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After noticing some patterns in delta epsilon proofs, I was able to prove a theorem regarding the factorability of quadratics. My question is, if this proof is correct, has it been documented before? And if it has been documented before are there any extensions of the theorem to other contexts?

Theorem: If $f$ is a quadratic with rational coefficients then $f(x)-f(q)$ is factorable $\forall q \in \mathbb{Q}$. (i.e. $\exists \lambda , r_1 , r_2, \in \mathbb{Q} $ $ st. f(x)-f(q)=\lambda (x-r_1)(x-r_2)$)

Proof: Fix $a,b,c \in \mathbb{Q}$ and consider $f(x):= ax^2+bx+c$. Then $\forall q \in \mathbb{Q} $ $f(x)-f(q)= ax^2+bx+c-(aq^2+bq+c)= ax^2+bx-aq^2-bq$. We will continue by reducing this quadratic into factored form $\lambda (x-r_1)(x-r_2)$ and demonstrating that $\lambda,r_1,r_2 \in \mathbb{Q}$. Here $r_1,r_2$ are given by the quadratic formula, and $a= \lambda$ which we already know to be rational. All that is left to show is that $$\frac{-b \pm \sqrt{b^2-4a(aq^2-bq)}}{2a} \in \mathbb{Q}$$ Which can be done by noticing that $\sqrt{b^2-4a(aq^2-bq)}=\sqrt{b^2+4abq+4a^2q^2}= \sqrt{(b+2aq)^2}=|b-2aq|$ $ \forall b,a,q \in \mathbb{R}$. Therefore the roots are rational $ \forall b,a,q \in \mathbb{Q}$ which was to be shown. $\square$

Corollary: $\forall a,b \in \mathbb{Q}$ $ \exists c \ne 0 \in \mathbb{Q}$ st. $ax^2+bx+c$ is factorable.

Proof: Apply the first result with any $q \in \mathbb{Q}$

I have also managed to generalize a form of this result to all functions, but it seems very useless. It should also be noted that it is the strongest version of the theorem that applies to higher degree polynomials that I have been able to prove.

Theorem B: For all continuous functions $f$ defined on some subset of the real line, and for all $r$ on that subset $f(x)-f(r)$ has a root of $r$.

Proof: Fix $f:D \to R$ and $r \in R$ st. $D,R \subset \mathbb{R}$. Then $f(x)-f(r)=0 \implies f(x)=f(r)$ which shows that r is a root of the function $f(x)-f(r)$ $ \square$

Remark: Theorem A I have been able to use widely in my tutoring because it allows for the creation of a factorable quadratic with extreme ease and gives a lot of control on what type of quadratic you obtain. The strongest version of A that applies to higher degrees seems to be that for even degrees $f(x)-f(r)$ has at least two real (or rational depending on your coefficients and choice of r) roots with at least one being r itself. I have not yet felt the need to write this into a formal proof because it seems not very useful, especially when compared with the quadratic case, but I suppose it is true. Theorem B seems both incredibly obvious and slightly non obvious, but I don’t think it’s that useful.

If anyone could offer more insight than what I have here that would be much appreciated.

Bill Dubuque
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EzTheBoss 2
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  • Theorem A: This is true and your proof looks correct, but it's not the "right" or the best proof. You should prove it like you proved Theorem B. Namely $f(x) - f(q)$ clearly has one rational root $x = q$. That means you can factor $f$ as $f = (x - q)g$ with some polynomial $g$. $g$ necessarily has degree 1. Hence $g = c (x-r)$. Analyzing the coefficients of $f = (x-q)c (x-r)$ yields $\lambda,r \in \mathbb{Q}$. Such resuls are not "documented", because for an experienced mathematician they are very easy to show; in a text you would easily claim them (and generalizations) without proof. –  May 19 '23 at 00:41
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    Corollary: It's not good style to write "trivial". Rather say something like "apply Theorem A with any $q \in \mathbb{Q}$". That being said, the result follows even easier. Take $c=0$; then $ax^2 + bx + c = ax^2 + bx = x(ax + b) = a(x-0)(x-b/a)$. –  May 19 '23 at 00:45
  • Theorem B: As noted before, this is indeed the better argument for why such functions $f(x) - f(r)$ have a root. It works in much more generality. And if $f$ is a polynomial, you can factor out the root. –  May 19 '23 at 00:47
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    Theorem A is proved by verifying that $f(x)-f(q)=(x-q)(ax+aq+b)$. – Karl May 19 '23 at 00:50
  • @JohannBirnick I have fixed the Corollary to read a lot better. And I agree on the notion that my proof is a bit complicated compared with the simpler one. – EzTheBoss 2 May 19 '23 at 01:08
  • Could you please post an example where this good work is useful? Thanks. – NoChance May 19 '23 at 01:56
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    Theorem A is very useful when tutoring high school students as it is often the case that more examples of factorable quadratics are needed. To find one using this method one can simply choose an arbitrary a,b, and q value and generate $f(x) = ax^2+bx-aq^2-bq$. This is where it because usable in my work, although I found this fact when playing around with epsilon delta proofs of quadratics. Outside of these two applications I don't know of any uses, though I wouldn't be surprised if there are some. – EzTheBoss 2 May 19 '23 at 02:30
  • By the Factor Theorem $, x-q\mid f(x)-f(q),$ which immediatey yields the sought result, see the linked dupes. – Bill Dubuque May 19 '23 at 06:26
  • @EzTheBoss2 There is a much easier way of finding factorizable quadratic polynomials. Just start with the factorization $\lambda (x-r_1)(x-r_2)$ and multiply our to find $\lambda x^2 - \lambda (r_1 + r_2) x + r_1 r_2$. –  May 19 '23 at 07:57
  • I feel like that method is harder to compute quickly, but it does give you control over what both roots are rather than just one. I have gotten so quick with my method that I can essentially just write one down because it involves remembering far fewer numbers at a time. $ax^2+bx$ is essentially just written out and then you plug a $q$ into is and subtract off the result. Whereas when starting with factored form you have to keep track of two numbers (both roots), and use a formula that is harder to remember. But both are valid. – EzTheBoss 2 May 19 '23 at 12:43
  • Thanks to everyone for their comments! I will check out the linked posts to learn more. – EzTheBoss 2 May 19 '23 at 12:49

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