# mathematical operations

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• Jan 11th 2010, 11:42 PM
Bingk
sorry :)
• Jan 11th 2010, 11:43 PM
Bingk
Quote:

Originally Posted by Drexel28
No. An operation, most commonly a binary operation is just a function $\displaystyle f:G^2\mapsto G$. You take two elements of a set and get one back. How could $\displaystyle \mathbb{R}^2$ be one dimensional?

um ... if the codomain is a set of ordered pairs (i.e. { (x,y) | x,y element of R } ... you get back ONE element from a set, so the one element could be an ordered pair, triple, or whatever you want ....

Just want to be sure, so you believe that "operation" can be changed with "function" (which I agree with), but not always vice-versa, right? If so, could you explain why it doesn't work the other way around?
• Jan 11th 2010, 11:44 PM
Drexel28
Quote:

Originally Posted by Bingk
um ... if the codomain is a set of ordered pairs (i.e. { (x,y) | x,y element of R } ... you get back ONE element from a set, so the one element could be an ordered pair, triple, or whatever you want ....

Just want to be sure, so you believe that "operation" can be changed with "function", but not vice-versa, right? If so, could you explain why it doesn't work the other way around?

Ok, it's late. So bear with me. A operation will always be a function, the difference is that in most cases the "function" is of the form $\displaystyle f:X\jmath\mapsto X$.
• Jan 11th 2010, 11:58 PM
Bingk
Quote:

Originally Posted by Drexel28
Ok, it's late. So bear with me. A operation will always be a function, the difference is that in most cases the "function" is of the form $\displaystyle f:X\jmath\mapsto X$.

hehehe ... not in my part of the world ;)

ah, okay ... I get what you're saying ... an operation is a particular type of function that is of the mentioned form :).

Um, sorry for kinda dragging this out, but if that is indeed the case, why do we sometimes distinguish some operations to be operations (addition, multiplication, etc.), whilst other times, we don't make this distinction (in particular, the floor FUNCTION is an operation, but we call it function instead of operation)?

I'm sure there are other named functions out are called functions, but can also be considered to be operations ... why is that? (basically same question as above) ... that's what I'm really curious about ... is there a particular "rule/idea" for which operations can be named operations?
• Jan 11th 2010, 11:59 PM
Drexel28
Quote:

Originally Posted by Bingk
hehehe ... not in my part of the world ;)

ah, okay ... I get what you're saying ... an operation is a particular type of function that is of the mentioned form :).

Um, sorry for kinda dragging this out, but if that is indeed the case, why do we sometimes distinguish some operations to be operations (addition, multiplication, etc.), whilst other times, we don't make this distinction (in particular, the floor FUNCTION is an operation, but we call it function instead of operation)?

I'm sure there are other named functions out are called functions, but can also be considered to be operations ... why is that? (basically same question as above) ... that's what I'm really curious about ... is there a particular "rule/idea" for which operations can be named operations?

The floor function is unary operation. We usually just call these functions, the others you mentioned where binary operations.
• Jan 12th 2010, 12:06 AM
Bingk
Quote:

Originally Posted by Drexel28
The floor function is unary operation. We usually just call these functions, the others you mentioned where binary operations.

ah ... so unary operations are usually called functions? that's basically what I was looking for, thanks :)
• Jan 12th 2010, 07:23 PM
Bacterius
Well, in the case of unary functions you need to introduce a new notation somehow, because using operators such as $\displaystyle +$, $\displaystyle \times$, would raise an obvious issue.
• Jan 12th 2010, 07:35 PM
Drexel28
Quote:

Originally Posted by Bacterius
Well, in the case of unary functions you need to introduce a new notation somehow, because using operators such as $\displaystyle +$, $\displaystyle \times$, would raise an obvious issue.

And what may I ask is the problem?
• Jan 12th 2010, 08:03 PM
Bacterius
I'm just trying to grasp what kind of issue Bingk is encountering when differentiating functions/operations.
• Jan 12th 2010, 10:52 PM
Drexel28
Quote:

Originally Posted by Bacterius
I'm just trying to grasp what kind of issue Bingk is encountering when differentiating functions/operations.

Differentiating functions and operators can somtimes be quite different. (Wink)
• Jan 12th 2010, 11:02 PM
Bacterius
Yeah :D
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