Is the set a function?

ryanlong304

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Is the set A={((x,y),(3y,2x,x+y)): x and y are real numbers} a function? If so, what is its domain, codomain, and range?

This seems like it would be a function to me. It's mapping R2 onto R3, and it does seem to satisfy the conditions of being a function.
So, I would expect for the domain to be R2 and the codomain to be R3. My main problem is trying to identify what the range would be. From what I understand, the first two coordinates (3y and 2x) can be any real numbers, but the third coordinate (x+y) is more restricted, right? The reason that I say this is because it would be impossible to get (0,0,1) as an output of the function. The only way of writing the output that then makes sense to me is something like: {(3k, 2c, c+k): c and k are real numbers}. This just seems like improper notation to me.

I would greatly appreciate any help that anyone can offer.
 
Is the set A={((x,y),(3y,2x,x+y)): x and y are real numbers} a function? If so, what is its domain, codomain, and range?

This seems like it would be a function to me. It's mapping R2 onto R3, and it does seem to satisfy the conditions of being a function.
It clearly is a function. It is a set of ordered pairs, every pair in \(\displaystyle R^2\) is a first term, and no two pairs have the same second term. \(\displaystyle R^2\) is the domain. \(\displaystyle R^3\) is the codomain.

The range is more tricky, clearly \(\displaystyle (1,1,0)\) is not in the range.
 
The definition of "function" is that if (x1, y) and (x2, y) are in the same function, then x1= x2. Here, we note that if 3y1= 3y2, 2x1= 2x2, and x1+ y1= x2+ y2, then from the first two equations x1= x2 and y1= y2.

To determine the range, suppose 3y= a, 2x= b, and x+ y= c. Then x+y= b/2+ a/3= c. That is the same as 2a+ 3b- 6c= 0. In an x,y, z coordinate system, 2x+ 3y- 6z= 0 is a plane containing the origin. That plane is the range.
 
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