How many values 1 < c < 100 are there such that for all irreducible proper fractions a/c and b/18 and m = gcd(c, 18), the fraction d/m = a/c + b/18 is reducible? (a, b, c, d are all natural numbers)
This obviously isn't a Geometry problem, so I must be listing it because Rapoport makes an error. We can see what the error is if we think about adding fractions -- to add a/c + b/18, we must first find a common denominator -- the LCM of the denominators. But Rapoport states that m is the GCD of the denominators instead of the LCM.
I know -- in yesterday's post I mentioned a GCF and LCM song and the importance of not confusing these two, but here Rapoport confuses them. (Maybe she needs to hear my song!)
OK, so let's restate this problem correctly without her error:
How many values 1 < c < 100 are there such that for all irreducible proper fractions a/c and b/18 and m = lcm(c, 18), the fraction d/m = a/c + b/18 is reducible? (a, b, c, d are all natural numbers)
Actually, this problem is quite difficult, and I don't have a complete proof yet. The statement of the problem is confusing due to quantifiers such as "for all." So let's restate it like this:
- First, you choose a number c between 1 and 100.
- Then I choose the numbers a and b.
- I add the fractions a/c and b/18 using the common denominator m = lcm(c, 18).
- If the answer can't be reduced, then I win. If the answer can be reduced, then you win.
- The question is, how many possibilities for c do you have that can guarantee a win for you?
Notice that I can't just choose any values for a and b. I must choose values that make the starting fractions a/c and b/18 both irreducible and proper.
For example, I can't choose b = 2, since 2/18 can be reduced. Neither can I choose b = 3 or b = 4, since we can reduce 3/18 and 4/18. But I can't choose b = 19 either, since even though 19/18 can't be reduced, it's an improper fraction. There are only six allowable values of b: 1, 5, 7, 11, 13, 17, leading to the fractions 1/18, 5/18, 7/18, 11/18, 13/18, 17/18.
The allowable values of a for me depend on what you choose for c. You must choose c first before I get to choose a, so that a/c is both irreducible and proper.
OK, so you can try a few values of c to see what happens:
- You choose c = 2. I can only choose a = 1 to make the fraction 1/2, which we then add to one of the six fractions 1/18, 5/18, 7/18, 11/18, 13/18, 17/18. The common denominator is 18 and so we write 1/2 = 9/18. Adding these to each of the six fractions listed above gives the following list: 10/18, 14/18, 16/18, 20/18, 22/18, 26/18. Each of these can be reduced. Thus if you choose c = 2, you win!
- You choose c = 3. I choose a = b = 1: 1/3 + 1/18 = 6/18 + 1/18 = 7/18 -- irreducible. Thus if you choose c = 3, I win.
- You choose c = 4. I choose a = b = 1: 1/4 + 1/18 = 9/36 + 2/36 = 11/36 -- irreducible. Thus if you choose c = 4, I win.
- You choose c = 5. I choose a = b = 1: 1/5 + 1/18 = 18/90 + 5/90 = 23/90 -- irreducible. Thus if you choose c = 5, I win.
You aren't doing too well, are you? You might now try c = 6 to see what happens. Notice that I have only two possibilities for a, giving the fractions 1/6 and 5/6. The common denominator is 18, and so my first fraction is either 3/18 or 15/18. In either case I get odd/18. And the six possibilities for my second fraction are also in the form odd/18, so odd/18 + odd/18 = even/18, which is reducible no matter which fractions I choose. Thus if you choose c = 6, you win!
In fact, you can easily see now that if you choose c = 18, you win as well. I now have the same six choices for a that I have for b, and all of them lead to odd/18 + odd/18 = even/18, which is reducible.
So you now have three guaranteed winning values for c so far -- 2, 6, and 18. But there are still many values less than 100 that are left to check -- and that will take forever.
But here's an idea: the smallest two values are 2 and 6. Maybe the winning values form an arithmetic sequence, such as 2, 6, 10, 14. The next value is 18, which we already know is a winner.
To be sure, let's try some random values and see what happens:
You choose c = 70, I choose a = 69, b = 11: 69/70 + 11/18 = 1006/630 = 503/315
You choose c = 42, I choose a = 1, b = 7: 1/42 + 7/18 = 52/126 = 26/63
You choose c = 42, I choose a = 41, b = 17: 41/42 + 17/18 = 242/126 = 121/63
You choose c = 71, I choose a = 30, b = 13: 30/71 + 13/18 = 1463/1278 irreducible
You choose c = 69, I choose a = 64, b = 13: 64/69 + 17/18 = 683/414 irreducible (414 = lcm(69, 18))
I could do more examples, but that's enough for now. It appears so far that our conjecture that the winning values of c are 2, 6, 10, 14, 18, ..., 90, 94, 98 is plausible.
And if that conjecture is true, then exactly 1/4 of the values between 1 and 100 are winners. That's 25 possible winners -- and of course, today's date is the 25th.
OK, I admit it -- the only reason it occurred to me to conjecture that every singly even value of c is a winner is because I already knew that today's date is the 25th. But that's not doing mathematics -- that's just checking the date and writing down anything that fits the date.
I can come up with a partial proof. The fact that the known values of c that do work (2, 6, 18) all lead to the pattern odd/18 + odd/18 = even/18 reducible, so this might work in the general case.
Let c = 2(2n + 1) -- 2 times an odd number, hence it's singly even. Then we write:
a/(2(2n + 1)) + b/18
= 9a/(18(2n + 1)) + b(2n + 1)/(18(2n + 1))
= (9a + b(2n + 1))/(18(2n + 1))
And both 9a and b(2n + 1) are odd, so their sum is even. (We know that a is odd otherwise a/c would already be reducible, so 9a is odd. And b is odd because it's either 1, 5, 7, 11, 13, or 17.) Thus the resulting fraction is even/even, which is reducible.
But there are other cases to consider. The LCM of 18 and 2(2n + 1) isn't always 18(2n + 1) -- for example, if n = 7, then the lcm(18,30) = 90, not 270. And all we've shown is that 2, 6, 10, 14, 18, 22, 26, and so on are winners -- not yet that the in-between values are losers.
A complete proof appears to be quite tedious. But this is all the time I wish to spend on this problem, especially considering that it's not Geometry.
OK, I have some more information on a possible reopening date for my new district. If Orange County remains off the watch list for two weeks, that takes us to Labor Day. A board meeting is regularly scheduled for later that week. And if all goes well, the date for starting the hybrid schedule could be Monday, September 21st. Since Monday at secondary schools under the hybrid plan are online for everyone, the first actual day in the classroom would be Tuesday, September 22nd.
I figured that the reopen date would be around that date. Recall that in the one district approved for the elementary waiver (Los Alamitos Unified), the first day is two weeks after the earliest date allowable under the waiver. So if the other Orange County districts add a similar two-week cushion after the fourteenth day off the watch list, we arrive at September 21st as the reopen date.
The two-week cushion also allays fears that the coronavirus might spike over Labor Day weekend, just as it did over Memorial Day weekend in California. The worry is that people might gather over the holiday weekend (at the beach, for example) and spread the disease.
If the schools do indeed reopen on September 21st, what does that mean for the calendar? Well, that corresponds to Day 24. Distance learning will have spanned the first five weeks of the year -- this is about halfway through the first quarter -- that is, one quaver. Thus the first quaver progress report will be distance learning only, while the first quarter progress report will be half distance, half hybrid. (I remind you that this is not the 4 x 3 district, so it's the final semester grade that matters.)
On the blog, we'll be midway through Chapter 2 of the U of Chicago text. I expect that in the actual text that my new district uses (the Glencoe text), the classes will also be about halfway through the second chapter.
Again, I'll continue to keep you posted on the reopening plans in my district.
This is what I wrote last year about today's lesson:
Lesson 0.7 of Michael Serra's Discovering Geometry is called "Islamic Art." This is in the Second Edition -- in the modern editions, "Islamic Art" is Lesson 0.6. Serra begins:
"Islamic art is rich in geometric forms. Islamic artists were familiar with geometry through the works of Euclid, Pythagoras, and other mathematicians of antiquity, and they used geometric patterns extensively in their art and architecture.
"Many of [Muhammad's] followers interpreted his words to mean that the representation of humans or animals in art was forbidden. Therefore, instead of using human or animal forms for decorations, Islamic artists used intricate geometric patterns."
As usual, the questions I derive from Serra's text instruct the student to create Islamic-style art. This art is based on tessellations.
There are a few interesting things in this lesson. First, Serra includes a sidebar called "Improving Reasoning Skills -- Bagels I." As it turns out, Bagels is an old 1980's computer game. I never played it on my old computer, but as a young child, I actually had an old toy (Speak & Math) which included a version of Bagels (called "Number Stumper"). Here's a link to a modern version of Bagels:
http://www.dst-corp.com/james/Bagels.html
During the Responsive Classroom training at my old school, the presenter actually suggested Bagels as an opening week activity. In her version of the game, the word "Bagels" was replaced with "Nada," but the words "Pico" and "Fermi" were retained (so she called the game "Pico, Fermi, Nada"). Again, I don't post any version of "Pico, Fermi, Bagels/Nada," but if you want, you can use it in your own classroom instead of the "Islamic Art" lesson.
I do however include Serra's project for this lesson, "Geometry in Sculpture." This isn't directly related to Islamic art, though. Instead, he writes about Umbilic Torus, a sculpture. It was created by Helaman Ferguson and used as a trophy for the Jaime Escalante award -- named, of course, for the world's most famous math teacher.
Here is the Blaugust prompt for today:
What’s the one thing in your school year you’re most looking forward to? A lesson, a unit, a field trip, a school tradition
This is a new prompt for me. The last time August 25th wasn't on the weekend was 2017 -- and once again, that was the year that Shelli didn't have a Blaugust challenge.
Well, the one thing I'm most looking forward to is obvious -- the reopening on September 21st. I'm a substitute teacher, and I haven't had any work or earned any money since March. I know that being in the classroom during the coronavirus will be tough, but once again I need the money.
Let me answer the rest of Shelli's prompt. The "unit" that I'm looking forward to is Chapter 2 (on definitions and if-then statements), since that's the unit we'll be on in late September. And the "lessons" that I'm looking forward to are the lessons we'll be covering that week -- 2-5, 2-6, and 2-7.
In case you're wondering, here are those actual lessons:
- Lesson 2-5: Good Definitions
- Lesson 2-6: Unions and Intersections of Figures
- Lesson 2-7: Terms Associated with Polygons
So far, I've been making no changes to my lessons from last year (even though it may be uncertain whether these lessons make sense in a distance environment). But once the schools reopen, I'll want to rewrite these lessons in order to simulate how they'd actually be taught in the hybrid classroom.
In particular, all three of these lessons might be recast as opening week activities, since these will be given during what is essentially an opening week in the classroom for our students. But these can't be the type of activity that requires a lot of touching and groups, since social distancing still will be required once the schools reopen. Fortunately, I have nearly a month to figure these lessons out.
As for the last two parts of Shelli's prompt -- "field trip" and "school tradition" -- I don't have any responses for these as a sub. Indeed, it's unlikely I'd be able to answer these even if I were a regular teacher either. Most schools are cancelling their field trips this year, and many school traditions are also events that must be cancelled during the virus.
Back at my old charter school, the field trip we all looked forward to was the LA County Fair. I described this trip in posts from that time (September 2016), and I even wrote a song about that trip ("Meet Me in Pomona"), which I still sing to this day.
As for school traditions, I don't recall whether we had really any. I was only there for one year, and the school itself was only around for five years. So there wasn't much time to establish traditions.
The Halloween celebration that year might have been a tradition. The idea was for our celebration to be a literary one -- students could dress up as any character from a book. But especially these days when publishing a book is much easier, almost any blockbuster movie or popular TV show has had books written about it. So it's quite easy for "must be a book character" to stray far from its original literary intent. I don't know how Halloweens were celebrated after I left the school.
It's ironic that I could have unwittingly been responsible for establishing a Pi Day celebration at my old school. I left the school just before Pi Day, but decided to surprise my sixth graders with a pizza party visit. Two years later, I found out that there was a scheduled Pi Day celebration there.
Today's Blaugust participant is Bear St. Michael:
https://questionsaboutmath.blogspot.com/2020/08/designing-psets-for-remote-learning.html
But the big idea was using weekly problem sets (PSETs). That's what I plan on doing this fall. So I want to take an opportunity here to reflect on what I think PSETs can do for remote learning specifically, and why.
Hey -- are these PSETs as in traditionalist PSETs? We already know that the traditionalists highly recommend that students regularly work on PSETs to improve their knowledge of math.
But of course, once we click on them, we see that St. Michael's PSETs don't look anything like what the traditionalists mean. His PSETs start with an underlying "big problem" which the students can discuss with each other throughout the week. The traditionalists' PSETs are usually stated as page (whatever), do problems 2 to (whatever) even.
Like the other Blaugust participants, St. Michael is trying to figure out how to adjust his assignments to work in a distance learning environment.
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