Niels Henrik Abel was a Norwegian mathematician born at the turn of the nineteenth century. As the title of the chapter implies, Abel was poor his entire life. According to Livio, Abel was the child of two parents who were often in trouble with the law. The family was so poor that they had to eat horse meat. But despite this difficult family life, the pastor was able to homeschool Abel until he became a teenager, at which time he was sent to the Cathedral School in Norway's capital.
As a teacher myself, I am intrigued by the way Livio contrasts two of Abel's math teachers at the Cathedral School. His first teacher, Hans Peter Bader, was a terrible teacher. Corporal punishment was common back then, and Bader frequently applied it to his students. In fact, Bader was fired after being accused of beating one of his students to death. Abel was deeply discouraged and his grades dropped dramatically.
Abel's other math teacher, after the firing of Bader, was Bernt Michael Holmboe. Holmboe was certainly a much better teacher than his predecessor. Abel excelled under his new teacher, who encouraged the youngster to go beyond the standard curriculum and study the great mathematicians like Euler, Newton, and Gauss. Holmboe wrote of his student, "With the most incredible genius he unites an insatiable interest in and ardor for mathematics, such that quite probably, if he lives, he will become one of the great mathematicians."
Of these two, Bader and Holmboe, which one do I want to emulate more? This is a no-brainer -- of course I wish to be more like Holmboe. But sometimes, I feel that my teaching style is more like Bader's than Holmboe's. Of course, I don't mean that I would implement corporal punishment, but sometimes I wonder whether I'm verbally attacking students if they say the wrong answer. I want to encourage my weaker students to study math -- not discourage the stronger students from learning, as Bader unwittingly did to Abel.
Livio writes that Abel was "whizzing through the standard curriculum," which is why Holmboe encouraged Abel to go above and beyond that curriculum. But many people believe that this would be impossible today, especially if "standard curriculum" means "Common Core." This is not an easy question to answer. The ideal answer would be that the test should be one that contains items that a prodigy like Abel could study on his own and omit those that require special Common Core training to answer. In the lower grades, it's obvious what should be included (i.e., traditionalist arithmetic) but in higher grades it's not as clear-cut. I also favor changing the test in such a way that rewards above grade-level students. But still, even with Common Core and all that it entails, I owe it to my above grade-level students to make them feel that they are learning in my class, and to inspire the other students to work harder to they can become more like the top students.
We can see many similarities between Abel's world in 1815 -- the year he entered Cathedral School -- and the world of many of our poorer students two centuries later. Abel's parents both lived unsavory lives -- just like many of our students' parents -- and he was often depressed whenever he was separated from his friends -- again, just like kids today. Yet Abel never used his poverty as an excuse not to study -- and neither should our students. Abel's story is one that I would like to tell my students someday in my own classroom.
Abel thought that he had discovered the Quintic Formula. It was a few years later, in 1823. when Abel proved the opposite -- that there is no Quintic Formula. Livio writes that Abel's proof is far too technical to include in his book. But he points out the basic idea -- Abel began by assuming that there is a Quintic Formula and shows that this assumption leads to a contradiction. In other words, Abel's proof is an indirect proof -- the sort of proof that we learn about in Section 13-4 of the U of Chicago math text. Livio himself uses the Latin phrase reductio ad absurdum -- "reduction to the absurd" -- which is formal name for indirect proof.
When one thinks about it, it makes perfect sense for Abel's proof to be indirect. How in math do we prove that something is impossible? The only real way is to assume that it's possible and show that this leads to a contradiction. Many of our proofs that lines are parallel were indirect, because we showed that it's impossible for the lines to intersect. Abel's proof is now often referred to as Abel's Impossibility Theorem. It's also often called the Abel-Ruffini Theorem -- after Paolo Ruffini, the mathematician who first mentioned the possibility that a Quintic Formula is impossible.
Compared to Abel's Impossibility Theorem, today's question is very simple. Question 13 on the PARCC Practice Test is on dilations:
The figure shows line AC and line PQ intersecting at point B. Lines A'C' and P'Q' will be the images of lines AC and PQ, respectively, under a dilation with center P and scale factor 2.
Which statement about the image of lines AC and PQ would be true under the dilation?
(A) Line A'C' will be parallel to line AC, and line P'Q' will be parallel to PQ.
(B) Line A'C' will be parallel to line AC, and line P'Q' will be the same line as PQ.
(C) Line A'C' will be perpendicular to line AC, and line P'Q' will be parallel to PQ.
(D) Line A'C' will be perpendicular to line AC, and line P'Q' will be the same line as PQ.
The necessary theorem is repeated several times through the first three sections of Chapter 12. A line and its image under a size transformation (dilation) are parallel. So we can already eliminate the two choices that mention "perpendicular." As it turns out, because the center of the dilation is P, that point P is a fixed point of the dilation, and any line passing through P must be an invariant line of the dilation -- that is, its image under the dilation is the line itself. Therefore the line P'Q' will be the same line as PQ, and so the correct answer is (B).
The idea that a line is parallel to its dilation image is emphasized in the Common Core Standards. It appears in the U of Chicago and a few questions in the first three sections of Chapter 12 mention it, but it is not as strongly highlighted as in the Common Core. In particular, the idea that a line passing through the center of the dilation is invariant doesn't appear in the U of Chicago text at all.
Some readers might point out that the U of Chicago text is therefore wrong. The theorem states that a line and its dilation and its image are parallel. Nowhere in the theorem does it state that the line does not pass through the center of the dilation -- so we can't rule out the possibility that the line passes through the center. Since such a line is mapped to itself while the theorem claims that the line is mapped to a parallel line, we've found a contradiction (there's that indirect proof again!) that proves that the theorem is incorrect.
As it turns out, this is not really a contradiction, because in the U of Chicago text, a line is considered to be parallel to itself. So a line passing through the center of the dilation actually is mapped to a parallel line -- namely the line itself.
Not everyone, of course, accepts this unusual definition of "parallel." In some ways, this is actually an inclusive definition of "parallel" in the same way that considering a parallelogram to be a trapezoid entails an inclusive definition of "trapezoid." From one perspective, it makes sense to define a line to be parallel to itself -- after all, parallel lines have the same slope, and a line clearly has the same slope as itself, so therefore a line is parallel to itself. Notice that the proof of this theorem that a line is parallel to its dilation image is a coordinate proof that merely shows that the line and its image have the same slope.
Indeed, we can prove the special case that a line passing through the center of the dilation is invariant as follows: we already know that the line and its image are parallel by the above theorem. Since the center of the dilation is a fixed point, the image must also contain the center. If two parallel lines have a point in common, then they must be identical. So the line and its image -- two parallel lines having the center in common -- must be identical. Therefore the line is an invariant line. QED
On the other hand, calling the U of Chicago's definition of parallel an "inclusive definition" seems a bit strange. Having two pairs of parallel lines is a special case of having one pair of parallel lines, so it makes sense to consider the parallelogram a trapezoid. But in what world can having every point in common be a special case of having no points in common?
If we use the so-called "inclusive" definition of parallel, then choice (B) is correct, but choice (A) remains correct as line PQ is parallel to itself. The correctness of choice (B) doesn't automatically eliminate choice (A), as we've seen PARCC questions with more than one correct answer. But the intended answer is for choice (B) to be correct, not (A) (otherwise, we would have seen disclaimers such as "select all that apply").
So the U of Chicago definition fails to distinguish between a correct answer (B) and an incorrect answer (A) on a very specific PARCC problem. Therefore, any course that claims to be a Common Core Geometry course must reject the U of Chicago definition. I will be rewriting any worksheet that uses the U of Chicago definition of parallel in time for next year.
But now it remains -- how do we prove that a line that doesn't pass through the center of the dilation is parallel to its image using the most common definition of parallel?
In Algebra I classes, teachers often avoid merely saying "Parallel lines have the same slope." Instead, they say that parallel lines have the same slope and different y-intercepts. We can use this to convert the proof from Section 12-1 of the U of Chicago into one that uses the common definition of parallel.
Let S_k be a dilation centered at the origin with scale factor k, and (0, b) be the y-intercept of a line not passing through the center. Since this line doesn't pass through the origin, b can't be 0. So the image of the point (0, b) must be (0, kb), which is a different point. So the lines do not have the same y-intercept, but the U of Chicago text already proves that they have the slope. The line and its image have the same slope and different y-intercepts, therefore they are parallel. QED
There are a few things to point out about this proof. First of all, we assumed that k can't be 1 -- otherwise b and kb would be equal after all. But a dilation with scale factor 1 is just the identity -- and we could argue that every point is the center of a dilation with scale factor 1. So there's no such thing as a line that doesn't pass through the center.
Another point to make is that Section 12-1 assumes that the center of the dilation is the origin. Does this proof still work if the center is the origin? Recall that this is a coordinate proof. In coordinate proofs we typically place the origin and axes at convenient locations. Well, the most convenient location for the origin is the center of the rotation. We do the same with coordinate proofs involving, say, quadrilaterals. Let's place the coordinates of a rectangle at (0, 0), (a, 0), (0, b), and (a, b) -- that is, we placed the origin at one of the vertices and the axes along two of the sides. Then we can prove that both diagonals have length sqrt(a^2 + b^2) and thus are congruent. This coordinate proof doesn't work only for rectangles with a vertex at the origin -- it works for all rectangles. And so it is with our dilation proof.
The real problem, though, is that we can't use a coordinate proof to properties of dilations because in Common Core, we use dilations to prove the properties of coordinates! We discussed this problem all throughout our study of Chapter 12 back in January. Let's briefly recall that proof -- which ultimately goes back to Dr. Hung-Hsi Wu.
We actually begin with the fact that a line passing through the origin is invariant. This follows basically from the definition of dilation -- if O is the center, then any point A is mapped to a particular point on ray OA. Then the image of any point on line OA is on ray OA -- which is part of line OA. So any point on a line passing through O is mapped to point on the same line. The image of line OA is the line itself.
Now let's show that a line not passing through O must be parallel to its image. Our earlier proof from January was a direct proof, but we can do it more elegantly with an indirect proof. Assume that the line and its image are not parallel -- that is, that they have a point in common. That is, there exists a point A on the line such that its image A' is also on the line (that's where the line and its image intersect). So the original line can now be written as line AA'. Notice that O is also on line AA' -- once again, it follows from first case above that the points O, A, and A' are collinear. But this contradicts the assumption that O is not on the original line. Since the assumption that the line and its image intersect leads to a contradiction, we conclude that the line and its image are parallel. QED
Speaking of indirect proofs, what about Abel's indirect proof? Livio writes down that Abel was finally rewarded with a professorship in Berlin -- a few days after he had died of tuberculosis. He was only 26 years old when he died. Today there is a major prize -- the Abel prize -- that is offered every year in Abel's home country of Norway to a top mathematician, similar to the Nobel Peace Prize (and is sometimes considered to be the analog of a Nobel Prize in mathematics). Just as Holmboe had written just a few years earlier, who knows how great a mathematician Abel would have been if he had lived.
In Algebra I classes, teachers often avoid merely saying "Parallel lines have the same slope." Instead, they say that parallel lines have the same slope and different y-intercepts. We can use this to convert the proof from Section 12-1 of the U of Chicago into one that uses the common definition of parallel.
Let S_k be a dilation centered at the origin with scale factor k, and (0, b) be the y-intercept of a line not passing through the center. Since this line doesn't pass through the origin, b can't be 0. So the image of the point (0, b) must be (0, kb), which is a different point. So the lines do not have the same y-intercept, but the U of Chicago text already proves that they have the slope. The line and its image have the same slope and different y-intercepts, therefore they are parallel. QED
There are a few things to point out about this proof. First of all, we assumed that k can't be 1 -- otherwise b and kb would be equal after all. But a dilation with scale factor 1 is just the identity -- and we could argue that every point is the center of a dilation with scale factor 1. So there's no such thing as a line that doesn't pass through the center.
Another point to make is that Section 12-1 assumes that the center of the dilation is the origin. Does this proof still work if the center is the origin? Recall that this is a coordinate proof. In coordinate proofs we typically place the origin and axes at convenient locations. Well, the most convenient location for the origin is the center of the rotation. We do the same with coordinate proofs involving, say, quadrilaterals. Let's place the coordinates of a rectangle at (0, 0), (a, 0), (0, b), and (a, b) -- that is, we placed the origin at one of the vertices and the axes along two of the sides. Then we can prove that both diagonals have length sqrt(a^2 + b^2) and thus are congruent. This coordinate proof doesn't work only for rectangles with a vertex at the origin -- it works for all rectangles. And so it is with our dilation proof.
The real problem, though, is that we can't use a coordinate proof to properties of dilations because in Common Core, we use dilations to prove the properties of coordinates! We discussed this problem all throughout our study of Chapter 12 back in January. Let's briefly recall that proof -- which ultimately goes back to Dr. Hung-Hsi Wu.
We actually begin with the fact that a line passing through the origin is invariant. This follows basically from the definition of dilation -- if O is the center, then any point A is mapped to a particular point on ray OA. Then the image of any point on line OA is on ray OA -- which is part of line OA. So any point on a line passing through O is mapped to point on the same line. The image of line OA is the line itself.
Now let's show that a line not passing through O must be parallel to its image. Our earlier proof from January was a direct proof, but we can do it more elegantly with an indirect proof. Assume that the line and its image are not parallel -- that is, that they have a point in common. That is, there exists a point A on the line such that its image A' is also on the line (that's where the line and its image intersect). So the original line can now be written as line AA'. Notice that O is also on line AA' -- once again, it follows from first case above that the points O, A, and A' are collinear. But this contradicts the assumption that O is not on the original line. Since the assumption that the line and its image intersect leads to a contradiction, we conclude that the line and its image are parallel. QED
Speaking of indirect proofs, what about Abel's indirect proof? Livio writes down that Abel was finally rewarded with a professorship in Berlin -- a few days after he had died of tuberculosis. He was only 26 years old when he died. Today there is a major prize -- the Abel prize -- that is offered every year in Abel's home country of Norway to a top mathematician, similar to the Nobel Peace Prize (and is sometimes considered to be the analog of a Nobel Prize in mathematics). Just as Holmboe had written just a few years earlier, who knows how great a mathematician Abel would have been if he had lived.
PARCC Practice EOY Exam Question 13
U of Chicago Correspondence: Section 12-2, Size Changes Without Coordinates
Key Theorem: Size Change Theorem
Under a size change:
(d) lines and their images are parallel.
Under a size change:
(d) lines and their images are parallel.
Common Core Standard:
Commentary: There are a few questions about lines being parallel to their images throughout the first three sections of Chapter 12 and in the SPUR section under Properties. The idea that a line passing through the center is unchanged doesn't appear in the U of Chicago at all.
CCSS.MATH.CONTENT.HSG.SRT.A.1.A
A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged.
A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged.
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