# 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. Clio Corvid