F(x-y) = f(x,y)

Is there a way to prove that in general a function of 2 variables, f(x,y), can be written as a function of one variable, f(x-y)?

I made a similar claim in an assignment and now I need to prove it.

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No, because that is not true at all.

>in general

No there's no way to prove something that's untrue is true. Do you think people use multivariable math for the fun of it?

Not at all. I believe people use multi-variable math for the sport of it.

There are reduction in variable techniques that people use though, so I just figured there might be a method in there that I'm forgetting

You're stuck.
Compute the area of a rectangular garden X meters by Y meters.
Now try to compute the area of the garden when you're given only one piece of information, Z, which is the result of some operation (addition, multiplication, exponentiation, etc.) performed upon X and Y.
In going from knowing 2 things to knowing 1 thing, information has been irrevocably lost.

Whatever prompted to you make the claim?
If your technique worked, 10 variables could be reduced to 9. Applied again, 9 could be reduced to 8. Ultimately, a description of the entire universe comes down to a single number -- which we already know is "42".

>There are reduction in variable techniques that people use though, so I just figured there might be a method in there that I'm forgetting
Sure. You can replace x and y by the tuple (x, y), and then unpack that again inside your function. But you can't recover x and y from x-y, as in OP.

what drugs were you on when you wrote that

Do:
f(x,y) = f'(xyn)
Don't do them times eachother but put them next to eachother and 'n' is for how much digits long the x and y are. So f(3, 70) = f'(03702).
Trust me, I study compsci.

well it's obviously true. Just find a way to encode the two variables as a string or even better a number and work with that

You could also do just f(x,y) =f(x-y) and redifine '-' not as an operator but as a separator just as the ','

Assuming subtraction was just a random example and not the only method you're allowed, you can do it using this:
plato.stanford.edu/entries/goedel-incompleteness/sup1.html
Integers can be represented unqiuely as the products of prime numbers so you can make a new function that that takes one number which is made up of the products of primes raised to the power of your arguments from your original function
e.g. If your original function is f(x,y)=x/y and your inputs are x=4 and y=2 then your input z for g(z)=f(x,y) would be 144 since 144 = 2*2*2*2*3*3 = 2^4 * 3^2 and because 144 is in fact the only number equal to the product of four twos and two threes that means g(144) will unambiguously equate to f(4,2) and get you the same answer of 2. You can even reverse the order and do f(2,4)=g(324)=.5 because only 324 is equal to the product of two twos and four threes. You could also extend this to a starting function of more than two arguments by going out to the next n primes and doing the same thing with exponent encoding.

for a bounded domain it is possible, because you can map every point of your 2D bounded domain to a line of length corresponding to the product of the length of each dimension, but I don't see any use for this kind of mapping since the 1D function will most likely not be analytic or even continuous

why would you need such a proof?

you can do it, but you loose all the nice properties like continuity.

functions are linear where f(x) + f(y) = f(x+y)
and Kf(x) = f(Kx)

What if it's a function on real numbers?

Prove real numbers exist.

i just checked them

alternate digits, so abc.efg... and uvw.xyz... becomes aubvcw.exfygz...

Real answer here: you obviously mean to ask what technique could there be to show that for some function f(x,y), if there exists some g(z) so that f(x,y)=g(x-y). Well since you say that you assume this and you want to prove it to be true, make some preliminary checks. Is f(x,x) a constant function? You can check by differentiating (if you can) and see if the result is 0. Do other kinds of substitutions that use the properties you know of f. For instance if f is symmetric in x,y then g is an even function and so on.

Once you are convinced that there is such a g. One way to prove the result is by showing that f(x,z-x) is constant in x. Which you can show by differentiating partially with respect to x. Can make reference to the chain rule and so on.

You can *sometimes* split f(x,y) into g(x-y)+h(x+y), essentially separating f into left and right movers.

>I made a similar claim in an assignment and now I need to prove it.
lol faggot

> x-y

Nigga you trippin