What is the 10th (base two) root of 10000 (base five)?
Well, 10 in base two is just two, so we're asking about the square root of 10000. Notice that the square root of 10000 in any base is 100 in that base -- that's because 10000 is just b^4 in base b, while 100 is just b^2, and sqrt(b^4) = b^2. Thus sqrt(10000) = 100 (base five) = 25. Therefore the answer is 25 -- and of course, today's date is the 25th.
Hold on a minute -- that's not a Geometry question, so why am I doing a Pappas question today? As these are the final weeks of school, I always celebrate by revisiting my most popular posts during the past year. My #1 post in hit count was a summer post on number bases. And so this is a good excuse to have another post on bases -- one of my favorite topics.
(By the way, today I subbed in an eighth grade U.S. History class. Because all classes just watch a video -- Civil War by Ken Burns -- and take notes, I won't do a "Day in the Life" today.)
The two bases that appear in this Pappas problem are two (binary) and five (quinary). Even though binary is ubiquitous in computers, it's so small that it's considered well below the human scale -- large numbers take too long to write in binary.
Quinary, at least, has its own thread at the Dozens Online website. (Recall that I have Dozens Online to thank for making number bases my most popular post of the year.)
https://www.tapatalk.com/groups/dozensonline/quick-kid-quinary-t728.html?sid=23ef46302163fa278d02da09d5411b8f
At this thread, it's pointed out that humans could have developed quinary as the dominant base over decimal, since we do have five fingers on a hand. But quinary might still be a bit too small to be considered human-scale -- and besides, it's odd. Since even bases are generally preferred, we count two hands and use decimal as the dominant base.
When I was young, my first scientific calculator, the TI-34, came with a manual. This calculator can convert numbers to binary as well as two binary-related bases, octal and hexadecimal. But in order to introduce the concept of number bases, the manual uses base five, the "handal" system.
{5} (default quinary/handal)
In this base, 10 is simply called "hand." The count goes as follows: one, two , three, four, hand, handy-one, handy-two, handy-three, handy-four, twandy (20). The other numbers also sound like decimal numbers with the syllable "hand" thrown in: thrandy (30), frandy (40), and of course, we must have one handred (100).
The manual didn't reach 1000. I suppose the easiest way to get my "hand" in is "one thouhand." Then 10000, following the decimal pattern, would be "hand thouhand." Perhaps it's better to change the Greek word "myriad" instead -- "one myrihand." So today's Pappas problem becomes:
The square root of one myrihand is one handred.
{a} (default decimal)
Since handal is such as small base, it has many simple divisibility rules. Divisibility by 2 and 4 is covered by the omega, while 3 and 6 are covered by the alpha. The SPD rule -- used for divisibility by 5 in dozenal, applies to 13 (twandy-three) in handal, since 26 = 2 * 13. Here SPD is based on one more than the square of the base (square-alpha). There is also SPD based on square-omega. It's useless in even bases since any factor of square-omega is available in either alpha or omega. But in odd bases, an extra power of two (beyond alpha or omega) is available in square-omega. In this case, square-omega gives us 8 (handy-three), since 24 = 3 * 8.
Quinary is small enough that SPD based on cube-alpha is also available. One more than 5^3 is 126, which factors as 2 * 7 * 9, so this gives us 7 (handy-two) and 9 (handy-four). For 7, we need to memorize 18 multiples of seven. This is fewer than the 29 multiples of five needed to do 5 in dozenal, but each multiple of seven has three quinary digits, so it's a wash. For 9, we only need to memorize 14 multiples of nine.
So quinary has feasible divisibility rules for factors 2-10. The useful numbers 2-6 have easier rules (divisor, alpha, omega) while the less useful 7-9 have more obscure rules (SPD).
But quinary has no simple rule for 11 (twandy-one). In fact, we can prove that 11 can't have a simple rule based on SPD for squares or cubes -- 11 can only have divisor, alpha, or omega rules. Thus decimal and dozenal are the smallest bases with an easy rule for 11 (along with undecimal of course).
The only possible SPD rule for 11 would be based on fifth powers. Since fifth powers are so large, this is infeasible for all bases -- except:
{2} (default binary)
It's possible to use SPD for eleven (1011) in binary. Only three multiples need to be memorized, and all are trivial: 00000, 01011, and 10110.
{a} (default decimal)
In ternary, 23 multiples of eleven must be memorized. This is fewer than 5 in dozenal, but the multiples contain five digits each, so it's difficult. Binary is below human-scale, but we can do 11 in bases 4, 8, and 16 by converting them to binary first.
OK, let's get back to music. Today we wish to code Mocha for 16EDL music:
The 16EDL scale:
Degree Ratio Note
16 1/1 white E
15 16/15 green F
14 8/7 red F#
13 16/13 ocher G
12 4/3 white A
11 16/11 amber B
10 8/5 green C
9 16/9 white D
8 2/1 white E
In 12EDL, it's easy to alternate between the 12:10:8 (that is, 6:5:4) and 11:9:7 chords. For 16EDL, we must get a little more creative. Our main chord is 16:13:11, and so our secondary chord might end up being something like 15:12:10 (a just major triad). Here's a possibility:
10 CLS
20 N=16
30 FOR A=0 TO 6
40 B=4
50 X=A-INT(A/2)*2
60 IF X=0 THEN D=16 ELSE D=15
70 PRINT D;
80 L=RND(B)
90 SOUND 261-N*D,4*L
100 IF L>1 THEN FOR I=1 TO L-1:PRINT " ";:NEXT I
110 B=B-L
120 IF B>0 THEN D=17-RND(9):GOTO 70
130 PRINT
140 NEXT A
150 PRINT 16
160 SOUND 261-N*16,16
So in this song, each measure starts with either 16 or 15 to represent the root of the 16:13:11 and 15:12:10 triads, and then random notes fill out each measure.
Yesterday, I wrote about converting the "Dren Song" to EDL. We might program it like this:
NEW
10 N=20
20 FOR V=1 TO 2
30 FOR X=1 TO 12
40 FOR Y=1 TO 7
50 READ D
60 IF V=1 THEN SOUND 261-N*D,4
70 IF V=2 THEN SOUND 261-N*(20-D),4
80 NEXT Y
90 FOR I=1 TO 400
100 NEXT I,X
110 RESTORE
120 NEXT V
130 DATA 12,11,10,12,13,12,11
140 DATA 13,12,11,13,12,11,10
150 DATA 12,11,10,12,13,12,11
160 DATA 13,12,11,13,12,11,12
170 DATA 10,9,8,10,11,10,9
180 DATA 12,11,10,12,13,12,11
190 DATA 10,9,8,10,11,10,9
200 DATA 11,10,9,11,10,9,10
210 DATA 10,9,8,10,9,8,7
220 DATA 11,10,9,11,10,9,8
230 DATA 10,9,8,10,9,8,7
240 DATA 9,8,7,9,8,7,8
This song begins with N=20, which gives us F minor. Another alternative is N=13, which would give us ocher C minor (to represent C# minor, the key in which I originally wrote the song).
The song consists of 24 lines, with the second 12 lines as the inversion of the first 12. The inversion is played by replacing D with 20 - D depending on whether it's the first (V=1) or second (V=2) verse.
Each line consists of seven quarter notes followed by a quarter rest (the FOR I loop).
What scale should we declare the song to be in. Here 12 is the root note, but it's not quite 12EDL because 13 is used. It's sort of like Bart Hopkin's mode 12 of the 16EDL scale. Since 15 isn't used, perhaps it's more accurate to call it mode 12 of the 14EDL scale. Then again, maybe we can simply call it 12EDL, add 13.
This just goes to show us that sometimes rather than having a span of one octave beginning and ending on the tonic, it's better to use a different sort of range. In this song, the range is 13-7, which 12 as the tonic.
As usual, with each EDL we play, I like to find EDO's we can convert it to. Let's run the EDO program I wrote a few years back, and input 16EDL. Here is the list of EDO's it outputs:
1, 2, 4, 6, 7, 10, 24, 31, 41, 53, 87, 130, ...
The first non-trivial EDO we see here is 24EDO. Unlike 19EDO and 27EDO which are tricky to play (especially the latter), 24EDO is easier because it's closely related to 12EDO.
http://xenharmonic.wikispaces.com/24edo
According to this link, 24EDO is a contorted 12EDO in the 5-limit -- that is, all 5-limit intervals (white, green, yellow) are the same in 24EDO as in 12EDO. The twelve new notes are quarter tones (half of a semitone). The quarter tones are used to give us 11 and 13. As it turns out, using the quarter tones for 7 (subminor and supermajor) is slightly more accurate than simply using 12EDO.
Here is some more information about 24EDO from the link, including Kite's colors. (As usual, all links will be dead after July.)
Basics
The 24edo system divides the octave into 24 equal parts of exactly 50 cents each. It is also known as quarter-tone tuning, since it evenly divides the 12-tone equal tempered semitone in two. Quarter-tones are the most commonly used microtonal tuning due to its retention of the familiar 12 tones and since it is the smallest microtonal equal temperament that contains all the 12 notes, and also because of its use in theory and occasionally in practice in Arabic music. It is easy to jump into this tuning and make microtonal music right away using common 12 equal software and even instruments - see this pageCombining ups and downs notation with color notation, qualities can be loosely associated with colors:
| quality | color | monzo format | examples |
|---|---|---|---|
| downminor | blue | {a, b, 0, 1} | 7/6, 7/4 |
| minor | fourthward white | {a, b}, b < -1 | 32/27, 16/9 |
| " | green | {a, b, -1} | 6/5, 9/5 |
| mid | jade | {a, b, 0, 0, 1} | 11/9, 11/6 |
| " | amber | {a, b, 0, 0, -1} | 12/11, 18/11 |
| major | yellow | {a, b, 1} | 5/4, 5/3 |
| " | fifthward white | {a, b}, b > 1 | 9/8, 27/16 |
| upmajor | red | {a, b, 0, -1} | 9/7, 12/7 |
The 11th harmonic, and most intervals derived from it, (11:10, 11:9, 11:8, 11:6, 12:11, 15:11, 16:11, 18:11, 20:11) are very well approximated in 24-tone equal temperament. The 24-tone interval of 550 cents is 1.3 cents flatter than 11:8 and is almost indistinguishable from it. In addition, the interval approximating 11:9 is 7 steps which is exactly half the perfect fifth.
Alternatively, ups and downs notation can be used. Here are the blue, green, jade, yellow and red triads:
| color of the 3rd | JI chord | notes as edosteps | notes of C chord | written name | spoken name |
|---|---|---|---|---|---|
| blue | 6:7:9 | 0-5-14 | C Ebv G | C.vm | C downminor |
| green | 10:12:15 | 0-6-14 | C Eb G | Cm | C minor |
| jade | 18:22:27 | 0-7-14 | C Ev G | C~ | C mid |
| yellow | 4:5:6 | 0-8-14 | C E G | C | C major or C |
| red | 14:18:27 | 0-9-14 | C E^ G | C.^ | C upmajor or C dot up |
Even though we've been using "d" for quarter-flat, the quarter-sharp symbol isn't easy for me to write in ASCII format. So let's follow the suggestion and use "^" for up/quarter-sharp, and so "v" will be down/"quarter-flat."
Here is the 16EDL scale in E, converted to 24EDO:
The 16EDL scale, converted to 24EDO:
Degree Ratio Note
16 1/1 E
15 16/15 F
14 8/7 Gv
13 16/13 G^
12 4/3 A
11 16/11 Bv
10 8/5 C
9 16/9 D
8 2/1 E
Since 24EDO agrees with 12EDO in the 5-limit, 24EDO tempers out both the syntonic comma and the Pythagorean comma. By tempering out the syntonic comma, 24EDO is a meantone tuning in which C-E is a major third. By tempering out the Pythagorean comma, G# and Ab are enharmonic, and B# and C are also enharmonic.
A YouTube search for 24EDO reveals the following video of a quarter-tone guitar. (No, I will not be playing my 16EDL songs on the 24EDO guitar this summer.)
OK, since I'm posting videos anyway, let me post the "Be Sharps" video from The Simpsons:
Here are the key parts I wish to emphasize in this video. The song "Coney Island Baby" begins at the 0:24 mark of the video. It starts with Seymour (bass), Homer (baritone), Clancy (lead), and Apu (tenor) singing a barbershop 7th chord, with each singing the word "Goodbye." The notes of this chord follow a 4:5:6:7 (otonal) pattern. The root of the chord is F, and so in septimal meantone, the notes are F-A-C-D#.
Then at 0:48, the quartet sings a D chord (the lyric here is "again"). In septimal meantone, a D barbershop chord is D-F#-A-B#, and so the "Be Sharps" really do sing the note B#. (Later Barney replaces Clancy as the lead, and D-F#-A-B# occurs near the middle of "Baby on Board," but there is no clean cut of this chord.)
Again 19EDO and 31EDO are good EDO's for septimal meantone. And 12EDO also supports septimal meantone, even though B# sounds as C. But 24EDO, while still meantone, isn't a septimal meantone EDO, since a 7/4 above D is now B quarter-sharp, not B#.
There are also higher EDO's that approximate the 13-limit. In fact, I can tie this to number bases -- you see, there was a debate on Dozens Online about the best EDO for 13-limit:
https://www.tapatalk.com/groups/dozensonline/dare-i-admit-good-musical-things-about-decimal-t400.html
The original poster, Tony, asks, "Dare I admit good musical things about decimal?" He then points out that half of 100 in decimal is 50, and 50EDO approximate the 13-limit well. Indeed, 50EDL is also a septimal meantone EDO.
But notice that 50EDO doesn't appear in the list of good EDO's for 16EDL. Instead, another EDO emerges as a good EDO for 13-limit -- 72EDO. (Admittedly 72EDO doesn't appear in the 16EDL list, but it does appear in the list for 14EDL, which is also 13-limit.)
Another poster, Ebbe, explains that 72EDO, unlike 50EDO, isn't a meantone scale (much less septimal meantone). Indeed, the step size for 50EDO is about the size of the syntonic comma, and so it's not surprising that the best EDO's in this range are no longer meantone. (But 72EDO nonetheless tempers out the Pythagorean comma, so we still have enharmonics G#=Ab and B#=C.) And 72 is half of the dozenal hundred (gross), and so it's "dozenal for the win again."
In 72EDO, six steps make a semitone and twelve steps make a whole tone. Thus each step of the scale can be called a twelfth-tone.
The major third above C is now one step below E -- that is, the scale maps the syntonic comma to a single step. The septimal comma is now two steps. The jade quarter-tone is now three steps (as 3/12 equals 1/4), and the ocher third-tone is now four steps (as 4/12 = 1/3). So we can see how ocher fits 72EDO better than 24EDO.
In fact, here are some of the notes of 72EDO, in sequence (listed as thirds above C)
Step Note Color
16 Eb-2 blue Eb (subminor 3rd above C)
17 Eb-1 yellow Eb
18 Eb white Eb
19 Eb+1 green Eb (minor 3rd above C)
20 Eb+2 red Eb (emerald E)
21 E-3 jade Eb (amber E, neutral 3rd above C)
22 E-2 blue E (ocher Eb)
23 E-1 yellow E (major 3rd above C)
24 E white E
25 E+1 green E
26 E+2 red E (supermajor 3rd above C)
Within each classical name, the rainbow is emerald-amber-blue-yellow-white-green-red-jade-ocher.
Here's one more recent music-related post at the Dozens Online website:
https://www.tapatalk.com/groups/dozensonline/why-the-golden-ratio-is-golden-t1875.html
The original poster Dan is writing about Phi, the Golden Ratio. In this post he ties Phi to music:
One practical use of "almost rational" irrational numbers is musical tuning. For example, the equal-tempered major third has a frequency ratio of the cube root of two (1.259921 decimal, 1.315188 dozenal), approximating the just intonation ratio 5/4. So, would a tuning based on setting the minor sixth to φ (or equivalently, setting the major fifth to (8/φ)^(1/4), or 691.7274 cents) sound dissonant?
(It's too bad this wasn't posted before Phi Day of the Century.) Anyway, we know that the ratio of consecutive Fibonacci numbers approaches Phi, so let's start with these:
Ratio Name
1/1 unison
2/1 octave
3/2 perfect fifth
5/3 major sixth
8/5 minor sixth
13/8 large tridecimal neutral sixth
21/13 small tridecimal neutral sixth
The largest Fibonacci numbers within Mocha range are 144 and 233, so playing Degree 233 (Sound 28) and Degree 144 (Sound 117) best approximates Phi in Mocha.
Amond EDO's, the aforementioned 72EDO approximates Phi well. If the root note is C, then Phi appears two steps above Ab (about 833 cents). We can refer to this note as either emerald A (to represent 13/8) or ocher-blue Ab (21/13).
The Xenharmonic site has its own page about applying Phi to music:
http://xenharmonic.wikispaces.com/Golden%20ratio
Phi taken as a musical ratio (ϕ*f where f=1/1) is about 833.1 cents. This is sometimes called "acoustical phi".
As the ratios of successive terms of the Fibonacci sequence converge on phi, the just intonation intervals 3/2, 5/3, 8/5, 13/8, 21/13, ... converge on ~833.1 cents.
"Logarithmic phi", or 1200*ϕ cents = 1941.6 cents (or, octave-reduced, 741.6 cents) is also useful as a generator, for example in Erv Wilson's "Golden Horagrams".
("Acoustical Phi" is to "Logarithmic Phi" as EDL is to EDO. Thus Acoustical Phi appears in Fibonacci EDL's like 34EDL, while Logarithmic Phi appears in Fibonacci EDO's like 34EDO.)
Question 19 of the SBAC Practice Exam is on solving equations:
Consider this solution to a problem:
Problem: -4(6 - y) + 4 = -4
Step 1: -24 - 4y + 4 = -4
Step 2: -20 - 4y = -4
Step 3: -4y = 16
Step 4: y = -4
In the first response box, enter the number of the step where the mistake is made.
In the second response box, enter the correct solution to the problem.
This is a strong first semester Algebra I problem. The original problem has lots of negative signs, and so the obvious error to search for is a sign error. And sure enough, there is a sign error right away -- -4 times -y should be 4y, not -4y. So the step that contains the mistake is step 1.
The solution is easy to correct. All we have to do is change all the signs on y:
Problem: -4(6 - y) + 4 = -4
Step 1: -24 + 4y + 4 = -4
Step 2: -20 + 4y = -4
Step 3: 4y = 16
Step 4: y = 4
Question 20 of the SBAC Practice Exam is on solving equations:
Consider a sequence whose first five terms are: -1.75, -0.5, 0.75, 2, 3.25
Which function (with domain all integers n > 1) could be used to define and continue this sequence?
A) f (n) = (7/4)(n - 1) - 5/4
B) f (n) = (5/4)(n - 1) - 7/4
C) f (n) = (7/4)n - 5/4
D) f (n) = (5/4)n - 7/4
Because this is an arithmetic sequence, this is also considered to be first semester Algebra I. The first thing we notice is that the terms are listed as decimals, but the choices are all fractions. So the students must convert between decimals and fractions. There is an embedded calculator available, but that assumes that the students know how to use it to make the conversion. I used the calculator (powered by Desmos!) to divide to convert fractions to decimals. Some calculators can convert decimals to fractions, but I don't see that option on this calculator. So the first step would be to divide to convert 5/4 (a major 3rd!) and 7/4 (a barbershop 7th!) to fractions.
We find out that the first term -1.75 is -7/4 and the common difference is -1.25 or 5/4. But we can't use the first term unless we use the (n - 1) version of the formula, f (n) = f (1) + (n - 1)d. Plugging in to this formula, we obtain f (n) = -7/4 + (n - 1)5/4, which is rewritten as f (n) = (5/4)(n - 1) - 7/4. So the correct answer is B).
Today is an activity day. Here the activities come from Chapter 6 of the U of Chicago Algebra I text, since this chapter matches these questions the best. Lesson 6-9 of the U of Chicago text is called "Subtracting Quantities" and emphasizes use of the Distributive Property with negatives, while Lesson 6-3 contains arithmetic sequences. The Exploration Question in the latter lesson is on programming in BASIC. Erase that music song using NEW so we can program Mocha to answer the activity question, or just skip that part of the activity.
SBAC Practice Exam Question 19
Common Core Standard:
Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
SBAC Practice Exam Question 20
Common Core Standard:
Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
Commentary: Distributing a negative sign correctly appears in Lesson 6-9 of the U of Chicago Algebra I text. But the method of finding explicit formulas in Lesson 6-3 -- finding the phantom "zeroth term," doesn't work for Question 10. Instead, students must know and be able to use the formula f (n) = f (1) + (n - 1)d, which most likely appears in later texts based on the Common Core, such as Glencoe and Edgenuity.
Monday is Memorial Day, and so my next post will be on Tuesday.
Monday is Memorial Day, and so my next post will be on Tuesday.
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