Sum of Three Cubes

I now have a new example of an existence proof to show my students.

Last year, mathematicians Andrew Booker and Andrew Sutherland found solutions to the following two equations: x^3 + y^3 + z^3 = 33 and x^3 + y^3 + z^3 = 42. The first was found by Booker alone; the latter was found by the collaboration of both mathematicians. These deceptively simple-looking equations were cracked with a lot of math and a lot of computational firepower. The solutions:

(8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³ = 33

$latex (–80,538,738,812,075,974)3 + 80,435,758,145,817,5153 + 12,602,123,297,335,6313 = 42$

At the time of this writing, that settles the existence of solutions of x^3 + y^3 + z^3 = n for all positive integers n less than 100. For now, the smallest value of n for which the existence of a solution is not known is n = 114.

For further reference, including links to the original articles by Booker and then Booker and Sutherland, please see:

Existence Proofs

Source: https://xkcd.com/1856/

How To Use Facebook Emoji to Respond to a Mathematical Proof

Source: https://www.facebook.com/MathWithBadDrawings/photos/a.822582787758549/2865881973428610/?type=3&theater

Goodbye Aberration: Physicist Solves 2,000-Year-Old Optical Problem

This was a nice write-up (with some entertaining interspersed snark) of the solution of the the Wasserman-Wolf problem concerning the construction of a perfect lens (like a camera lens). Some quotes:

[L]enses are made from spherical surfaces. The problem arises when light rays outside the center of the lens or hitting at an angle can’t be focused at the desired distance in a point because of differences in refraction.

Which makes the center of the image sharper than the corners…

In a 1949 article published in the Royal Society Proceedings, Wasserman and Wolf formulated the problem—how to design a lens without spherical aberration—in an analytical way, and it has since been known as the Wasserman-Wolf problem…

The problem was solved in 2018 by doctoral students in Mexico. For those fluent in Spanish, the university press release can be found here. As an added bonus, here’s the answer:

 

My Favorite One-Liners: Part 119

Source: https://www.facebook.com/MathematicalMemesLogarithmicallyScaled/photos/a.1605246506167805/3119457221413385/?type=3&theater

Adding by a Form of 0 (Part 4)

In my previous post, I wrote out a proof (that an even number is an odd number plus 1) that included the following counterintuitive steps:

2k = (2k - 1) + 1 = ([2k - 1 - 1] + 1) + 1

A common reaction that I get from students, who are taking their first steps in learning how to write mathematical proofs, is that they don’t think they could produce steps like these on their own without a lot of coaching and prompting. They understand that the steps are correct, and they eventually understand why the steps were necessary for this particular proof (for example, the conversion from 2k-1 to [2k - 1 -1]+1 was necessary to show that 2k-1 is odd).

Not all students initially struggle with this concept, but some do. I’ve found that the following illustration is psychologically reassuring to students struggling with this concept. I tell them that while they may not be comfortable with adding and subtracting the same number (net effect of adding by 0), they should be comfortable with multiplying and dividing by the same number because they do this every time that they add or subtract fractions with different denominators. For example:

\displaystyle \frac{2}{3} + \frac{4}{5} = \displaystyle \frac{2}{3} \times 1 + \frac{4}{5} \times 1

= \displaystyle \frac{2}{3} \frac{5}{5} + \frac{4}{5} \times \frac{3}{3}

= \displaystyle \frac{10}{15} + \frac{12}{15}

= \displaystyle \frac{22}{15}

In the same way, we’re permitted to change 2k-1 to 2k-1 + 0 to 2k -1 - 1 + 1.

Hopefully, connecting this proof technique to this familiar operation from 5th or 6th grade mathematics — here in Texas, it appears in the 5th grade Texas Essential Knowledge and Skills under (3)(H) and (3)(K) — makes adding by a form of 0 in a proof somewhat less foreign to my students.

Adding by a Form of 0 (Part 3)

As part of my discrete mathematics class, I introduce my freshmen/sophomore students to various proof techniques, including proofs about sets. Here is one of the examples that I use that involves adding and subtracting a number twice in the same proof.

Theorem. Let A be the set of even integers, and define

B = \{ n: n = m+1 for some odd integer m\}

Then A = B.

Proof (with annotations). Before starting the proof, I should say that I expect my students to use the formal definitions of even and odd:

  • An integer n is even if n = 2k for some integer k.
  • An integer n is odd if n = 2k+1 for some integer k.

To prove that A = B, we must show that A \subseteq B and B \subseteq A. The first of these tends to trickiest for students.

Part 1. Let n \in A. By definition of even, that means that there is an integer k so that n = 2k.

To show that n \in B, we must show that n = m + 1 for some odd integer m. To this end, notice that n = (n-1) + 1. Thus, we must show that n - 1 is an odd integer, or that n -1 can be written in the form 2k+1. To do this, we add and subtract 1 a second time:

n = 2k

= (2k - 1) + 1

= ([2k - 1 - 1] + 1) + 1

= ([2k-2] + 1) + 1

= (2[k-1] + 1) + 1.

By the closure axioms, k-1 is an integer. Therefore, 2[k-1] + 1 is an odd number by definition of odd, and hence $n \in B$.

The above part of the proof can be a bit much to swallow for students first learning about proofs. For completeness, let me also include Part 2 (which, in my experience, most students can produce without difficulty).

Part 2. Let n \in B, so that n = m + 1 for some odd integer m. By definition of odd, there is an integer k so that $m = 2k+1$. Therefore, n = (2k+1) + 1 = 2k+2 = 2(k+1). By the closure axioms, k +1 is an integer. Therefore, n is even by definition of even, and so we conclude that n \in A.

\square

For what it’s worth, this is the review problems for which I recorded myself talking through the solution for the benefit of my students.

In my opinion, the biggest conceptual barriers in this proof are these steps from Part 1:

2k = (2k - 1) + 1 = ([2k - 1 - 1] + 1) + 1.

These steps are undeniably awkward. Back in high school algebra, students would get points taken off for making the expression more complicated instead of simplifying the answer. But this is the kind of jump that I need to train my students to do so that they can master this technique and be successful in their future math classes.

Adding by a Form of 0 (Part 2)

Often intuitive appeals for the proof of the Product Rule rely on pictures like the following:

The above picture comes from https://mrchasemath.com/2017/04/02/the-product-rule/, which notes the intuitive appeal of the argument but also its lack of rigor.

My preferred technique is to use the above rectangle picture but make it more rigorous. Assuming that the functions f and g are increasing, the difference f(x+h) g(x+h) - f(x) g(x) is exactly equal to the sum of the green and blue areas in the figure below.

In other words,

f(x+h) g(x+h) - f(x) g(x) = f(x+h) [g(x+h) - g(x)] + [f(x+h) - f(x)] g(x),

or

f(x+h) g(x+h) - f(x+h) g(x) + f(x+h) g(x) - f(x) g(x).

This gives a geometrical way of explaining this otherwise counterintuitive step for students not used to adding by a form of 0. I make a point of noting that we took one term, f(x+h), from the first product f(x+h) g(x+h), while the second term, g(x), came from the second product f(x) g(x). From this, the usual proof of the Product Rule follows:

[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h}

\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}

\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}

\displaystyle = \lim_{h \to 0} f(x+h) \frac{g(x+h) - g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) - f(x) }{h} g(x)

= f(x)g'(x) + f'(x) g(x)

For what it’s worth, a Google Images search for proofs of the Product Rule yielded plenty of pictures like the one at the top of this post but did not yield any pictures remotely similar to the green and blue rectangles above. This suggests to me that the above approach of motivating this critical step of this derivation might not be commonly known.

Once students have been introduced to the idea of adding by a form of 0, my experience is that the proof of the Quotient Rule is much more palatable. I’m unaware of a geometric proof that I would be willing to try with students (a description of the best attempt I’ve seen can be found here), and so adding by a form of 0 becomes unavoidable. The proof begins

\left[\left( \displaystyle \frac{f}{g} \right)(x) \right]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h)}{ g(x+h)} - \frac{f(x)}{ g(x)}}{h}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{ g(x) g(x+h)}}{h}

= \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x) - f(x) g(x+h)}{ h g(x) g(x+h)}.

At this point, I ask my students what we should add and subtract this time to complete the derivation. Given the previous experience with the Product Rule, students are usually quick to chose one factor from the first term and another factor from the second term, usually picking f(x) g(x). In fact, they usually find this step easier than the analogous step in the Product Rule because this expression is more palatable than the slightly more complicated f(x+h) g(x). From here, the rest of the proof follows:

[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x) + f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} + \frac{f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} - \frac{f(x) g(x+h) - f(x)g(x)}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{[f(x+h) - f(x)] g(x)}{h} - \frac{f(x) [g(x+h) - g(x)]}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) - f(x) }{h} g(x) - f(x) \frac{ g(x+h) - g(x)}{h }}{g(x) g(x+h)}

= \displaystyle \frac{ f'(x) g(x) - f(x) g'(x)}{g(x)^2}

P.S.

  • The website https://mrchasemath.com/2017/04/02/the-product-rule/ also suggests an interesting pedagogical idea: before giving the formal proof of the Product Rule, use a particular function and the limit definition of a derivative so that students can intuitively guess the form of the rule. For example, if g(x) = x^2:

Adding by a Form of 0 (Part 1)

Adding by a form of 0, or adding and subtracting the same quantity, is a common technique in mathematical proofs. For example, this technique is used in the second step of the standard proof of the Product Rule in calculus:

[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h}

\displaystyle = \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x+h) g(x) + f(x+h) g(x) - f(x) g(x)}{h}

\displaystyle = \lim_{h \to 0} \left[ \frac{f(x+h) g(x+h) - f(x+h) g(x)}{h} + \frac{f(x+h) g(x) - f(x) g(x)}{h} \right]

\displaystyle = \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x+h) g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) g(x) - f(x) g(x)}{h}

\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}

\displaystyle = \lim_{h \to 0} f(x+h) \frac{g(x+h) - g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) - f(x) }{h} g(x)

= f(x)g'(x) + f'(x) g(x)

Or the proof of the Quotient Rule:

\left[\left( \displaystyle \frac{f}{g} \right)(x) \right]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h)}{ g(x+h)} - \frac{f(x)}{ g(x)}}{h}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{ g(x) g(x+h)}}{h}

= \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x) - f(x) g(x+h)}{ h g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x) + f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} + \frac{f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} - \frac{f(x) g(x+h) - f(x)g(x)}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{[f(x+h) - f(x)] g(x)}{h} - \frac{f(x) [g(x+h) - g(x)]}{h }}{g(x) g(x+h)}

= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) - f(x) }{h} g(x) - f(x) \frac{ g(x+h) - g(x)}{h }}{g(x) g(x+h)}

= \displaystyle \frac{ f'(x) g(x) - f(x) g'(x)}{g(x)^2}

This is a technique that we expect math majors to add to their repertoire of techniques as they progress through the curriculum. I forget the exact proof, but I remember that, when I was a student in honors calculus, we had some theorem that required an argument of the form

|x - y| = |x - A + A - B + B - C + C - D + D - E + E - F + F - y|

\le |x - A| + |A - B| + |B - C| + |C - D| + |D - E| + |E - F| + |F - y|

\le \displaystyle \frac{\epsilon}{7} + \frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7}

= \epsilon

But while this is a technique that expect students to master, there’s no doubt that this looks utterly foreign to a student first encountering this technique. After all, in high school algebra, students would simplify something like x - A + A - B + B - C + C - D + D - E + E - F + F - y into x-y. If they were to convert x-y into something more complicated like x - A + A - B + B - C + C - D + D - E + E - F + F - y, they would most definitely get points taken off.

In this brief series, I’d like to give some thoughts on getting students comfortable with this technique.

Wason Selection Task: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on the Wason Selection Task.

Part 1: Statement of the problem.

Part 2: Answer of the problem.

Part 3: Pedagogical thoughts about the problem, including variants.