mathematical operations

**hebby** · January 5th 2010, 09:33 PM

I think the last one is the type im looking for, could someone show me how that might work with a number like

209383404?

Thanks

**HallsofIvy** · January 6th 2010, 03:31 AM

You have been given two equivalent expressions.
$\lfloor x\rfloor$ , called the "floor" or "integer part" (for positive numbers) gives the largest integer larger than or equal to x.

$\{x\}$ , called the "decimal part" is $x-\lfloor{x}\rfloor$ .

If x= 209383404, then x/1000= 209383.404. It's "floor" is 209383.

Or if x= 209383404, x/1000= 209383.404. It's "decimal part" is .404 and then 209383.404- .404= 209383.

**Bingk** · January 6th 2010, 07:55 AM

Hi ... Yeah, I know that it's essentially the same, but it goes back to my question about the floor function being an operation (i.e. I'm pretty sure it is also an operation, but I'm wondering if there's any significance in it being called a function rather than an operation)

Also, I asked if my proposal was okay, because I think it fails the requirement of two operations ...

hebby, this is how it would work:

$404 \equiv 209383404 \ mod \ 1000$

So,
$\frac{209383404 - 404}{1000} = 209383$

**Drexel28** · January 6th 2010, 08:07 AM

Originally Posted by Bingk

Hi ... Yeah, I know that it's essentially the same, but it goes back to my question about the floor function being an operation (i.e. I'm pretty sure it is also an operation, but I'm wondering if there's any significance in it being called a function rather than an operation)

You would agree that $+$ is an operation? But, why isn't viewing this plus the same thing as viewing the function $f:\mathbb{R}^2\mapsto \mathbb{R},\text{ }(x,y)\mapsto x+y$ ?

**Bingk** · January 6th 2010, 02:01 PM

Hi ... I actually had a response typed out before I lost electricity (and thus internet connection) ...

That's kind of my point, why is it called an operation instead of a function? Why floor function instead of floor operation? Similarly, why addition operation, instead of addition function? Is there any particular reason for this terminology? I know the main difference between an operation and a function is that an operation maps to only one "dimension", whilst functions can be mapped to more.

In any case, this is kinda heading off-topic ...

hebby, depending on how much calculus you've learnt, you should be familiar with the floor function, and I think that would be the solution. Otherwise, well ... you'll have to think of something

**Drexel28** · January 6th 2010, 02:34 PM

Originally Posted by Bingk

Hi ... I actually had a response typed out before I lost electricity (and thus internet connection) ...

That's kind of my point, why is it called an operation instead of a function? Why floor function instead of floor operation? Similarly, why addition operation, instead of addition function? Is there any particular reason for this terminology? I know the main difference between an operation and a function is that an operation maps to only one "dimension", whilst functions can be mapped to more.

In any case, this is kinda heading off-topic ...

hebby, depending on how much calculus you've learnt, you should be familiar with the floor function, and I think that would be the solution. Otherwise, well ... you'll have to think of something

I'll get back more in detail later but [tex]f:/mathbb{R}^2/mapsto/mathbb{R}^2[\math] surely is a function

**Bingk** · January 9th 2010, 10:13 PM

Um, there's something wrong with your latex, but I'm guessing you're saying that f:R^2 -> R^2 ... and yeah, that is a function, and I did say that functions can be mapped to more than one dimension. It's operations that are limited to R^n -> R.

I was just wondering (since my math is limited), why the distinction between operation and function, and why in particular some functions are distinguished as operations, whilst others are not (even though they are also operations) ... such as the floor function ....

**Bacterius** · January 9th 2010, 10:41 PM

Originally Posted by Bingk

Um, there's something wrong with your latex, but I'm guessing you're saying that f:R^2 -> R^2 ... and yeah, that is a function, and I did say that functions can be mapped to more than one dimension. It's operations that are limited to R^n -> R.

I was just wondering (since my math is limited), why the distinction between operation and function, and why in particular some functions are distinguished as operations, whilst others are not (even though they are also operations) ... such as the floor function ....

An operation is a very basic calculation based on an axiom : addition, multiplication, division, modulus, substraction, etc ... They can be used on their own or as parts of functions to accomplish something. Example : $4 + 4$ , $3 \times 9$ , $7 \equiv 3 \pmod{4}$ ... An operation can be viewed as a simple function.

A function is a combination of operations arranged in order to achieve a particular goal. Take the factorial as an example : $n! = 2 \times 3 \times 4 \times ... \times n$ . This is a combination of $n - 1$ multiplications, arranged in order to obtain the factorial of a number.

A function is not necessarily defined as $f(x)$ or $g(x, y)$ or any other letter. $n!$ is a function. $\sqrt{x}$ is a function. $e^x$ is a function. Etc ...

Do you understand better now ?

**Bingk** · January 9th 2010, 10:55 PM

yup ... makes more sense ... but if that is the case, going back to hebby's original question, then I don't think we could use the floor function ....

For your convenience ... hebby is asking how you can "cut off" the last three digits (on the right), of some integer, in two mathematical operations ...

i.e. 123456 becomes 123

**Bacterius** · January 9th 2010, 11:10 PM

I don't know if this is possible in two operations. I can do it in three, but I can't find a way to do it in two. Anyway, what is the point of this kind of question ? If one can find a way to do it somehow, one will do it this way until further improvement, don't you think ? He won't just be looking for something that takes two operations ...

**Drexel28** · January 10th 2010, 12:19 PM

Originally Posted by Bingk

Um, there's something wrong with your latex, but I'm guessing you're saying that f:R^2 -> R^2 ... and yeah, that is a function, and I did say that functions can be mapped to more than one dimension. It's operations that are limited to R^n -> R.

I was just wondering (since my math is limited), why the distinction between operation and function, and why in particular some functions are distinguished as operations, whilst others are not (even though they are also operations) ... such as the floor function ....

Define a binary operation on $\mathbb{R}^2$ by $f:\mathbb{R}^2\times\mathbb{R}^2\mapsto\mathbb{R}^ 2$ given by $f\left(\left(x,y\right),\left(z,w\right)\right)=\l eft(x+z,z+w\right)$

**Drexel28** · January 10th 2010, 12:19 PM

Originally Posted by Bacterius

An operation is a very basic calculation based on an axiom : addition, multiplication, division, modulus, substraction, etc ... They can be used on their own or as parts of functions to accomplish something. Example : $4 + 4$ , $3 \times 9$ , $7 \equiv 3 \pmod{4}$ ... An operation can be viewed as a simple function.

A function is a combination of operations arranged in order to achieve a particular goal. Take the factorial as an example : $n! = 2 \times 3 \times 4 \times ... \times n$ . This is a combination of $n - 1$ multiplications, arranged in order to obtain the factorial of a number.

A function is not necessarily defined as $f(x)$ or $g(x, y)$ or any other letter. $n!$ is a function. $\sqrt{x}$ is a function. $e^x$ is a function. Etc ...

Do you understand better now ?

I would disagree with this. Simplicity of a function is irrelevant.

**CaptainBlack** · January 10th 2010, 01:52 PM

Originally Posted by hebby

hey thanks, but how would the N divided by 1000 work? as you will end up with a number with decimals, how would you remove those?

Don't make floating posts. A thread is not a two person conversation you need to quote what you are responding to or the thread will make no sense.

The quote button is on the right bottom of every post.

CB

**Bingk** · January 11th 2010, 02:23 PM

Originally Posted by Drexel28

Define a binary operation on $\mathbb{R}^2$ by $f:\mathbb{R}^2\times\mathbb{R}^2\mapsto\mathbb{R}^ 2$ given by $f\left(\left(x,y\right),\left(z,w\right)\right)=\l eft(x+z,z+w\right)$

Hi, just wondering, would this be considered an operation then: $f:\mathbb{R}\times\mathbb{R}\mapsto\mathbb{R}^2$ given by $f\left(x,y\right)=\left(x,y\right)$ ?

Cuz from what you gave, I can still see how it's kind of "one dimensional" (the only dimension is $R^2$ )

Since you disagree with Bacterius, how would you distinguish operations from functions? Just curious as to how different people's views are

**Drexel28** · January 11th 2010, 02:36 PM

Originally Posted by Bingk

Hi, just wondering, would this be considered an operation then: $f:\mathbb{R}\times\mathbb{R}\mapsto\mathbb{R}^2$ given by $f\left(x,y\right)=\left(x,y\right)$ ?

Cuz from what you gave, I can still see how it's kind of "one dimensional" (the only dimension is $R^2$ )

Since you disagree with Bacterius, how would you distinguish operations from functions? Just curious as to how different people's views are

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

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