We know that addition, subtraction, multiplication, and divisision are binary operations.
Question:
Is raising a number to a power also a binary operation?
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?
Now, instead of , I reverse it to in . For this, I think 0 is an identity of since and .
Again, if I change it to in . Will it be correct to say the 1 is an identity of since there is no integer less that 1?
My thinking is that if [tex]x>1, then and if , nothing has changed .
Yah?
If by you mean all the integers, then is not the identity, because , not The identity would have to be , which is not in unless you consider the "extended integers". I've never seen that, though I have seen the extended reals.
Now, if you consider 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.
Details can sure be important! However, you have to be able to see the big picture as well. They're both important....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.
You're welcome for whatever help I was able to provide. Have a good one!