Mathematical Rigor in the Context
of Quadratic Polynomials
Branko Ćurgus

Introduction

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Proofs in mathematics are based on logic. Logic is a science that studies forms of rigorous reasoning. To help students with the basics of logic, I wrote the essay "Brief Review of Mathematical Logic." An important topic in this essay, and logic in general, is the universal quantifier and the existential quantifier. As mathematics is a participatory activity, students need a familiar context in which quantifiers appear. There are short examples in my essay. Looking for a significant example that would resonate with students, I stumbled upon a theorem about quadratic functions: a characterization of such a function to be nonnegative. I state and prove this theorem below. A drawback of this proof is that it is long. Since the theorem looks useful, I give all the details of the proof. A reader can enjoy each item of the long proof separately. I hope that each item of the proof provides a good example of how proofs with quantifiers are constructed.

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A Theorem and its proof

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Parts of the theorem below are probably present in high school algebra. I hope that presenting a rigorous version of it will help you understand the nature of proofs in mathematics.

Theorem.  For all real numbers $a, b, c$ the following equivalence holds \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0 \qquad \Leftrightarrow \qquad \bigl(a \geq 0\bigr) \land \bigl(c \geq 0\bigr) \land \bigl(b^2 - 4ac \leq 0\bigr). \]

Having fun in trying to prove small pieces of this theorem is a good practice at proofs.

Proof. The proof of this theorem consists of two parts. In Part 1 we prove the converse implication $\Leftarrow$ by considering two cases. In Part 2 we prove the direct implication $\Rightarrow$.

• Part 1. We first prove the implication: For all real numbers $a, b, c$ we have \[ \bigl(a \geq 0\bigr) \land \bigl(c \geq 0\bigr) \land \bigl(b^2 - 4ac \leq 0\bigr)\qquad \Rightarrow \qquad \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0. \] Let $a, b, c$ be arbitrary real numbers. Assume that the hypothesis of the preceding implication is satisfied. That is we assume \[ \bigl(a \geq 0\bigr) \land \bigl(c \geq 0\bigr) \land \bigl(b^2 - 4ac \leq 0\bigr). \] We will do a proof by cases. Since we assume that $a\geq 0$ there are two exclusive cases for $a:$ Case 1 $a = 0,$ Case 2 $a\gt 0.$
  • Case 1. Assume \[ \bigl(a = 0\bigr) \land \bigl(c \geq 0\bigr) \land \bigl(b^2 - 4ac \leq 0\bigr). \] Since we assume $a = 0$ and $b^2 - 4ac \leq 0$, we deduce that $0 \geq b^2 - 4ac = b^2 - 4\cdot 0 \cdot c = b^2.$ Thus $b^2 \leq 0.$ Since we always have $b^2 \geq 0,$ we deduce that $b^2 = 0.$ Conseqently, $b=0.$ Let $x \in \mathbb{R}$ be arbitrary. We calculate \[ ax^2 + bx+ c = 0\cdot x^2 + 0\cdot x+ c = c. \] Since $c\geq 0,$ this proves that \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0 \] in this case. The implication stated at the beginning of Part 1 is proved in this case.
  • Case 2. Assume \[ \bigl(a \gt 0\bigr) \land \bigl(c \geq 0\bigr) \land \bigl(b^2 - 4ac \leq 0\bigr). \] Let $x \in \mathbb{R}$ be arbitrary. The following calculation is inspired by the completion of a square process that you might have seen in an algebra class. Since we assume $a > 0$ we have \begin{align*} ax^2 + bx + c \ \ & = \ \ a \left( x^2 + \frac{b}{a} x + \frac{c}{a} \right) \\ & = \ \ a \left( \left( x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} \right) \\ & = \ \ a \left( \left( x + \frac{b}{2a}\right)^2 + \frac{-(b^2-4ac)}{4a^2} \right) \\ & = \ \ a \left( x + \frac{b}{2a}\right)^2 + \frac{-(b^2-4ac)}{4a} . \end{align*} Hence, for an arbitrary $x \in \mathbb{R}$ we have \begin{equation}\label{eq:p1b1} \tag{*} ax^2 + bx + c = a \left( x + \frac{b}{2a}\right)^2 + \frac{-(b^2-4ac)}{4a}. \end{equation} Since we assume $a > 0$ and $-(b^2 -4ac) \geq 0$ we have \[ a \left( x + \frac{b}{2a}\right)^2 \geq 0 \quad \text{and} \quad \frac{-(b^2-4ac)}{4a} \geq 0 . \] Therefore, \begin{equation} \label{eq:p1b2} \tag{**} a \left( x + \frac{b}{2a}\right)^2 + \frac{-(b^2-4ac)}{4a} \geq 0. \end{equation} From \eqref{eq:p1b1} and \eqref{eq:p1b2} we deduce \[ ax^2 + bx + c \geq 0 . \] Since $x \in \mathbb{R}$ was arbitrary we have proved that \[ \forall x \in \mathbb{R} \quad ax^2 + bx + c \geq 0. \] This completes the proof of the implication stated at the beginning of Part 1.
• Part 2. Here we prove the implication: For all real numbers $a, b, c$ we have \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0\qquad \Rightarrow \qquad \bigl(a \geq 0\bigr) \land \bigl(c \geq 0\bigr) \land \bigl(b^2 - 4ac \leq 0\bigr). \] This implication is proved in three parts
  • Part 2a. In this part we prove the implication: For all real numbers $a, b, c$ we have \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0\qquad \Rightarrow \qquad a \geq 0 . \] Proof. We prove the contrapositive of the preceding implication. That is we prove: For all real numbers $a, b, c$ we have \[ a\lt 0 \qquad \Rightarrow \qquad \exists\mkern+1.5mu x \in \mathbb{R} \quad ax^2 + b x+ c \lt 0. \] Let $a, b, c \in \mathbb{R}.$ Assume $a\lt 0.$ We do a proof by cases for $b$ and $c.$ We distinguish three cases: Case 1 $b\neq 0$ and $c$ arbitrary, Case 2 $b=0$ and $c\leq 0,$ Case 3 $b=0$ and $c \gt 0.$ It is clear that these three cases include all possibilities for $b$ and $c.$
    • Case 1. $b\neq 0$ and $c$ arbitrary. Set \[ x_0 = - \frac{1}{b} - \frac{c}{b} \] and calculate, taking into account that $a(x_0)^2 \leq 0,$ \[ a(x_0)^2 + b x_0 + c = a (x_0)^2 - 1 - c + c = a (x_0)^2 - 1 \leq -1 \lt 0. \]
    • Case 2. $b = 0$ and $c \leq 0.$ Set \[ x_0 = 1 \] and calculate, taking into account that $c \leq 0,$ \[ a(x_0)^2 + b x_0 + c = a + c \leq a \lt 0. \]
    • Case 3. $b = 0$ and $c \gt 0.$ Set \[ x_0 = 1 - \frac{c}{a} \] and calculate, taking into account that $-c \lt 0$ and $(c^2)/a \lt 0$ \[ a(x_0)^2 + b x_0 + c = a \left(1 - 2 \frac{c}{a} + \frac{c^2}{a^2} \right) + c = a - c + \frac{c^2}{a} \lt a \lt 0. \]
    In each case we proved the contrapositive. This completes the proof of Part 2a.
  • Part 2b. In this part we prove the implication: For all real numbers $a, b, c$ we have \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0\qquad \Rightarrow \qquad c \geq 0. \] Assume \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0. \] Since $0\in\mathbb{R},$ this assumption implies that \[ a\cdot 0^2 + b\cdot 0+ c = c \geq 0. \] Thus, $c \geq 0$ and the stated implication is proved.
  • Part 2c. In this part we prove the implication \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0\qquad \Rightarrow \qquad b^2 - 4ac \leq 0. \] Assume \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + b x+ c \geq 0. \] In Part 2a we proved that the preceding assumption implies $a \geq 0.$ Therefor, we will consider two cases: Case 1 $a = 0,$ and Case 2 $a > 0.$
    • Case 1. Assume $a = 0$. In this case the assumption reads \begin{equation*} \forall x \in \mathbb{R} \quad bx + c \geq 0. \end{equation*} Next we can prove that the following implication holds \begin{equation*} \forall x \in \mathbb{R} \quad bx + c \geq 0 \qquad \Rightarrow \qquad b = 0. \end{equation*} As in Part 2a, the contrapositive is easier to prove: \begin{equation*} b \neq 0 \qquad \Rightarrow \qquad \exists\mkern+1mu x \in \mathbb{R} \quad bx + c \lt 0 . \end{equation*} To prove the preceding implication we set \[ x_0 = -\frac{1+c}{b} \] and calculate \[ b\left(-\frac{1+c}{b} \right) + c = -1-c+c = -1 \lt 0. \] Thus, we proved that $b=0.$ Therefore \[ b^2 - 4 a c = 0^2-4\cdot 0 \cdot c = 0 \leq 0. \] This proves the implication stated at the beginning of Part 2c in this case.
    • Case 2. Assume $a \gt 0$. Since we assume \begin{equation*} \forall x \in \mathbb{R} \quad ax^2 + bx + c \geq 0 , \end{equation*} substituting \[ x = - \frac{b}{2a} \in \mathbb{R} \] in the preceding quadratic expression we get \[ a \left(- \frac{b}{2a}\right)^2 + b \left(- \frac{b}{2a}\right) + c \geq 0. \] Simplifying the left-hand side in the preceding inequality leads \[ -\frac{b^2 - 4ac}{4a} \geq 0. \] Since $4a > 0$, the preceding inequality implies \[ b^2 - 4 a c \leq 0. \] This proves the implication stated at the beginning of Part 2c in this case.

This completes the proof of Part 2 and the proof of the theorem. QED

This is a long proof consisting of several parts. If you read it carefuly, I hope, you will agree that it provides a rigorous proof of the theorem. I finished the proof with QED, which is one common way to finish a mathematical proof.

Comment. One could reasonably complain that the proof I presented above is too complicated. I could not think of a more straightforward proof. I always encourage students to try to improve existing proofs. So, the above proof is a good candidate for improvement. There might exist proofs on the internet that are simpler than the one I presented above. I did not look for them. Below I will give a different proof for Part 2a. Recall that the above proof for Part 2a considers three cases. I always put some effort into finding proof that does not involve cases. The proof below is the result of my efforts for Part 2a. Proof for Part 2a. Recall that in Part 2a we considered the following implication: For all real numbers $a, b, c$ we have \[ a\lt 0 \qquad \Rightarrow \qquad \exists\mkern+1.5mu x \in \mathbb{R} \quad ax^2 + b x+ c \lt 0. \] Below is a proof that avoids using the cases explicitly. Let $a, b, c \in \mathbb{R}.$ Assume $a\lt 0.$ Set \[ d = \max\{b,c,0\}. \] In words $d$ is the largest of the three numbers $b,$ $c,$ and $0.$ Since $d$ is the largest of the three numbers we have \[ b-d \leq 0, \qquad c-d \leq 0, \qquad 0 \leq d. \] Now we define $x_0$ as \[ x_0 = 1 - \frac{d}{a}. \] and calculate (please recall that $b-d \leq 0,$ $c-d \leq 0,$ and $d(d-b)/a \leq 0$) \begin{align*} a (x_0)^2 + b x_0 + c & = a \left( 1 - 2 \frac{d}{a} + \frac{d^2}{a^2} \right) + b\left( 1 - \frac{d}{a} \right) + c \\ & = a - 2 d + \frac{d^2}{a} + b - \frac{bd}{a} + c \\ & = a + (b-d) + (c-d) + \frac{d (d-b)}{a} \\ & \leq a. \end{align*} Since $a \lt 0,$ we have proved that for this specific number $x_0$ given in terms of $a, b, c$ we have \[ a (x_0)^2 + b x_0 + c \lt 0. \] This proves the implication.

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Five Related Theorems

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Together with the theorem that we proved in the preceding item one should state several more related theorems. I list all that I could think of here.

Theorem.  For all real numbers $a, b, c$ we have \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + bx + c \leq 0 \qquad \Leftrightarrow \qquad (a \leq 0) \ \land \ (c \leq 0) \ \land \ (b^2 - 4 a c \leq 0). \]

Theorem.  For all real numbers $a, b, c$ we have \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + bx + c = 0 \qquad \Leftrightarrow \qquad (a = 0) \ \land \ (c = 0) \ \land \ (b= 0). \]

Theorem.  For all real numbers $a, b, c$ we have \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + bx + c \gt 0 \qquad \Leftrightarrow \qquad \bigl(a\gt 0 \land b^2 - 4 a c \lt 0\bigr)\lor \bigl(a=0\land b=0 \land c \gt 0 \bigr). \]

Theorem. For all real numbers $a, b, c$ we have \[ \forall\mkern+1mu x \in \mathbb{R} \quad ax^2 + bx + c \lt 0 \qquad \Leftrightarrow \qquad \bigl(a\lt 0 \land b^2 - 4 a c \lt 0\bigr)\lor \bigl(a=0\land b=0 \land c \lt 0 \bigr). \]

Theorem.  For all real numbers $a, b, c$ we have \[ \Bigl( \exists\mkern+1mu s \in \mathbb{R} \quad as^2 + bs + c \lt 0 \Bigr) \bigwedge \Bigl( \exists\mkern+1mu t \in \mathbb{R} \quad at^2 + bt + c \gt 0 \Bigr) \qquad \Leftrightarrow \qquad b^2 - 4 a c \gt 0. \]

Providing proofs for these theorems is more entertainment for math enthusiasts.

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