We know that addition, subtraction, multiplication, and divisision are binary operations.

Question:

Is raising a number to a power also a binary operation?

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- Aug 4th 2010, 12:23 PMnoviceMore 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? - Aug 4th 2010, 12:27 PMAckbeet
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?

- Aug 4th 2010, 12:36 PMnovice
Yes sir, it does make sense since binary operation is a function $\displaystyle \mathbb{R} \times\mathbb{R} \rightarrow \mathbb{R}$.

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

Does it mean that $\displaystyle min(\delta, 1)=\delta$ too is a binary operation? - Aug 4th 2010, 12:41 PMAckbeet
Here you have to be careful. Are you talking about the function $\displaystyle \min(x,y)$? Or the function $\displaystyle f(\delta)=\min(\delta,1)$? The first is binary, the second unary.

- Aug 4th 2010, 12:59 PMnovice
Oh, yes, I was indeed very careless.

- Aug 4th 2010, 04:48 PMnovice
Now, instead of $\displaystyle \min(x,y)$, I reverse it to $\displaystyle x*y= \max(x,y)$ in $\displaystyle \mathbb{Z}$. For this, I think 0 is an identity of $\displaystyle (\mathbb{Z},*)$since $\displaystyle \max(x,0)=x$ and $\displaystyle \max(0,y)=y$.

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

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

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

Now, if you consider $\displaystyle \mathbb{N},$ 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. - Aug 4th 2010, 07:00 PMnovice
Thanks again sir. You have taught me that mathematics is all about little details. Hmm, $\displaystyle -\infty \not \in \mathbb{Z}$--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.

- Aug 5th 2010, 02:50 AMAckbeetQuote:

...mathematics is all about little details...

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!