*Mathematics and the Physical World*has the title of "Numbers, Known and Unknown." In this chapter, Kline writes all about unknown numbers -- that is, algebra.

"Algebra begins with the known and ends with the unknowable." -- Anonymous

And of course, the author of quote about the unknown is, well, unknown! But Kline begins his chapter writing about a mathematician whose very well-known indeed:

"When Karl Friedrich Gauss, one of the greatest mathematicians of all time, was in elementary school, his teacher, harried by endless clerical work, assigned to the class the problem of finding the sum of all the numbers from 1 to 100."

We've talked about Gauss before on the blog, but not in the context of this problem. But then again, the story that Kline tells here is so famous that many Algebra II teachers still tell it today in their classrooms today. In short, Gauss was able to add the numbers from 1 to 100 quickly, not using arithmetic, but algebra. The sum of the naturals from 1 to

*n*is

*n*(

*n*+ 1)/2, and this formula is given to students to show how useful algebra is.

If "math" is a subject that many students hate, then "algebra" is the reason that they hate it. Kline talks a little about the origin of the word

*algebra*:

"The historical associations of the word

*algebra*almost substantiate the sordid character of the subject. The word comes from the title of a book written by the ninth-century Arabian mathematician Al-Khowarizmi.... Figuratively,

*al-jebr*meant restoring the balance of an equation in transposing terms.... [In Moorish Spain], in the form

*algebrista*, it came to mean a bonesetter, or a restorer of broken bones.... Thus it might be said that there is a good historical basis for the fact that the word

*algebra*stirs up disagreeable thoughts."

So we have this word "algebra," a foreign-sounding Arabic word, used to describe this mixture of numbers and letters that might as well be a foreign language which, as Kline writes, "many people complain about the need to learn...."

J.K. Rowling, the author of the

*Harry Potter*series, writes that "the fear of a name increases fear of the thing itself." This is why, as I wrote a month and a half ago, I'm not sure whether I want to use the name "Algebra III" for the governor's proposed senior math course in California. In fact, I once read about a college algebra course where the professor was able to motivate the students to learn algebra simply by avoiding the word "algebra" until just before the final!

I also mentioned the following back when our state still had a high school exit exam, the CAHSEE:

-- The CAHSEE tested up to eighth grade math.

-- The CAHSEE tested up to Algebra I.

The first sentence sounds as if the CAHSEE requirements were too

*low*-- why should students have to master only

*eighth*grade math in order to graduate from the

*twelfth*grade? And indeed, this argument was used by those arguing that the CAHSEE should have been more challenging. On the other hand, the second sentence sounds as if the CAHSEE requirements were too

*high*-- why should students have to master a subject that hardly any adults use in order to graduate? And indeed, this argument was used by those arguing that the CAHSEE should have been easier.

But at the time the CAHSEE existed, our state standards recommended eighth grade Algebra I. And so the two statements above were actually

*equivalent*! Yet it was easy to sway opinion from "CAHSEE is too easy" to "CAHSEE is too hard" simply by using a seven-letter Arabic-derived word.

Let's get back to Kline. He writes about how the ancient Greek mathematician Archimedes used algebra to prove that King Hiero's gold crown was a fake. I wrote about Archimedes earlier this month regarding his discovery of the sphere volume formula -- that day, I posted a YouTube video containing a

*Square One TV*song about Archimedes. One line from the song mentions "the king's gold crown."

Kline also shows how algebra is used to solve the Isoperimetric Problem. That day, I stated that among all rectangles with a given perimeter, the square has the greatest area -- but I didn't provide a proof of that statement. Well, Kline shows us how to prove this. Here is his proof: let

*x*be the side of a rectangle whose perimeter is 100, but isn't a square, so

*x*is not 25. Then Kline writes:

(25 -

*x*)^2 > 0

625 - 50

*x*+

*x*^2 > 0

625 -

*x*(50 -

*x*) > 0

That is, the area of the square of perimeter 100 (that is, 625) minus the area of the rectangle with side

*x*and perimeter 100 is greater than 0 -- so the square is larger. Afterwards, Kline proves that the area of a circle with circumference is still greater than this -- hinting at the full Isoperimetric Theorem.

Since "algebra" is an Arabic word, and here in California seventh graders first learn about that part of the world, it makes since to introduce seventh graders to algebra.

Question 5 of the PARCC Practice Test is on reflections and rotations in the coordinate plane:

5. Triangle

*ABC*is graphed in the

*xy*-coordinate plane, as shown.

(Here are the coordinates of

*ABC*:

*A*(3, 2),

*B*(4, 6),

*C*(6, 4).)

Part A

Triangle

*ABC*is reflected across the

*x*-axis to form triangle

*A'B'C'*. What are the coordinates of

*C'*after the reflection?

A. (-6, 4)

B. (3, -2)

C. (4, -6)

D. (6, -4)

Part B

Triangle

*ABC*in the

*xy*-coordinate plane will be rotated 90 degrees counterclockwise about point

*A*to form triangle

*A"B"C"*. Which graph represents

*A"B"C"*?

A. (

*A"*(3, 2),

*B"*(7, 1),

*C"*(5, -1))

B. (

*A"*(-4, 2),

*B"*(-5, 6),

*C"*(-7, 4))

C. (

*A"*(3, 2),

*B"*(-1, 3),

*C"*(1, 5))

D. (

*A"*(-3, 4),

*B"*(-2, 0),

*C"*(0, 2))

Part A should be straightforward. To reflect over the

*x*-axis, we switch the sign of

*y*-- that is, the point (

*x*,

*y*) reflected over the

*x*-axis is (

*x*, -

*y*). Since the preimage

*C*has coordinates (6, 4), the image

*C'*has coordinates (6, -4), which is (D). We can see what the common student errors for Part A will be just by looking at the wrong choices. Choice (A) occurs if students switch the sign of

*x*instead of

*y*, choice (B) occurs if the students find

*A'*instead of

*C'*, and choice (C) occurs if the students find

*B'*instead of

*C'*.

Part B is trickier, because the center of the rotation is not the origin. One way to eliminate choices is to note that since

*A*is the center of rotation, the image of

*A*must be

*A*itself, so that enables us to eliminate some choices right off the bat. Of the remaining choices, we see that (A) appears to be rotated clockwise while only (C) is rotated counterclockwise, so choice (C) is correct.

The U of Chicago text covers reflections in Chapter 4 and rotations in Lesson 6-3. But we do know of one deficiency with the way transformations are covered in the text -- coordinates are emphasized on the PARCC, but not in the text. I consider this to be a wasted opportunity by the authors of the U of Chicago text -- at least the simplest reflections (those where the mirror is either axis) could be completed on the coordinate plane.

Rotations centered at points other than the origin are usually not taught in any text -- that is, we aren't given a simple formula as (

*x*,

*y*) -> (

*x*, -

*y*) is for

*x*-axis reflections. It is possible to use algebra -- going back to the topic of today's Kline chapter -- to find the answer. We first notice that the rotation 90 degrees counterclockwise, but centered at the origin, is:

(

*x*,

*y*) -> (-

*y*,

*x*)

Then the same rotation, only centered at (

*a*,

*b*), has the form:

(

*a*+

*x*,

*b*+

*y*) -> (

*a*-

*y*,

*b*+

*x*)

Replacing

*x*with

*x*-

*a*and

*y*with

*y*-

*b*gives us:

(

*x*,

*y*) -> (

*a*+

*b*-

*y*,

*b*-

*a*+

*x*)

But this is error-prone and difficult to remember. It also depends on the evident, yet unproved, assertion that a rotation centered at the point (

*a*,

*b*) is the composite of a translation mapping (

*a*,

*b*) to the origin, followed by the same rotation centered at the origin, followed by a translation mapping the origin back to (

*a*,

*b*).

Instead, I once noticed that on the PARCC, in every problem where we are asked to rotate a triangle, the center is either the origin or one of the vertices of the triangle. This helps us greatly, since the rotation always maps its center to itself. To rotate the other two vertices, it's probably better just to look at the vectors from the center to each of the other two vertices (which are sides of the triangle we're trying to rotate) and just rotate them by counting squares in the new direction (just as we do to find the slope of perpendicular lines). This is assuming, of course, that the correct image triangle isn't evident from looking at the graphs (for example, in a problem where graphs aren't provided).

**PARCC Practice EOY Question 5**

**U of Chicago Correspondence: Lesson 6-3, Rotations**

**Key Theorem: Formulas for Reflection and Rotation**

**The reflection over the**

*x*-axis maps (*x*,*y*) to (*x*, -*y*), and the rotation 90 degrees clockwise around the origin maps (*x*,*y*) to (-*y*,*x*).

**Common Core Standard:**

CCSS.MATH.CONTENT.HSG.CO.A.2

Represent transformations in the plane using, e.g., transparencies and geometry software; describe transformations as functions that take points in the plane as inputs and give other points as outputs.

**Commentary: The U of Chicago text covers transformations, but doesn't emphasize the coordinate plane enough. Today's worksheet provides the students with some much-needed practice with transformations on the coordinate plane.**

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