Question

Is the set \(\theta=\\{((x, y),(3 y, 2 x, x+y)): x, y \in \mathbb{R}\\}\) a function? If so, what is its domain and range? What can be said about the codomain?

Short answer

Yes, the set forms a function. Its domain is the entire plane of real pairs (\(\mathbb{R}^2\)) and its range is the set of all triplets of real numbers (\(\mathbb{R}^3\)). The codomain is assumed to be all triplets of real numbers (\(\mathbb{R}^3\)).

Step-by-step solution

Verify if it's a function

To verify if the set is a function, we need to ensure that for every (x, y) in the domain, there's only one output in the co-domain. The mapping given is ((x, y),(3 y, 2 x, x+y)). It's clear that for each (x, y), there's only one output (3y, 2x, x+y). So, yes, this set is a function.

Identify the domain

The domain of a function is the set of all inputs for the function. In this case, our function's inputs are all (x, y) pairs where x and y are real numbers. Hence, the domain for our function is the entire real plane, denoted by \(\mathbb{R}^2\).

Identify the range

The range of a function consists of all output values. In this case, for each (x, y) in the domain (which is the set of all real pairs), there's an output (3y, 2x, x+y). Because y, x, and x+y can have any real value, the range is all triplets of real numbers, which we can denote as \(\mathbb{R}^3\).

Identify the codomain

In this case, the codomain was not defined in our original function. By default, the codomain is assumed to be the same set that the range is taken from. As identified in step 3, our range is \(\mathbb{R}^3\), so we'll assume our codomain is all triplets of real numbers (\(\mathbb{R}^3\)). However, it should be noted that if there were restrictions on the output, the codomain might be different from the range.