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### Prove that the function f:ZxZ -> ZxZ defined by f(x,y) = (2x+y,x+y) is a bijetion.

Here, we must remember the definition of a bijection. To be bijective, a function must be both injective and surjective. For a function to be injective, we must have that f(a) = f(b) implies that a = b. For a function to be surjective, we must have that for all elements of the range (right hand side), there is an element of the domain (left hand side) that is sent to it by the function.

So, let's start by showing injectivity. We take two pairs (p,q) and (r,s) such that f(p,q) = f(r,s).

We have 2p + q = 2r + s (1) and p + q = r + s (2). I have labelled them (1) and (2) for convenience. Now we look at (1) - (2), and get p = r. Applying this to equation (2), q = s. Therefore, (p,q) = (r,s), our function is injective.

Next, we must consider surjectivity. Let us take a point (a,b) in Z x Z. If 2x + y = a and x + y = b, then as before we can find that x = a - b and y = 2b - a. Therefore, for any a and b in Z x Z, we have found values of x and y that map to it. Since a and b are integers, x and y must also be integers. Therefore our function is surjective as required.

Combining the two results, we have that the function is bijective.

1 year ago

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