One-One, Onto and Bijective Functions

1. Write domain and codomain first. 2. To prove one-one, put f(a) = f(b) and simplify until a = b. To disprove, give two different inputs with the same image. 3. To prove onto, put y = f(x), solve for x, and verify x belongs to the domain. 4. Combine the two results to decide bijective.

Use a mapping diagram or graph with axes. For graph-based checking, label x-axis, y-axis, graph y = f(x), and a horizontal line y = c. The horizontal-line test supports one-one decisions.

Claiming f(x) = x^2 is onto from R to R because x = square root of y, while ignoring that negative y values in the codomain have no real preimage.

A function f: A to B is one-one if different elements of A have different images in B. It is onto if every element of B has a preimage in A. It is bijective if it is both one-one and onto.

A function f: A to B is one-one if different elements of A have different images in B. It is onto if every element of B has a preimage in A. It is bijective if it is both one-one and onto. One-one test: f(a) = f(b) implies a = b. Onto test: for every y in codomain B, there exists x in domain A such that f(x) = y. Bijective: one-one and onto. Horizontal-line test: any horizontal line meeting the graph more than once means not one-one. The function f: R to R defined by f(x) = 2x + 3 is one-one because f(a) = f(b) gives 2a + 3 = 2b + 3, so a = b. It is onto because for any y in R, y = 2x + 3 gives x = (y - 3)/2, which is real.

One-one depends on uniqueness of input for each output, while onto depends on whether the whole codomain is covered. The codomain must be stated before testing onto. A graph may help for real-valued functions: a horizontal line cutting the graph more than once indicates the function is not one-one. One-one test: f(a) = f(b) implies a = b. Onto test: for every y in codomain B, there exists x in domain A such that f(x) = y. Bijective: one-one and onto. Horizontal-line test: any horizontal line meeting the graph more than once means not one-one. Check f: R to R, f(x) = x^3 + 1. One-one: assume f(a) = f(b). Then a^3 + 1 = b^3 + 1, so a^3 = b^3, hence a = b. Therefore f is one-one. Onto: take any y in R. Put y = x^3 + 1. Then x^3 = y - 1, so x = cube root of (y - 1), which is real. Hence every real y has a real preimage, so f is onto. Therefore f is bijective. Often asked as a proof for a defined function, or as a graph/mapping-based classification. Long-answer questions may ask students to prove bijectivity before finding an inverse. Claiming f(x) = x^2 is onto from R to R because x = square root of y, while ignoring that negative y values in the codomain have no real preimage.