Exactly One Way Out

A canteen sells 3 drink flavours and 2 cup sizes. Every possible order is a pairing of one flavour with one size: (mango, small), (mango, large), (guava, small), (guava, large), (lychee, small), (lychee, large). This complete list of pairings is called the Cartesian product of the flavour set and the size set. If A and B are two sets, the Cartesian product A x B is the set of all ordered pairs (a, b) with a from A and b from B: A x B = {(a,b) : a in A, b in B}. Since there are 3 flavours and 2 sizes, there are 3 x 2 = 6 possible orders, matching the general rule: if A has p elements and B has q elements, then A x B has exactly pq elements. Order matters inside each pair: (mango, large) is a completely different order from (large, mango), which does not even make sense, since sizes and flavours come from different sets in a fixed order.

Let A = {1, 2, 3, 4, 5, 6, 7}, and define a relation R from A to A by R = {(x, y) : y = 2x, x, y both in A}. Checking each value of x: x=1 gives y=2 (in A, keep it); x=2 gives y=4 (keep); x=3 gives y=6 (keep); x=4 gives y=8, which is NOT in A, so this pair is dropped; similarly x=5,6,7 all give y-values outside A. So R = {(1,2), (2,4), (3,6)}, a small subset of the full 49-pair Cartesian product A x A. The domain of R is the set of first elements actually used, {1,2,3}. The range is the set of second elements that actually appear, {2,4,6}. The codomain is the entire target set that was declared in advance, all of A = {1,...,7}, whether or not every element of it shows up. Here the range {2,4,6} is a proper subset of the codomain {1,...,7}: plenty of codomain elements, like 1, 3, 5 and 7, are never actually hit.

A relation f from A to B is a function if every single element of A has exactly one image in B: not two, not none. Consider the relation 'y is a square root of x' from A = {4, 9, 16} to B = {2, 3, 4, -2, -3, -4}. Here 4 is linked to BOTH 2 and -2 (since both square to 4): one element of the domain, two images. This fails the function test entirely. Contrast this with the relation 'y is half of x' from A = {2, 4, 6, 8} to B = {1, 2, 3, 4}: R = {(2,1), (4,2), (6,3), (8,4)}. Every element of A has exactly one image, so this IS a function. It does not matter if different elements of the domain happen to share the same image (that is perfectly fine); what breaks the function rule is a single domain element pointing to more than one image, or a domain element with no image at all.

A handful of functions recur so often they get their own names. The identity function f(x) = x sends every input straight back to itself. The modulus function f(x) = |x| strips the sign: |x| equals x when x is at least 0, and equals -x when x is negative, so f(-6) = 6 and f(6) = 6. The signum function reports only the SIGN of x, as 1 if x is positive, 0 if x is exactly 0, and -1 if x is negative, so signum(-8) = -1 regardless of how negative x is. The greatest integer function, written [x], gives the greatest integer that is less than or equal to x: [3.7] = 3, but [-2.3] = -3 (not -2), since -3 is the greatest integer that is still less than or equal to -2.3. Polynomial functions (like f(x) = x^3 - x + 2) and rational functions (one polynomial divided by another) round out the catalog, each with its own typical graph shape.

Real functions can be combined arithmetically. If f(x) = 3x + 1 and g(x) = x^2 - 4, then (f+g)(x) = f(x) + g(x) = x^2 + 3x - 3, and (f-g)(x) = f(x) - g(x) = -x^2 + 3x + 5. The product (fg)(x) is defined pointwise as f(x) times g(x): (fg)(x) = (3x+1)(x^2-4) = 3x^3 + x^2 - 12x - 4. This is worth pausing on, because the SAME-LOOKING notation (fg)(x) can mean something completely different elsewhere: when combining two functions by feeding one's output into the other, fg(x) instead means f(g(x)), the composite. Here, in the algebra-of-functions context, (fg)(x) is plain multiplication of the two output values, nothing more. The quotient (f/g)(x) = f(x)/g(x) is defined wherever g(x) is not zero; here that means x cannot be 2 or -2, since those make x^2-4 equal to zero.

A rational function is undefined wherever its denominator is zero, and finding the domain means finding, and excluding, exactly those values. Consider f(x) = (x+2) / (x^2-9). Since x^2-9 factors as (x-3)(x+3), the denominator is zero exactly when x = 3 or x = -3, so the domain of f is all real numbers except 3 and -3, written R minus {-3, 3}. Everywhere else, the function is perfectly well defined. This kind of check, factor the denominator, find its zeroes, exclude them, applies to any rational function, and is a genuinely different skill from finding a function's range, which asks what values the function actually outputs rather than what inputs it can accept.

How many different relations can exist from a set A with 2 elements to a set B with 3 elements? Every relation is some subset of A x B, and A x B here has 2 x 3 = 6 elements, so the number of possible relations is the number of subsets of a 6-element set, which is 2^6 = 64 (the very same 2^n subset-counting idea seen when counting subsets of a set directly). Only a small fraction of those 64 relations will actually qualify as functions. Cartesian products also show up far beyond finite lists: R x R = {(x, y) : x, y both in R} represents every coordinate point in a flat, two-dimensional plane, and R x R x R represents every point in three-dimensional space, laying the exact groundwork that coordinate geometry builds on.