Find the image of a two variable function

[Synapse0x]

So on my first course of Algebra in university I have been asked to find the image of the following function:
f : Z x Z -> Z, f(x, y) = x² - y²

In high school I worked only with single variable functions and I could have found the image by drawing its graph, using limits or using some specific shortcuts on well known functions (like the quadratic one).
I really have no idea how should I find the image of this one. Actually I don't even know how to plot it, I think its graph is 3D. Can somebody point me in the right direction?

 

[caffeinemachine]

So on my first course of Algebra in university I have been asked to find the image of the following function:
f : Z x Z -> Z, f(x, y) = x² - y²

In high school I worked only with single variable functions and I could have found the image by drawing its graph, using limits or using some specific shortcuts on well known functions (like the quadratic one).
I really have no idea how should I find the image of this one. Actually I don't even know how to plot it, I think its graph is 3D. Can somebody point me in the right direction?

You are not required to sketch the graph of the function.

The image of a function is a subset of the target space of the function.

You need to say what is $Z$ exactly?

As of now, my answer would be $\text{Im}(f)=\{x^2-y^2: (x,y)\in Z\times Z\}$.

 

[Synapse0x]

Yes, in other words what is Z exactly.

Your current answer makes sense but how should I state it more accurately. I mean what your wrote is not wrong but it would be the same as saying that the image of g, where g : R -> R, g(x) = x² - 4x + 1, is {x² - 4x + 1 : x in R} instead of [-3, +infinity)

 

[caffeinemachine]

Yes, in other words what is Z exactly.

Your current answer makes sense but how should I state it more accurately. I mean what your wrote is not wrong but it would be the same as saying that the image of g, where g : R -> R, g(x) = x² - 4x + 1, is {x² - 4x + 1 : x in R} instead of [-3, +infinity)

I think by $Z$ you mean the set of all the integers.

In that case, the image of $f$ is $\{n\in\mathbb Z: n \text{ is odd }\}\cup \{n\in \mathbb Z: 4 \text{ divides } n\}$.

To see this, first note that $f(y+1,y)=2y+1$.

Thus all the odd integers are in the image.

Now suppose $f(x,y)$ is even.

Then $2$ divides $(x-y)(x+y)$.

But then $4$ divides $(x-y)(x+y)$ (why?)

So if an even integer is in the image, then it have to be divisible by $4$.

Finally note that $f(y+2,y)=4(y+1)$.

Thus all the integers which are divisible by $4$ are in the image.

 

[blue_raver22]

I would suggest approaching this problem by first understanding what the term "image" means in the context of a two variable function. The image of a function is the set of all possible output values, or the range of the function. In other words, it is the set of all values that the function can produce for its given inputs.

To find the image of this particular function, we can start by considering the domain of the function, which is the set of all possible input values. In this case, the domain is the set of all integer pairs (x, y).

Next, we can analyze the function itself. In this case, f(x, y) = x2 - y2. This function takes in two integer inputs and produces a single integer output. We can start by plugging in some simple inputs to see what outputs we get. For example, if we input (0, 0), we get an output of 0. If we input (1, 0), we get an output of 1. If we continue to plug in different inputs, we can start to see a pattern emerge.

We can also use algebraic techniques to further understand the function. For example, we can rewrite the function as f(x, y) = (x+y)(x-y). This shows us that the output is dependent on both the sum and difference of the inputs. We can also see that the output will always be an integer, since we are dealing with integers in the domain.

To plot the function, we can use a 3D graph, where the x and y axes represent the input values and the z axis represents the output values. However, since the function is defined for all integer pairs, it would be impossible to plot every single point on the graph. Instead, we can plot a few key points and connect them to get an idea of the overall shape of the function.

In summary, to find the image of this two variable function, we can analyze the domain and function itself to understand the range or set of all possible output values. We can also use algebraic techniques and visualization tools to gain a deeper understanding of the function and its behavior.