Domain of multivariable function - definition

[Poetria]

Homework Statement

(Choose all options below that correctly complete the following sentence.)
The domain of a multivariable function z=f(x,y) is:
-a subset of the real numbers
-a union of two subsets of the real numbers
-a subset of the xy-plane
-the union of the set of all values x where f(x,y) exists with the set of all values where f(x,y) exists
-the set of all ordered pairs of points (x,y) so that f(x,y) exists

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Relevant Equations

I think that only the last definition is complete: "the set of all ordered pairs of points (x,y) so that f(x,y) exists"

E.g. all real numbers could be a domain but not necessarily, etc. Am I right?

 

[Gaussian97]

Homework Statement:: (Choose all options below that correctly complete the following sentence.)
The domain of a multivariable function z=f(x,y) is:
-a subset of the real numbers
-a union of two subsets of the real numbers
-a subset of the xy-plane
-the union of the set of all values x where f(x,y) exists with the set of all values where f(x,y) exists
-the set of all ordered pairs of points (x,y) so that f(x,y) exists

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Relevant Equations:: I think that only the last definition is complete: "the set of all ordered pairs of points (x,y) so that f(x,y) exists"

E.g. all real numbers could be a domain but not necessarily, etc. Am I right?

If the real numbers were the domain, given any real number (e.g 0), you should be able to compute its image. If you have a function $f(x,y)$ how do you compute the image of 0? Or the image of any other real number?

 

[Poetria]

Let the function (x,y) be z=(x+y)/x. Well, I can try to compute it: z=(x+0)/0, but it will be undefined.
I guess the image of function is its range?

 

[Gaussian97]

Let the function (x,y) be z=(x+y)/x. Well, I can try to compute it: z=(x+0)/0, but it will be undefined.
I guess the image of function is its range?

The image of a function for an element $a$ is the value of $f(a)$. You are concentrating on the fact that 0 is undefined because is in the denominator, which makes you ignore the real problem. What is the image of your function for the real number 1 which has no problem with division?
Or if you define another function $z=f(x,y)=x+y$, take any real number, what is its image?

 

[Delta2]

I agree with your choice of last option but I think there is one more option that is also correct. Which do you think it is?

 

[Poetria]

I agree with your choice of last option but I think there is one more option that is also correct. Which do you think it is?

I thought that this one "the union of the set of all values x where f(x,y) exists with the set of all values where f(x,y) exists" may be correct, but I wasn't sure.
Is the union of two such sets an equivalent of ordered pairs?

The image of a function for an element $a$ is the value of $f(a)$. You are concentrating on the fact that 0 is undefined because is in the denominator, which makes you ignore the real problem. What is the image of your function for the real number 1 which has no problem with division?
Or if you define another function $z=f(x,y)=x+y$, take any real number, what is its image?

Well, if z=f(x,y)=x+y, I can take x=1, but I would also need the value of y.
z=1+y

 

[Gaussian97]

I thought that this one "the union of the set of all values x where f(x,y) exists with the set of all values where f(x,y) exists" may be correct, but I wasn't sure.
Is the union of two such sets an equivalent of ordered pairs?Well, if z=f(x,y)=x+y, I can take x=1, but I would also need the value of y.
z=1+y

Therefore the image of a real number makes no sense. So, can the real numbers (or a subset of real numbers like "the union of the set of all values x where f(x,y) exists with the set of all values where f(x,y) exists") be the domain of the function?

 

[Delta2]

I thought that this one "the union of the set of all values x where f(x,y) exists with the set of all values where f(x,y) exists" may be correct, but I wasn't sure.
Is the union of two such sets an equivalent of ordered pairs?

No this is not the option I had in mind and also to answer your very last question, the union of two such sets is not an equivalent of a set of ordered pairs.

For example take the function $$f(x,y)=\frac{1}{x-2}+\frac{1}{y-3}$$. The set of where is defined with respect to x let's call it $A_x=\mathbb{R}-\{2\}$, while the set where is defined with respect to y is $A_y=\mathbb{R}-\{3\}$. The union of $A_x$ and $A_y$ is the whole $\mathbb{R}$, but the function is not defined in the whole $\mathbb{R}$ it is only defined for $$A=\{(x,y)\in \mathbb{R^2}:x\neq 2 , y\neq 3\}$$

 

[Delta2]

Sorry i had some typos in my last post, I think now all typos fixed.

 

[Poetria]

No this is not the option I had in mind and also to answer your very last question, the union of two such sets is not an equivalent of a set of ordered pairs.

For example take the function $$f(x,y)=\frac{1}{x-2}\frac{1}{y-4}$$. The set of where is defined with respect to x let's call it $A_x=\mathbb{R}-\{2\}$, while the set where is defined with respect to y is $A_y=\mathbb{R}-\{3\}$. The union of $A_x$ and $A_y$ is the whole $\mathbb{R}$, but the function is not defined in the whole $\mathbb{R}$ it is only defined for $$A=\{(x,y)\in \mathbb{R^2}:x\neq 2 , y\neq 3\}$$

Great explanation. :) Many thanks.
In that case I think the only option is this one "a subset of the xy-plane". My reservation was that there is no mention about if the function exists.

 

[Delta2]

In that case I think the only option is this one "a subset of the xy-plane". My reservation was that there is no mention about if the function exists.

yes that's the option i had in mind. The way I read this option is "A subset of the xy plane where the function exists (or is well defined)"

 

[Poetria]

yes that's the option i had in mind. The way I read this option is "A subset of the xy plane where the function exists (or is well defined)"

Thank you so much. :) I got it. :)

 

[Gaussian97]

Actually, there's no need to add any extra information, the phrase
"The domain of a multivariable function z=f(x,y) is a subset of the xy-plane"
is perfectly fine as it is.

 

[Poetria]

Actually, there's no need to add any extra information, the phrase
"The domain of a multivariable function z=f(x,y) is a subset of the xy-plane"
is perfectly fine as it is.

As Keith Devlin said, it's about mathematical thinking. It's clear to you but it wasn't clear to me. I suppose the term 'subset' is the key. It implicates a subset where the function is well-defined.