Quadratic Method: Existence Logic

Multiple prior references with similar work are structured by requiring the initial non-elementary fact that every quadratic has two roots, or that for any two numbers$b$and$c$, there definitely exist a pair$(u,v)$of numbers such that$u+v=b$and$uv=c$. For example, this requirement is made by any derivation that cites Viète's Relations (which state that two roots definitely exist and then gives equations for their sum and product), and also made by any derivation that starts by asserting that every standard-form quadratic equation is equivalent to the system of equations$u+v=b$and$uv=c$and then makes substitutions and deductions using those equations. To understand why the direction of logical deduction is important, and how it interacts with the existence requirement, it is helpful to consider the following pair of examples.

Example Problem 1

Find all$(u,v)$that satisfy the system of equations:$$\begin{array}{rl}u+v& =6\\ u-v& =2\\ 3u+v& =14\end{array}$$

"Solution" to Problem 1

Add the first and second equations to get$$\begin{array}{rl}2u& =8\\ u& =4\end{array}$$Substitute this into the third equation to get$$\begin{array}{rl}12+v& =14\\ v& =2\end{array}$$Therefore,$(u,v)=(4,2)$is the solution.

Example Problem 2

Find all$(u,v)$that satisfy the system of equations:$$\begin{array}{rl}u+v& =6\\ u-v& =2\\ u+3v& =7\end{array}$$

"Solution" to Problem 2

Add the first and second equations to get$$\begin{array}{rl}2u& =8\\ u& =4\end{array}$$Substitute this into the third equation to get$$\begin{array}{rl}4+3v& =7\\ v& =1\end{array}$$Therefore,$(u,v)=(4,1)$is the solution.

Contradiction

Unfortunately, only one of these "Solutions" produces an answer that works. In Problem 1, if we substitute$(u,v)=(4,2)$into all 3 equations, then all equations hold. However, in Problem 2, if we substitute$(u,v)=(4,1)$into all 3 equations, then the first two equations fail ($4+1\ne 6$and also$4-1\ne 2$). This is one reason why when we solve systems of equations in this way, it is important to have a final step of substituting the solutions back into all of the equations to check that they actually work.

Why do these methods of reasoning sometimes work and sometimes fail? The reason is because they all proceed by "forward" deduction, in which we collect true statements, and one step after another (whether the next sentence or the next equation in a sequential stack of equations) uses the entirety of the current collection of true statements to make one new deduction that must also be true. We then run the logical procedure until we narrow down the possibilities for$u$and$v$down to at most one option for each. This logical flow amounts to saying that at the start, we don't have any understanding of what$u$or$v$could be (they could be anything), and we take step after step narrowing down the possibilities for$u$and$v$until we are down to at most one option, after which we conclude that we're done.

The logical error is in concluding that if there is at most one option, then there is always exactly one option. There could be zero options that work. Why does Solution 1 work while Solution 2 doesn't? It's because it just so happens that in Problem 1 a solution does exist. So, once the options for$(u,v)$have been narrowed down to one remaining option, that must indeed be it. On the other hand, in Problem 2 no solution exists. So, logically Solution 2 has narrowed down the possibilities to: "if a solution exists at all,$u$must be$4$and$v$must be$1$". That is compatible with the fact that no solution exists, so it is incorrect to stop there and claim$(4,1)$is the solution.

Resolution

Methods that use forward deductions like the above can be made mathematically correct by inserting an initial required fact that a solution definitely exists. Then, the narrowing-down procedure becomes legitimate because it ultimately narrows down the possibilities to at most one precise choice, and using the initial required fact (a solution exists), that single remaining possibility must be it.

In the context of quadratic equations, the fact that a pair of numbers definitely exists with sum$b$and product$c$is true, but not elementary. The Babylonians and Greeks didn't even believe it to be true in general, because it often involves imaginary numbers. For example, the Babylonians and Greeks didn't think that there were two numbers that add to$0$and multiply to$1$. (Today, we know that$\pm i$work.) Instead, this is a consequence of the Fundamental Theorem of Algebra for quadratic polynomials, which states that every quadratic has exactly 2 roots. The easiest proof of that fact is by the quadratic formula, but then it is not as useful to rely on a different proof of the quadratic formula as the first step in proving the quadratic formula. Several people in the past did that as a way not to prove the quadratic formula, but to come up with an easy way to re-deduce it, often en route to a more sophisticated Algebraic result. A record of discussion around Venezuelan mathematicians Francisco Duarte and Luis Báez Duarte, who recorded this in 1999 in the context of the cubic formula, is resolved in this pair of recent Spanish-language articles: part 1 and part 2.

Example of Difficult-to-Reverse Use of Existence

In general, one cannot naively reverse the logical direction of a proof which starts with an initial existence requirement. In the context of quadratic equations, the following independently interesting proof attributed to Thomas Harriot (1560–1621) demonstrates the difficulty. Consider a general quadratic equation$x^2+Bx+C=0$.

1. There definitely exist two numbers$r$and$s$such that the quadratic can be factorized as$$x^2+Bx+C=(x-r)(x-s)$$
2. By comparing coefficients in the expansion,$$r+s=-B$$
3. By comparing coefficients in the expansion,$$rs=C$$
4. Squaring both sides of Step 2,$$(r+s{)}^2=B^2$$
5. Taking Step 4 and subtracting the quadruple of Step 3,$$(r+s{)}^2-4rs=B^2-4C$$
6. Simplifying the left side of Step 5,$$(r-s{)}^2=B^2-4C$$
7. Taking the square root of both sides of Step 6,$$r-s=\pm \sqrt{B^2-4C}$$
8. Adding Step 2 and Step 7, and then dividing by 2,$$r=\frac{-B\pm \sqrt{B^2-4C}}{2}$$
9. Subtracting Step 8 from Step 2,$$s=\frac{-B\mp \sqrt{B^2-4C}}{2}$$
10. There are two sign choices that arise in Step 7. Either choice of sign propagates through Steps 8 and 9 to produce the same unordered pair of numbers$r$and$s$:$$\frac{-B\pm \sqrt{B^2-4C}}{2}$$
11. Step 1 (definite existence of the unordered pair) together with Step 10 (only one remaining possibility for the unordered pair) imply that the two numbers in Step 10 are the two precise numbers in the factorization in Step 1; by the zero-product property, they are the complete set of roots.

If one wishes to remove the initial assumption (Step 1), it is not possible to naively reverse the deductive steps. For a clear obstacle, observe that while Step 2 obviously implies Step 4, unfortunately Step 4 does not imply Step 2. Indeed, Step 4 implies that$r+s=\pm B$, with no immediate control on whether it is$+B$or$-B$, even though Step 2 requires$-B$exclusively. This example shows that in general, in order to remove an initial existence assumption, care must be taken to assemble the logical steps.