CCSS.MATH.CONTENT.K.CC.A.1

Count to 100 by ones and by tens.

CCSS.MATH.CONTENT.1.NBT.A.1

Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.

So now counting to 100 is a kindergarten, rather than first grade standard. For some strange reason, the counting target for first graders is now 120. Considering that the counting target for second graders is 1000, both before and under Common Core, I wonder why 200 wasn't chosen as a target between 100 and 1000, rather than 120.

Here is an article from about a year ago which discusses the change in the counting to 100 standard, from first grade under California standards to kindergarten under Common Core:

http://edsource.org/2014/common-core-standards-bring-dramatic-changes-to-elementary-school-math-2#.VMcUjv7F8VU

There's a long comment thread at the end of this article. Many of the issues discussed on this blog appear in this thread, including traditionalism vs. progressive constructivism (and since the topic is early elementary math, I agree with the traditionalist), the delay by a year of the memorization of basic math facts, and even high school Integrated Math. (The commenter appears mainly concerned with the

*transition*from a traditionalist to an integrated pathway.)

To maintain consistency with the Common Core, kindergarten teachers would now celebrate the 100th day of school, while first grade teachers should wait until the 120th day of school. The only problem with this is that the 120th day of school marks the end of the second trimester -- so it's often the day of a (district-mandated) test, not a party.

Now let's return to geometry. It's now time for the slopes of parallel and perpendicular lines. In Section 3-4 of the U of Chicago text, we see the following theorem:

Parallel Lines and Slopes Theorem:

Two nonvertical lines are parallel if and only if they have the same slope.

Of course, Common Core now expects us to prove this theorem. The text writes:

"A proof requires quite a bit of algebra, and is omitted."

Actually, I don't see that much algebra needed for the proof at all. In fact, the proof of the theorem is very similar to yesterday's proof of the well-definition of slope. This time, instead of the four points (

*x*_1,

*y*_1), (

*x*_2,

*y*_2), (

*x*_3,

*y*_3), and (

*x*_4,

*y*_4) all being on the same line, we place the first two points on the first of two parallel lines and the other two on the other line.

But now we're given the two

*parallel lines*, not the points so

*we*are allowed to choose any points on the line that we wish. Since we want to set up corresponding angles, we begin by selecting a line to be the transversal and then let (

*x*_1,

*y*_1) and (

*x*_3,

*y*_3) both lie on this transversal. Any line may be chosen as the transversal as long as it intersects both given parallel lines.

Well, since we're given that the two parallel lines aren't vertical (as vertical lines have no slope), we can let any vertical line be the transversal. In fact, why don't we choose the easiest vertical line to work with -- the

*y*-axis? These means that our two points (

*x*_1,

*y*_1) and (

*x*_3,

*y*_3) are now the two points (

*x*_1, 0) and (

*x*_3, 0). Notice that these two chosen points are where the two lines intersect the

*y*-axis -- that is, they are the

*y*-intercepts. So let's change those

*x*'s to

*b*'s -- and so the two points in our proof will be labeled (

*b*_1, 0) on the first line and (

*b*_2, 0) on the second line.

Notice that by letting the transversal be the

*y*-axis and two of our points have 0 as their second coordinate, we're demonstrating the hallmarks of a good coordinate proof. We want to place as many key lines and segments along the coordinate axes as possible.

Now let's consider our second points on each line. Since our first points have 0 as the

*y*-coordinate, let's let the second points have 1 as the

*y*-coordinate. So these points are (

*x*_1, 1) and (

*x*_2, 1).

Now just as we did for the Distance Formula and Slope Well-Definition proofs, we add the points

*E*(

*x*_1, 0) and

*F*(

*x*_2, 0) to form two right triangles. Notice that these two right triangles aren't merely similar, as they were in the Slope Well-Definition proof -- they are in fact

*congruent*.

Given:

*k*| |

*l*,

*A*(

*b*_1, 0),

*B*(

*x*_1, 1) on

*k*,

*C*(

*b*_2, 0),

*D*(

*x*_2, 1) on

*l*

Prove: slope of

*k*= slope of

*l*

*Statements Reasons*

1.

*k*| |

*l*, etc. 1. Given

2. Angle

*BAE*=

*DCF*2. Corresponding Angles Consequence

3. Angle

*AEB*=

*CFD*3. All right angles are congruent.

4.

*BE*=

*DF*4. All segments of unit length are congruent.

5. Tri.

*AEB*,

*CFD*cong. 5. AAS Congruence Theorem

6.

*AE*=

*CF*6. CPCTC

7. slope

*k*=

*AE*/1 =

*AE*, 7. Definition of slope

slope

*l*=

*CF*/1 =

*CF*

8. slope

*k*= slope

*l*8. Substitution

We see that there are several cop-outs in the above proof to avoid excessive formalism. Technically speaking, "all segments of unit length (that is, of length 1) are congruent" is not a valid reason in a formal proof. We actually need four steps here -- the first to prove that

*BE*= 1, the second to prove that

*DF*= 1, the third to prove that

*BE*=

*DF*, and final to prove that these equal

*lengths*imply that the segments are

*congruent*. Arguably, we'd need to do the same with the right angles. Also, several steps are skipped in finding the slopes of both lines -- it may be instructive to calculate the slope by plugging the values into the formula to show that the slope of

*k*is

*x*_1 -

*b*_1, while the slope of

*l*is

*x*_2 -

*b*_2, and these are the values proved equal in the CPCTC step. (Notice that the theorem to be proved is stated as "if and only if," so we need to prove the converse -- that is, if the slopes are equal, then the lines are parallel. A proof of the converse requires SAS rather than AAS. For the converse, our given lines have the same slope -- so we can choose the familiar letter

*m*to denote that slope!)

But there is one more subtlety that we need to discuss. We assumed that

*AE*=

*x*_1 -

*b*_1 in the proof above, but actually by the definition of 1D distance,

*AE*= |

*x*_1 -

*b*_1|, not

*x*_1 -

*b*_1. That is, we've ignored the

*sign*of the slope in the above proof! Without this, it could be the fact that parallel lines have slopes with opposite sign. Even the proof of the well-definition of slope from yesterday isn't really complete since there was no mention of any sign.

What we actually need here is that extra postulate that Dr. Wu gave us -- Plane Separation. Our lines have constant slope because we showed that, if we start at a point on the line and can move

*v*units up and

*w*units right to arrive back on the line, then the same is true no matter where we start. There is no point where we can move

*w*units right, but we have to go

*v*units

*down*instead of up in order to get back on the line.

The reason is that set of all points

*w*units right of the line is a

*translation*image of the line, and as a translation image, it is

*parallel*to its pre-image. So it can't cross from one half-plane defined by the original line to the other. In particular, it can't cross from the half-plane containing the line

*v*units above the starting line to the half-plane containing the line

*v*units below the starting line. And so it's the Plane Separation Postulate that ultimately guarantees that slope is well-defined and that parallel lines have equal slope.

As Wu often points out, this is a subtle point that is ignored in most math classes. The problem, of course, is that we're about to jump to the case where the sign of the slope makes a huge difference -- the slope of

*perpendicular*lines.

Now here's a mini-activity that I like to begin with to illustrate the slopes of parallel and perpendicular lines. I choose two students and ask them to draw a line that is parallel to my given line, which has a predetermined slope of, let's say, 2/3. I inspect the two lines and declare the student whose slope more clearly represents going up 2 and right 3 units to be the winner.

Then I ask the students to draw a line that is

*perpendicular*to my given line. This will be harder for someone who hasn't learned the formula yet, and so it's likely that

*both*lines will be far away from a truly perpendicular line. And so I come in and draw my

*own*perpendicular line. To do this, I point out that since

*perpendicular*means 90 degrees, I rotate my paper 90 degrees and then redraw the original line on the rotated paper. I make sure that my new line also has rise of 2 and run of 3. Then I rotate the paper -90 degrees to retain the original position, and notice that the rise of 2 rotates to a run of 2, and the run of 3 rotates to a rise of -3. So the new slope is -3/2 -- the opposite reciprocal of the original slope. And then I declare

*myself*to be the winner!

I don't include a formal proof of the Perpendicular Lines and Slopes Theorem on the worksheet, but we can figure it out -- instead of a translation, we use a rotation. We rotate the given line

*AB*90 degrees about some point (say the origin) to obtain a new line

*A'B'*, and we see that its slope is the opposite reciprocal of the original slope, since rise and run become run and negative rise. Then, by the Two Perpendiculars Theorem, any other line

*CD*that is perpendicular to

*AB*must be parallel to

*A'B'*, and so has that same slope as

*A'B'*(which is the opposite reciprocal of that of

*AB*). It may be possible to skip a step here and simply rotate about the point where the two given lines

*AB*and

*CD*intersect -- then the image of line

*AB*would be line

*CD*itself (but

*A*doesn't necessarily rotate into either

*C*or

*D).*

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