meant, "There is no natural number less than 1." I would agree with your identity in this case.
• August 4th 2010, 07:00 PM
novice
Thanks again sir. You have taught me that mathematics is all about little details. Hmm, $-\infty \not \in \mathbb{Z}$--that's interesting. I was tempted to ask why, but I know it belongs to another thread, and perhaps it's just beyond my scope of understanding at this point.
• August 5th 2010, 02:50 AM
Ackbeet
Quote:

...mathematics is all about little details...
Details can sure be important! However, you have to be able to see the big picture as well. They're both important.

I once had a math professor who was a mathematical physicist. He was one of the most engaging lecturers I've ever had. He said that the way to get good at mathematics is to work in a high-energy physics lab. You have these 10,000 volt wires running around. If you make a mistake, you're dead. He said that's the way you should approach math.

Take it for what it's worth. (Nod)

You're welcome for whatever help I was able to provide. Have a good one!