Unlearning, or Proof by Contradiction

Sometimes, we have to unlearn the things we initially learned. And I don't mean this in the sense of having been deliberately deceived. Rather, I mean that to some extent, there are actually many situations in life that involve necessary lies—or believing things that are wrong for perfectly rational reasons. Sometimes it is only after we have consumed and digested such a falsehood that we can see the truth at all. Really, this form of learning is not unlike some parts of math.

Consider a mathematical proof in which we begin by assuming that something is one way. But by the end of the proof, we may realize, through contradiction, that it's actually another way.

Let us take the number 2 and generously hypothesize that the square root of 2 is actually rational. If this assumption were true, we should be able to prove it with an equation. Let the square root of 2 be the lowest form of $\frac{p}{q}$.

Since squares of even numbers are even, and squares of odd numbers are odd, it follows that in order to get back to two, we would have to raise both p and q to the power of 2, like this:

\[ 2 = \frac{p^2}{q^2} \]

If we multiply both sides by $q^2$, we can get rid of the denominator. Now we get $ p^2 = 2q^2 $. From here, we must infer that $p$ is an even number.

With our generous assumption that $p$ is even, let us substitute $p$ for $2r$, where r is an integer. Let us test our hypothesis with the following equation, simplifying both sides:

\[ (2r)^2 = 2q^2 \] \[ 4r^2 = 2q^2 \]

Uh oh. Now we've hit a snag. From here, if we attempt to divide by two on both sides, we get:

\[ 2r^2 = q^2 \]

And we cannot further divide or simplify our equation. At least, we can not do so within the realm of rational numbers!

How can this be? Remember, our initial hypothesis that the square root of 2 was rational rested on the assumption that $\frac{p}{q}$ was in its lowest form. But now here we see that $2r^2$ is equal to $q^2$. In other words, both $p$ and $q$ are divisible by 2—which contradicts our original claim that $\frac{p}{q}$ was in lowest terms.

This means they still share a common factor. Thus, neither side is in its lowest form. Proof by contradiction that the square root of two is not rational after all.

James Garfield's Pythagorean Proof

Today I learned James Garfield, who once worked as a lawyer, Civil War General, and served as the 20th President of the United States, was math savvy and published a novel Pythagorean theorem proof.^[1]

\[ \text{Area}_{\text{trapezoid } ACED} = \frac{1}{2} \cdot (AC + DE) \cdot CE = \frac{1}{2} \cdot (a + b) \cdot (a + b) = \frac{(a + b)^2}{2} \] \[ \begin{aligned} \text{Area}_{\text{trapezoid } ACED} &= \text{Area}_{\Delta ACB} + \text{Area}_{\Delta ABD} + \text{Area}_{\Delta BDE} \\ &= \frac{1}{2}(a \times b) + \frac{1}{2}(c \times c) + \frac{1}{2}(a \times b) \end{aligned} \] \[ (a + b) \times \frac{1}{2}(a + b) = \frac{1}{2}(a \times b) + \frac{1}{2}(c \times c) + \frac{1}{2}(a \times b) \] \[ a^{2} + b^{2} = c^{2} \]

Small Pieces

We can take this in smaller pieces. First, we can find the area of the right-angled trapezoid with the following equation:

\[ \text{Area}_{\text{trapezoid}} = \frac{1}{2} \cdot (a + b) \cdot (a + b) = \frac{(a + b)^2}{2} \]

We can find the area of each of the two outer triangles with the following:

\[ \text{Area}_{\text{triangle}} = \frac{ab}{2} \]

And the area of the inner triangle with:

\[ \text{Area}_{\text{inner triangle}} = \frac{c^2}{2} \]

Proof

Reducing, we can go to the end, beginning with our substituted and now simplified area equation demonstrated above:

\[ \frac{(a + b)^2}{2} = 2 \cdot \frac{ab}{2} + \frac{c^2}{2} \]

Then we expand \( (a + b)^2 \) on the left hand side. And our equation on the right can also be simplified since we're both multiplying and dividing \( ab \) by 2:

\[ \frac{a^2 + 2ab + b^2}{2} = ab + \frac{c^2}{2} \]

Multiply both sides by 2 to eliminate denominators:

\[ a^2 + 2ab + b^2 = 2ab + c^2 \]

Lastly, subtract \( 2ab \) from both sides:

\[ a^2 + b^2 = c^2 \]

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Footnotes

1. Mathematical Treasure: James A. Garfield’s Proof of the Pythagorean Theorem ↩︎

Euler's Equation

Euler's number, e, the constant 2.71828, is the base of the natural logarithms. Given n approaching infinity, Euler's number is the limit of:

\begin{align*}\displaystyle{\displaylines{(1 + 1/n)n}}\end{align*}

It's used frequently abroad across the sciences. It can also be elegantly expressed as an infinite series, like so:

\begin{align*} {\displaystyle e=\sum \limits _{n=0}^{\infty }{\frac {1}{n!}}=1+{\frac {1}{1}}+{\frac {1}{1\cdot 2}}+{\frac {1}{1\cdot 2\cdot 3}}+\cdots .} \end{align*}

Separately, the imaginary unit i, \({\displaystyle {\sqrt {-i}}}\), represents the imaginary solution to the quadratic equation, x² + 1 = 0. The value can also be used to extend real numbers to complex numbers.

And π is pi, the irrational number we all know and love, roughly approximate to 3.14159, representing the ratio of the circle's circumference to its diameter.

While it isn't absolutely understood, we can join the three numbers in a seemingly bizarre proof that just works.

\( {\displaystyle e^{i\pi }=-1} \)