Proof: The rationality of the y-intercept

Theorem: Given a quadratic function with rational roots, the $y$-intercept is rational if and only if the stretch is rational.

Proof: If $f$ is a quadratic function with rational roots $m$ and $n$ and vertical stretch $a$, then $$f(x) = a(x – m)(x – n) \\ = a(x^2 – (m+n)x + mn) \\ = ax^2 – a(m+n)x + amn.$$ The $y$-intercept is the value of $f(x)$ when $x = 0$, that is, $f(0) = amn$.

A rational number times a rational number is always rational. Therefore, $mn$ is rational, since $m$ and $n$ are rational. If $a$ is rational, then $amn$ is rational, and hence the $y$-intercept is rational.

A rational number times an irrational number is always irrational. Therefore, if $a$ is irrational, $amn$ is irrational, and hence the $y$-intercept is irrational. □

I’m offering this as an example of a simple proof. For students used to proofs from Geometry class, with their columns, I think it’s important to realize that two-column proofs, the sort that high school students so often despise, do not resemble what is generally considered a proof in mathematics beyond high school.

We can also add some interesting corollaries. Here’s one:

Corollary: Given a quadratic function with integer roots, the y-intercept is an integer if (but not only if) the stretch is an integer.

Proof: As above, the $y$-intercept is $amn$. The product of integers is always an integer, so if $a$, $m$, and $n$ are integers, then so is the $y$-intercept.

However, it is possible for a non-integer times an integer to be an integer. For instance, if $a = \frac{1}{3}$ and $mn = 3$, then $amn = 1$. Therefore, an integer $y$-intercept is not enough to prove that $a$ is an integer. □

These are sometimes called “paragraph proofs”, and are the standard way of writing proofs in mathematics (including the square to indicate the end of the proof.