But I will write that instead of a conference period, this teacher has a drama class. This class watches a YouTube video of Peter Pan starring Mary Martin, to prepare for their own performance of it.
All five English classes are currently reading John Steinbeck's The Pearl. I read this short novel as a young student, but not until sophomore year. This time, the students are to answer five "Think" questions on Chromebooks, with three sentences required for each response. About one student in each class fails to answer enough questions, or answers with fewer than three sentences. These are the guys I end up putting on my bad list.
Today on her Mathematics Calendar 2019, Theoni Pappas writes:
Which of the triplets below is a Pythagorean triplet?
(1) 1/sqrt(2), 1/sqrt(2), 2
(2) 2, 3, 5
(3) 4.5, 20, 20.5
(4) 2, 3, 6
(5) 14, 4, 16
A Pythagorean "triplet" (or triple) is a set of three numbers that satisfies the theorem of that name. So we could try plugging in all five choices into the equation a^2 + b^2 = c^2.
But notice that three of the five choices are wastes of time. Indeed, choices (1), (2), and (4) all fail to satisfy the Triangle Inequality. In other words, there is no triangle at all with sides of those lengths, much less a right triangle. Whenever we're given a multiple choice Pythagorean triple problem, it might be worth plugging them into Triangle Inequality first before trying the Pythagorean Theorem.
By the way, I even tried Triangle Inequality with choice (1) despite its irrational sides. I know just by looking that 1/sqrt(2) squared is less than 1, so the sum of the squares of two legs is less than two and much less than 2^2 = 4. (In fact, for those of us familiar with 45-45-90 triangles, the hypotenuse of a triangle with leg 1/sqrt(2) must be 1, not 2.)
Let's try the other two choices:
a^2 + b^2 + c^2
14^2 + 4^2 = 16^2
196 + 16 = 256
212 = 256
false
a^2 + b^2 + c^2
4.5^2 + 20^2 = 20.5^2
20.25 + 400 = 420.25
420.25 = 420.25
true
Thus the desired triple is 4.5-20-20.5. Therefore the correct answer is (3) -- and of course, today's date is the third.
Notice that this is the time of the month when multiple choice problems appear on the calendar. After all, Pappas couldn't have given a multiple choice problem three days ago unless she wants it to have 31 possible choices, as the date was the 31st. The date restriction necessarily limits multiple choice problems to the first week of the month.
It's time to review for the Chapter 13 Test. I keep changing how I would give this test, but I do have some review worksheets from last year and beyond, so I can post them today.
And this is what I wrote last year about today's review worksheet -- which includes a story about indirect proofs (including irrationality, since I just wrote about 1/sqrt(2) earlier):
Here is a link to a common indirect proof that sqrt(2) is irrational:
http://www.math.utah.edu/~pa/math/q1.html
Neither the U of Chicago nor Glencoe gives the proof outright. But both hint at it -- I just mentioned the U of Chicago's square root proofs. The Glencoe text asks the students to prove that if the square of a number is even, then it is divisible by four. As we can see at the above link, this fact is directly mentioned in the irrationality proof.
I remember once reading the proof of the irrationality of sqrt(2) in my textbook back when I was an Algebra I student. Until then, I had always heard that sqrt(2) was irrational, but I never realized that it was something that could be proved. So I was fascinated by the proof. Naturally, the text only included this as an extra page between the main sections, so it was something that the teacher skipped and most students probably ignored.
The irrationality of sqrt(2) has an interesting history. It goes back to Pythagoras -- he was one of the first mathematicians to use sqrt(2), since his famous Theorem could be used to show that the diagonal of a square has length sqrt(2). The website Cut the Knot, which has many proofs of the Pythagorean Theorem, also contains many proofs of the irrationality of sqrt(2):
http://www.cut-the-knot.org/proofs/sq_root.shtml
Now there is a famous story regarding sqrt(2) and Pythagoras. At the following link, we see that Pythagoras was the leader of a secret society, or Brotherhood:
http://nrich.maths.org/2671
Now Pythagoras and his followers believed that only natural numbers were truly numbers. Not even fractions were considered to be numbers, but simply the ratios of numbers -- numberhood itself was reserved only for the natural numbers. In some ways, this attitude resembles that of algebra students today -- when the solution of an equation is a fraction, they often don't consider it to be a real answer, even though modern mathematics considers fractions to be numbers. (The phrases real number and imaginary number reflect a similar attitude about 2000 years after Pythagoras -- that some numbers aren't really numbers.) So of course, the idea that there were "numbers" that weren't the ratio of natural numbers at all was just unthinkable.
Pythagoras and his followers must have spent years searching for the correct fraction whose square is 2, but to no avail. Finally, one of his followers, Hippasus, discovered the reason that they were having such bad luck finding the correct fraction -- because there is no such fraction! And, as the story goes, Pythagoras was so distraught, afraid that the secret that sqrt(2) was irrational would be revealed, that he ordered to have poor Hippasus drowned at sea!
But as I said, nowadays students simply complain when they have a fractional, or worse irrational, answer to a problem. No one has to drown any more just because of irrational numbers.
Question 10 on my test review, therefore, is actually the final step of that proof, since that's the step where the contradiction occurs. They are given a triangle with sides of length 3 and 8, and two angles each 40 degrees (one of which is opposite the side of length 3). The students are to use the Converse of the Isosceles Triangle Theorem to show that the missing side must also be of length 3, and then the Triangle Inequality to show that 3 + 3 must be greater than 8, a contradiction.
When I wrote this problem, I had trouble deciding how difficult I wanted my indirect proof to be. For example, I considered giving 100 as the measure of the angle opposite the side of length 8, and give only one 40-degree angle instead. Then the students would have to use the Triangle Angle-Sum Theorem to find the missing angle as 40 degrees before applying the Isosceles Converse.
Or, to go even further, we can derive a contradiction without making the angle isosceles at all. For example, we could make the angle opposite the 8 side to be, say, 90 degrees instead of 100. Then the missing angle would be 50 instead of 40. If the triangle is drawn so that 50 degrees is opposite the 3 side, then by the Unequal Angles Theorem, the missing side would be less than 3, so the sum of the two legs would still be less than the longest side.
But this might confuse the students even more -- especially if the 90-degree angle is marked with a box (to indicate right angle) rather than "90." A right triangle might lead a student to use the Pythagorean Theorem to find the missing leg. Although this still eventually leads to contradiction -- the missing side would be sqrt(55), which isn't less than 3 -- that irrational side length might still cause some students to drown.
And so I wrote my Question 10 on the review so that it will actually help the students prepare for the corresponding question on the test. I balance out this tough question with some easier questions about logic (converse, inverse, etc.). Hopefully the test won't be too hard for the students.
The first question on today's review worksheet is on Lesson 13-8 -- yesterday's lesson, including both exterior angles and Logo.
Here is a link to a common indirect proof that sqrt(2) is irrational:
http://www.math.utah.edu/~pa/math/q1.html
Neither the U of Chicago nor Glencoe gives the proof outright. But both hint at it -- I just mentioned the U of Chicago's square root proofs. The Glencoe text asks the students to prove that if the square of a number is even, then it is divisible by four. As we can see at the above link, this fact is directly mentioned in the irrationality proof.
I remember once reading the proof of the irrationality of sqrt(2) in my textbook back when I was an Algebra I student. Until then, I had always heard that sqrt(2) was irrational, but I never realized that it was something that could be proved. So I was fascinated by the proof. Naturally, the text only included this as an extra page between the main sections, so it was something that the teacher skipped and most students probably ignored.
The irrationality of sqrt(2) has an interesting history. It goes back to Pythagoras -- he was one of the first mathematicians to use sqrt(2), since his famous Theorem could be used to show that the diagonal of a square has length sqrt(2). The website Cut the Knot, which has many proofs of the Pythagorean Theorem, also contains many proofs of the irrationality of sqrt(2):
http://www.cut-the-knot.org/proofs/sq_root.shtml
Now there is a famous story regarding sqrt(2) and Pythagoras. At the following link, we see that Pythagoras was the leader of a secret society, or Brotherhood:
http://nrich.maths.org/2671
Now Pythagoras and his followers believed that only natural numbers were truly numbers. Not even fractions were considered to be numbers, but simply the ratios of numbers -- numberhood itself was reserved only for the natural numbers. In some ways, this attitude resembles that of algebra students today -- when the solution of an equation is a fraction, they often don't consider it to be a real answer, even though modern mathematics considers fractions to be numbers. (The phrases real number and imaginary number reflect a similar attitude about 2000 years after Pythagoras -- that some numbers aren't really numbers.) So of course, the idea that there were "numbers" that weren't the ratio of natural numbers at all was just unthinkable.
Pythagoras and his followers must have spent years searching for the correct fraction whose square is 2, but to no avail. Finally, one of his followers, Hippasus, discovered the reason that they were having such bad luck finding the correct fraction -- because there is no such fraction! And, as the story goes, Pythagoras was so distraught, afraid that the secret that sqrt(2) was irrational would be revealed, that he ordered to have poor Hippasus drowned at sea!
But as I said, nowadays students simply complain when they have a fractional, or worse irrational, answer to a problem. No one has to drown any more just because of irrational numbers.
Question 10 on my test review, therefore, is actually the final step of that proof, since that's the step where the contradiction occurs. They are given a triangle with sides of length 3 and 8, and two angles each 40 degrees (one of which is opposite the side of length 3). The students are to use the Converse of the Isosceles Triangle Theorem to show that the missing side must also be of length 3, and then the Triangle Inequality to show that 3 + 3 must be greater than 8, a contradiction.
When I wrote this problem, I had trouble deciding how difficult I wanted my indirect proof to be. For example, I considered giving 100 as the measure of the angle opposite the side of length 8, and give only one 40-degree angle instead. Then the students would have to use the Triangle Angle-Sum Theorem to find the missing angle as 40 degrees before applying the Isosceles Converse.
Or, to go even further, we can derive a contradiction without making the angle isosceles at all. For example, we could make the angle opposite the 8 side to be, say, 90 degrees instead of 100. Then the missing angle would be 50 instead of 40. If the triangle is drawn so that 50 degrees is opposite the 3 side, then by the Unequal Angles Theorem, the missing side would be less than 3, so the sum of the two legs would still be less than the longest side.
But this might confuse the students even more -- especially if the 90-degree angle is marked with a box (to indicate right angle) rather than "90." A right triangle might lead a student to use the Pythagorean Theorem to find the missing leg. Although this still eventually leads to contradiction -- the missing side would be sqrt(55), which isn't less than 3 -- that irrational side length might still cause some students to drown.
And so I wrote my Question 10 on the review so that it will actually help the students prepare for the corresponding question on the test. I balance out this tough question with some easier questions about logic (converse, inverse, etc.). Hopefully the test won't be too hard for the students.
The first question on today's review worksheet is on Lesson 13-8 -- yesterday's lesson, including both exterior angles and Logo.
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