Difference of Two Powers (Part 5)

In this series of posts, I’ve explored ways that students can discover the formula for the difference of two squares and the difference of two cubes:

$x^2 - y^2 = (x-y) (x+y)$

$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ .

If students have understood the origins of these two formulas, then it’s not much of a stretch for students to guess the formula for $x^4 -y^4$ . A geometric derivation requires four-dimensional visualization which is beyond of what can be reasonably expected of high school students. Still, students can look at the above two formula and guess that $x-y$ is a factor of $x^4-y^4$ , and that the second factor would contain $x^3$ and $y^3$ :

$x^4 - y^4 = (x-y)(x^3 + \hbox{~~~something~~~} + y^3)$ .

From this point forward, it’s a matter of either using long division to find the quotient of $x^4-y^4$ or else just guessing (and confirming) the nature of the $\hbox{something}$ .

Once students recognize that the answer is

$x^4 - y^4 = (x-y)(x^3 + x^2 y + x y^2 + y^3)$ ,

then the factorings of $x^5 - y^5$ , $x^6 - y^6$ , etc. become obvious.

Difference of Two Cubes (Part 4)

Here’s the formula for the difference of two cubes:

$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$

The formula isn’t terribly complicated; however, the factoring on the right-hand side is hardly the first thing that a student would guess if only given the left-hand side to simplify. The formula of course can be confirmed by multiplying out the right-hand side, but that’s really cheating. It’d be nice to have a way for students to develop the right-hand side, as opposed to merely confirming that the right-hand side is correct.

To this end, I suggest using base-10 blocks, a common manipulative found in elementary classrooms. The figure below shows a 10×10 cube with a 3×3 cube removed.

A (hopefully interesting) challenge for students would be how to build this figure only using the materials found in a typical base-10 kit, and also building it with as few pieces as possible. I think that most high school students, after some thought, can solve this puzzle by using 7 flats (for the bottom 7 layers), 21 rods, and 63 units. This of course provides the correct answer, as

$7 \times 100 + 21 \times 10 + 63 \times 1 = 963 = 10^3 - 3^3$ .

After finding the correct answer, students should give this picture some deeper thought. If we let $x = 10$ and $y = 3$ , then

$7 \times 100 = (x-y) x^2$ .

This makes sense on physical grounds: the volume of the “base” of 7 layers is $7 \times 10$ , and the $7$ came from the fact that the top 3 layers are incomplete.

Likewise, the 21 rods can be thought of as

$21 \times 10 = 7 \times 3 \times 10 = (x-y) y x$ .

Again, this makes sense just looking at the picture, as the 21 rods makes a solid that is 3 units high ( $y$ ), 10 units long ( $x$ ), and 7 units wide ( $x-y$ ).

Finally, the 63 units can be thought of as

$63 = 7 \times 3 \times 3 = (x-y) y^2$ .

Indeed, these 63 units form a solid with a square base of side 3 and a length of 7.

Adding them together, we find

$7 \times 100 + 21 \times 10 + 63 \times 1 = (x-y) x^2 + (x-y) xy + (x-y) y^2 = (x-y) (x^2 + xy+ y^2)$ ,

which is indeed the formula for the difference of two cubes. Now that students have discovered the formula for themselves, the formula can then be confirmed using the distributive law.

As a postscript, it should be possible to use two different colors of base-10 blocks (to represent positive and negative numbers) so that students can derive the formula

$x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ .

However, I don’t personally own two different colored base 10 kits, so I haven’t had time to think through how to do this.

Difference of Two Cubes (Part 3)

In my experience, students who have reached the level of calculus or higher have completely mastered the formula for the difference of two squares:

$x^2 -y^2 = (x-y)(x+y)$ .

However, these same students almost never know that there even is a formula for factoring the difference of two cubes $x^3 -y^3$ , and it’s a rare day that I have a student who can actually immediately recall the formula correctly. I suppose that this formula is either never taught in Algebra II or (more likely) students immediately forget the formula after it’s been taught since there’s little opportunity for reinforcing this formula in more advanced courses in mathematics.

I recently came across an interesting pedagogical challenge: Is there an easy way, using commonly used classroom supplies, for teachers to guide students to explore and discover the formula for the difference of two cubes in the same way that they can discover the formula for the difference of two squares? (The cheap way is for students to just multiply out the factored expression to get $x^3 -y^3$ , but that’s cheating since they shouldn’t know what the answer is in advance.)

I came up with a way to do this, and I’ll present it in tomorrow’s post. For now, I’ll leave a thought bubble for anyone who’d like to think about it between now and then.

Difference of Two Squares (Part 2)

In yesterday’s post, I discussed a numerical way for students in Algebra I to guess for themselves the formula for the difference of two squares.

There is a also well-known geometric way of deriving this formula (from http://proofsfromthebook.com/2013/03/20/representing-the-sum-and-difference-of-two-squares/)

The idea is that a square of side $b$ is cut from a corner of a square of side $a$ . By cutting the remaining figure in two and rearranging the pieces, a rectangle with side lengths of $a+b$ and $a-b$ can be formed, thus proving that $a^2 - b^2 = (a+b)(a-b)$ .

Again, this is a simple construction that only requires paper, scissors, and a little guidance from the teacher so that students can discover this formula for themselves.

Difference of Two Squares (Part 1)

In Algebra I, we drill into student’s heads the formula for the difference of two squares:

$x^2 - y^2 = (x-y)(x+y)$

While this formula can be confirmed by just multiplying out the right-hand side, innovative teachers can try to get students to do some exploration to guess the formula for themselves. For example, teachers can use some cleverly chosen multiplication problems:

$9 \times 11 = 99$

$19 \times 21 = 399$

$29 \times 31 = 899$

$39 \times 41 = 1599$

Students should be able to recognize the pattern (perhaps with a little prompting):

$9 \times 11 = 99 = 100 - 1$

$19 \times 21 = 399 = 400 - 1$

$29 \times 31 = 899 = 900 - 1$

$39 \times 41 = 1599 = 1600 - 1$

Students should hopefully recognize the perfect squares:

$9 \times 11 = 99 = 10^2 - 1$

$19 \times 21 = 399 = 20^2 - 1$

$29 \times 31 = 899 = 30^2 - 1$

$39 \times 41 = 1599 = 40^2 - 1$ ,

so that they can guess the answer to something like $59 \times 61$ without pulling out their calculators.

Continuing the exploration, students can use a calculator to find

$8 \times 12 = 96$

$18 \times 22 = 396$

$28 \times 32 = 896$

$38 \times 42 = 1596$

Students should be able to recognize the pattern:

$8 \times 12 = 10^2 - 4$

$18 \times 22 = 20^2 - 4$

$28 \times 32 = 30^2 - 4$

$38 \times 42 = 40^2 -4$ ,

and perhaps they can even see the next step:

$8 \times 12 = 10^2 - 2^2$

$18 \times 22 = 20^2 - 2^2$

$28 \times 32 = 30^2 - 2^2$

$38 \times 42 = 40^2 -2^2$ .

From this point, it’s a straightforward jump to

$(10-2) \times (10+2) = 10^2 - 2^2$

$(20-2) \times (20+2) = 20^2 - 2^2$

$(30-2) \times (30+2) = 30^2 - 2^2$

$(40-2) \times (40+2) = 40^2 -2^2$ ,

leading students to guess that $(x-y)(x+y) = x^2 -y^2$ .

Conditional Statements

Source: http://xkcd.com/1652/

From the Life of Fred

From the Life of Fred Pre-Algebra Course:

“A question for English majors: Suppose you wanted to say that the digit 9 followed by a decimal point is the same as a plain 9 without a decimal point. You write something like “9. is the same as 9.” To my eye, that seems a bit strange. Or, once, when I was in high school, I wrote in an essay the sentence: Rocky owed Sylvia $2.. The first dot was for the decimal, and the second dot was the period. The teacher marked it wrong. Okay, English majors, here is your multiple-choice question: Is English harder than math? Here are your choices: ☐yes or ☐yes.”

Source: https://thehomelibraryonline.wordpress.com/2013/08/15/life-of-fred-pre-algebra-course/

Avocado’s number

Source: https://www.facebook.com/GeekParents/photos/a.299515633482231.53052.299266843507110/591134594320332/?type=3&theater

Computer Cracks 200 Terabyte Math Proof

Here’s a cute problem, called the Boolean Pythagorean Theorem problem. Here are the first few Pythagorean triples:

$3^2 + 4^2 = 5^2$

$6^2 + 8^2 = 10^2$

$5^2 + 12^2 = 13^2$

$9^2 + 12^2 = 15^2$

$8^2 + 15^2 = 17^2$

$12^2 + 16^2 = 20^2$

$7^2 + 24^2 = 25^2$

$15^2 + 20^2 = 25^2$

$10^2 + 24^2 = 26^2$

$20^2 + 21^2 = 29^2$

$18^2 + 24^2 = 30^2$

$16^2 + 30^2 = 34^2$

$21^2 + 28^2 = 35^2$

$12^2 + 35^2 = 37^2$

$15^2 + 36^2 = 39^2$

$27^2 + 36^2 = 45^2$

$9^2 + 40^2 = 41^2$

$27^2 + 36^2 = 45^2$

OK, let’s have some fun with this. Let’s write every multiple of 5 (5, 10, 15, 20, 25, 30, 35, 40, 45) in boldface:

$3^2 + 4^2 = {\bf 5}^2$

$6^2 + 8^2 = {\bf 10}^2$

${\bf 5}^2 + 12^2 = 13^2$

$9^2 + 12^2 = {\bf 15}^2$

$8^2 + {\bf 15}^2 = 17^2$

$12^2 + 16^2 = {\bf 20}^2$

$7^2 + 24^2 = {\bf 25}^2$

${\bf 15}^2 + {\bf 20}^2 = {\bf 25}^2$

${\bf 10}^2 + 24^2 = 26^2$

${\bf 20}^2 + 21^2 = 29^2$

$18^2 + 24^2 = {\bf 30}^2$

$16^2 + {\bf 30}^2 = 34^2$

$21^2 + 28^2 = {\bf 35}^2$

$12^2 + {\bf 35}^2 = 37^2$

${\bf 15}^2 + 36^2 = 39^2$

$27^2 + 36^2 = {\bf 45}^2$

$9^2 + {\bf 40}^2 = 41^2$

$27^2 + 36^2 = {\bf 45}^2$

For nearly all of these equations, there is one number that’s in boldface and one that’s not. However, there’s one that is all in one typeface: ${\bf 15}^2 + {\bf 20}^2 = {\bf 25}^2$ .

So here’s a question: is it possible to divide the integers so that every Pythagorean triple (not just the small ones listed above) has at least one number in boldface and another that’s not?

This May, it was proved that it’s impossible. The proof is very brute-force (from https://cosmosmagazine.com/mathematics/computer-cracks-200-terabyte-maths-proof):

The team found all triples could be multi-coloured in integers up to 7,824. As soon as they hit 7,825, it became impossible.

But to prove a solution doesn’t exist, you need to try all possibilities. There are more than $10^{2300}$ ways to colour all those integers, so the scientists used a few mathematical tricks to reduce the number of combinations to trial to just under one trillion.

Two days later, with 800 processors at the University of Texas Stampede supercomputer crunching all possibilities in parallel, the team had their answer – no.

There is no way to colour the integers 1 to 7,825 in a way that leaves all Pythagorean triples multi-coloured, the team reported in arXiv.

I had to read this news article a couple of times to appreciate this: a supercomputer ran for two days on a supercomputer (without parallelization, computation time was 51,000 hours), producing an output file of 200 terabytes, comparable “to the size of the entire digitized text held by the US Library of Congress.” Wow.

Sphere Packing Solved in Higher Dimensions

I enjoyed reading this bit of mathematical news: https://www.quantamagazine.org/20160330-sphere-packing-solved-in-higher-dimensions/

The opening paragraphs:

In a pair of papers posted online this month, a Ukrainian mathematician has solved two high-dimensional versions of the centuries-old “sphere packing” problem. In dimensions eight and 24 (the latter dimension in collaboration with other researchers), she has proved that two highly symmetrical arrangements pack spheres together in the densest possible way.

Mathematicians have been studying sphere packings since at least 1611, when Johannes Kepler conjectured that the densest way to pack together equal-sized spheres in space is the familiar pyramidal piling of oranges seen in grocery stores. Despite the problem’s seeming simplicity, it was not settled until 1998, when Thomas Hales, now of the University of Pittsburgh, finally proved Kepler’s conjecture in 250 pages of mathematical arguments combined with mammoth computer calculations.