The concept of integrated math is a highly contentious issue. Technically speaking, integrated math is not directly related to Common Core. For years, schools could choose an integrated pathway, where high school students learn a little algebra and geometry every year, or a more traditional pathway, with a first year of algebra, one year of geometry, a second year of algebra, and then finally one year of pre-calculus mathematics.
Now the Common Core, in its high school mathematics standards, did not want to prescribe either an integrated or a traditional pathway. So the Common Core divides its high school standards into courses along both pathways. Nevertheless, many districts and even entire states (including Utah) use Common Core as a reason to switch to integrated math. My state, California, does not prescribe integrated math, but many districts chose an integrated pathway anyway. As this state is a Smarter Balanced state, it doesn't matter whether districts choose traditional or integrated math because students test in high school only once -- at the end of junior year, by when they would have finished all three years' worth of standards anyway. PARCC states, I believe, will have separate end-of-course exams for both pathways.
Because of this, many people use integrated math as a reason to oppose Common Core. Since Utah is an integrated math state, I found the following argument mentioned on a Utah anti-Core website:
http://www.utahnsagainstcommoncore.com/what-is-the-problem-with-common-core-math-in-utah-what-is-the-solution/
“The Common Core standards claim to be ‘benchmarked against the international standards’ but this phrase is meaningless. They are actually two or more years behind international expectations by eighth grade, and only fall further behind as they talk about grades 8-12. Indeed, they don’t even fully cover the material in a solid geometry course, or in the second year algebra course.”
“While the difference between these standards and those of the top states at the end of eighth grade is perhaps somewhat more than one year, the difference is more like two years when compared to the expectations of the high-achieving countries — particularly most of the nations of East Asia.”
So the argument here, given by Dr. Jim Milgram of Stanford, is that Common Core math -- especially the integrated math of Utah -- is far inferior to the standards of East Asian nations. (I notice that the name Dr. Hung-Hsi Wu appears on this page as well, even though Wu wrote essays on how to implement Common Core Standards.)
But here's the irony -- most nations other than the United States have an integrated pathway. Let's look at a nation whose standards are often held in high regard, Singapore:
http://www.moe.gov.sg/education/syllabuses/sciences/files/maths-secondary.pdf
We look at the standards for Secondary 3-4 (corresponds roughly with 9th-10th grade US), and we see sections on Numbers and Algebra, Geometry and Measurement, and Statistics and Probability, all integrated together. There are no separate courses called Algebra I, Geometry, or Algebra II.
And this is how it goes in most nations. The United States is an outlier -- we are the only major country that separates the courses. In short, by implementing integrated math, our schools are becoming MORE like those in East Asia and other nations, not LESS.
And so, though my opinion on Common Core is mixed, my opinion on integrated math isn't. I have absolutely nothing against Integrated Math. In particular, I do not see integrated math as a valid reason to oppose Common Core.
I sub at schools that use the integrated pathway. But I choose to focus this blog on geometry, since this is where Common Core math differs from its pre-Common Core variant the most. Because of this, I write this blog from the perspective of a geometry class on the traditional pathway.
Section 6-3 of the U of Chicago text is on rotations -- and we already covered this section early when I wanted to use rotations to develop the Parallel Consequences. So instead, I use today's lesson to give the students more practice with these transformations.
I begin with pairs of figures -- for simplicity, I just used letters. Students are to identify whether the second is the image of the first under a reflection, rotation, or translation.
But for the reflections, students should draw in the reflecting line. To find this mirror is easy -- just find two corresponding points and find their perpendicular bisector.
For rotations and translations, students can draw in the two reflecting lines such that the composite of the two reflections is that rotation or translation. This is tricky, but the key is to look at the Two Reflection Theorems for Rotations and Translations.
Notice that these theorems are similar. In both cases, the magnitude of the rotation or translation is exactly two times something -- either the angle at which the mirrors intersect for the former, and the distance between the mirrors for the latter. Inspecting the proofs, we see the reason for this is that the expression x + x + y + y appears in both proofs.
Also, for both theorems, the pair of mirrors whose composite is a given rotation or translation is not unique in either case. For rotations, all that matters is that the first mirror passes through the center of the rotation. So the first mirror can be any line through that point. Then the second line is chosen so that the angle between the mirrors is half the magnitude of the rotation.
Translations are even more counter-intuitive regarding the two parallel mirrors. The only requirement for the first mirror is that it be perpendicular to the direction (vector) of the translation. As the U of Chicago text writes:
"A surprising thing about the solution to the Example is that line m could be anywhere in the plane so long as it is perpendicular to line AA". It does not have to be close to the preimage."
(emphasis mine)
That is, for a horizontal translation of right two units, the first mirror could be any vertical line, and then the second line would be one unit to the right of it. The two mirrors could be x = 0 and x = 1, or x = 100 and x = 101, or even x = 1000000 and x = 1000001. The first reflection may carry the image very, very far away from the preimage, yet the second reflection will bring the image back right next to the original preimage -- two units to the right, to be exact!
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