Domain of multivariable function - definition

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  • #1
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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?
 

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  • #2
Gaussian97
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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

------
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?
 
  • #3
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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?
 
  • #4
Gaussian97
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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?
 
  • #5
Delta2
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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?
 
  • #6
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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
 
  • #7
Gaussian97
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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?
 
  • #8
Delta2
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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 lets 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 $$A=\{(x,y)\in \mathbb{R^2}:x\neq 2 , y\neq 3\}$$
 
  • #9
Delta2
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Sorry i had some typos in my last post, I think now all typos fixed.
 
  • #10
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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 lets 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.
 
  • #11
Delta2
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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 thats 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)"
 
  • #12
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yes thats 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. :)
 
  • #13
Gaussian97
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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.
 
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  • #14
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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.
 

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