Category: Math Posts

Ellipses

Olly / / Math Posts

Today I did more exercises from the review of conics. Most of them concerned analyzing and graphing ellipses. It was still second year algebra all over again, but I was a little less bored than by the exercises with parabolas yesterday. I think I may have been bucked up by this intriguing 3Blue1Brown video about ellipses that happened to come up on my YouTube homepage last night:

I also did a little extracurricular reading today about the Goldbach conjecture and the related ternary Goldbach conjecture. The latter was recently proven by a mathematician called Harald Helfgott. I have not yet absorbed even the overall method of the proof, but I did understand that the abstract proof only applies to numbers greater than a constant $C$ that is very large in human terms yet small enough that all numbers less than $C$ can be checked by a computer. I thought that was interesting.

A Little Frustration

Olly / / Math Posts

Today I read the review of conics and did the first section of exercises. Conics was my least favorite of the topics covered by my math education. Although I’m sure we learned other things, my memory of second year algebra is of endless repetitive graphing of parabolas, ellipses, and hyperbolas. The exercises for this section proved a bit like that, unfortunately. Find vertex, focus, directrix and graph…find vertex, focus, directrix and graph…

Furthermore, near the end of my study time I found that I had been doing the exercises correctly and producing accurate graphs, but consistently transposing $x$ and $y$ when writing down the coordinates of points. That’s something I recall from the past, as well.

Revenge of the Squares, Part 2

Olly / / Math Posts

Today I finished the 46 exercises from the review of analytic geometry, then rounded out my study time with more exercises from Analysis with an Introduction to Proof.

I had expected the proof from Revenge of the Squares, Part 1 to require showing that the figure must be a parallelogram. I managed it without that, but afterward I started thinking about how I would have proved that fact, had I needed it. The results are below.

Lemma: If, when two lines are intersected by another line, a pair of alternate interior angels is equal, then the first two lines are parallel.

Assume that two lines, $A$ and $B$, are not parallel.

Let them both be intersected by a third line, $C$. Since $A$ and $B$ are not parallel, they intersect, and $A$, $B$, and $C$ enclose a triangle.

Three Lines

Let $x$, $y$, and $z$ be the interior angles of the triangle, as shown in the diagram. Let the interior angle alternate to $x$ be called $k$ and the interior angle alternate to $y$ be called $j$.

Since the sum of the measures of the interior angles of a triangle is 180 degrees, $x+y+z=180$. Thus, $x=180-y-z$.

Furthermore, since $x$ and $j$ are complimentary supplementary, $x+j=180$, meaning that $x=180-j$.

Therefore, $$180-j=180-y-z$$ $$-j=-y-z$$ $$j=y+z$$

Since $z\neq 0$, it follows that $j\neq y$.

By similar reasoning, $k\neq x$.

Hence, if two lines are not parallel, then, when they are both intersected by a third line, neither pair of alternate interior angles is equal.

By contrapostition, it follows that if, when two lines are intersected by another line, a pair of alternate interior angels is equal, then the first two lines are parallel.

Proposition: A convex quadrilateral with each pair of opposite sides equal is a parallelogram (i.e. each pair of opposite sides is also parallel).

Consider a convex quadrilateral $ABCD$, where $AB=CD$ and $BC=AD$.

Parallelogram Before
Parellelogram After

Draw $\overline{BD}$.

Since $AB=CD$, $BC=AD$, and $\overline{BD}$ is shared, $\triangle ABD\cong\triangle BCD$ by the side-side-side property.

Therefore, $\angle CBD\cong\angle ADB$ and $\angle BDC\cong\angle ABD$.

It follows, by the lemma, that $\overline{AB}\parallel\overline{CD}$ and $\overline{AD}\parallel\overline{BC}$.

Thus, $ABCD$ is a parallelogram.

SSS again. Proving that remains on the to-do list.

Revenge of the Squares, Part 1

Olly / / Math Posts

I intended to split my study time today between analytic geometry review and logic review, but ended up getting drawn into some exploration as well. The result is that I have two proofs about quadrilaterals ready to post. The first of the two appears below. It’s a proof of my hunch from the post Squares Again that if a convex, four-sided, equilateral polygon has one right angle, then it has four right angles. It turns out that I could prove this without knowing that such a figure must be a parallelogram. (It must, though. That’s the proof for tomorrow.)

Lemma: The angles opposite (i.e. subtended by) the equal sides of an isosceles triangle are also equal.

Triangle Before 1
Triangle After

Consider the isosceles triangle $\triangle ABC$ with $AB=AC$.

Draw a line bisecting $\angle CAB$ and extend it to intersect $\overline{BC}$. Let the point of intersection be called $D$.

Now, by the side-angle-side property, $\triangle ACD\cong\triangle ABD$, since $AB=AC$, $\angle CAD\cong\angle BAD$, and $\overline{AD}$ is shared.

Therefore, $\angle ACD\cong\angle ABD$. These are the angles opposite the equal sides of the original isosceles triangle, so the lemma is proven.

Proposition: If a convex, four-sided, equilateral polygon has one (interior) right angle, then it has four (interior) right angles.

Let $ABCD$ be an convex, four-sided, equilateral polygon, and let $\angle BCD$ be a right angle.

Square Before
Square After

Divide the polygon by drawing $\overline{BD}$.

Note that, since $\overline{BD}$ is shared and all other line segments are equal, $\triangle ABD\cong\triangle BCD$ by the side-side-side property.

Hence, since $\angle BCD$ is a right angle and the angles of congruent triangles are equal, $\angle BAD$ is also a right angle.

Similarly, $\angle ADB\cong\angle CBD$.

Furthermore, by the lemma, since $\triangle ABD$ is isosceles, $\angle ADB\cong\angle ABD$, and since $\triangle ACD$ is isosceles, $\angle BDC\cong\angle CBD$.

Therefore, $\angle ADB\cong\angle ABD\cong\angle BDC\cong\angle CBD$.

Let the measure of those angles be called $x$. Since the sum of the measures of the angles of a triangle is 180 degrees, either triangle gives us the relationship $90+2x=180$, so $x=45$.

The measures of $\angle ADC$ and $\angle ABC$ are both $2x=2(45)=90$. Thus they are also right angles.

Hence, if a convex, four-sided, equilateral polygon has one right angle, then it has four right angles.

This proof uses SSS again—I actually thought of it while trying to work out a proof of SSS—as well as SAS, through the lemma1. It also uses the fact that the measures of the angles of a triangle sum to 180 degrees. I’m pretty sure that theorem is only a couple of steps from the parallel postulate, though.

My original version of this proof didn’t make use of the lemma about isosceles triangles, but instead argued that the two congruent triangles could be mapped onto each other by either rotation or reflection, and that was why all of the acute angles had to be equal. That was kind of a nifty argument, but not easy to formalize.

Stay tuned tomorrow for the proof about parallelograms.

  1. To do. ↩︎

Perpendicular Lines

Olly / / Math Posts

Today I spent all my study time on exploration and proofs, as opposed review exercises. My first goal was to prove Theorem 2 from yesterday:

Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1m_2=-1$; that is, their slopes are negative reciprocals:$$m_2=-\frac{1}{m_1}$$

I succeeded in that, and also came up with an argument for the converse of the Pythagorean Theorem, which is needed for the proof.

Argument for the converse of the Pythagorean Theorem

Proposition: Given a triangle with side lengths $a$, $b$, and $c$, if $a^2+b^2=c^2$, then the triangle is a right triangle.

Congruent Triangles

Let $\triangle ABC$ be a triangle such that $CA=\sqrt{AB^2+BC^2}$.

Construct a right triangle $\triangle EFG$ such that $EF=AB$ and $FG=BC$.

By the Pythegorean Theorem, $GE=\sqrt{EF^2+FG^2}$. Furthermore, $\sqrt{EF^2+FG^2}=\sqrt{AB^2+BC^2}=AC$.

Thus, $EF=AB$, $FG=BC$, and $GE=AC$, meaning that by the side-side-side property, $\triangle EFG$ is congruent to $\triangle ABC$.

The angles of congruent triangles are equal. Therefore, since $\triangle EFG$ is a right triangle, $\triangle ABC$ is also a right triangle.

The use of the SSS property here arose from a hint I found on the Internet. I would like to be able to prove that property, but have had no luck so far.1 It isn’t obvious to me why it should be true that having all sides equal guarantees congruence for triangles when it does not for other polygons.

An abbreviated proof of Theorem 2

Proposition: Two lines with slopes $m$ and $n$ are perpendicular if and only if $mn=-1$; that is, their slopes are negative reciprocals:$$n=-\frac{1}{m}$$

Let $y=mx+b$ and $y=nx+c$ be two lines such that $m\neq n$.

By Theorem 1 (from yesterday), since $m\neq n$, the two lines are not parallel. Thus, they intersect.

Let $E$ be the point at which the lines intersect.

We know from the proof of Theorem 1 that the coordinates of $E$ are $\left(\frac{c-b}{m-n},\frac{mc-nb}{m-n}\right)$.

Let $F\neq E$ be point on $y=mx+b$ with coordinates $(x_F, y_F)$ and let $G\neq E$ be point on $y=nx+c$ with coordinates $(x_G, y_G)$.

Furthermore, let $x_F=x_G=\frac{c-b}{m-n}-1$.

Perpendicular Lines

By plugging $\frac{c-b}{m-n}-1$ into both equations, we can find the values of $y_F$ and $y_G$ in terms of $m$, $n$, $b$, and $c$.

Using the formula below, we can then find the distance between each pair of points:

$$P_1P_2=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$ where $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$.

The algebra yields the following: $FG=|m-n|$, $EF=\sqrt{1+m^2}$, and $EG=\sqrt{1+n^2}$.

Now, assume that the lines $y=mx+b$ and $y=nx+c$ are perpendicular. This means that they meet at a right angle. Thus $\triangle EFG$ is a right triangle, and by the Pythagorean Theorem

$$\left(\sqrt{1+m^2}\right)^2+\left(\sqrt{1+n^2}\right)^2=\left(|m-n|\right)^2$$$$m^2+n^2+2=m^2-2mn+n^2$$$$2=-2mn$$$$mn=-1$$

Therefore, if the two lines, $y=mx+b$ and $y=nx+c$, are perpendicular, then $mn=-1$.

Instead, assume that $mn=-1$. In that case, $$\left(\sqrt{1+m^2}\right)^2+\left(\sqrt{1+n^2}\right)^2=m^2+n^2-2mn=(m-n)^2\text{.}$$

It follows, by the converse of the Pythagorean Theorem, that $\triangle EFG$ is a right triangle. Since the right angle of a right triangle is opposite the hypotenuse, $\angle FEG$ is a right angle. Hence the lines $y=mx+b$ and $y=nx+c$ are perpendicular.

Therefore, if the slopes of two lines, $y=mx+b$ and $y=nx+c$, have a product of $-1$, then the lines are perpendicular.

I had some other ideas today, as well, including regarding the question about squares from my post Squares Again. Expect those in the next few days.

  1. To do. ↩︎

Parallel Lines

Olly / / Math Posts

Today I worked on some exercises in Analysis with an Introduction to Proof, then read the review of analytic geometry I had queued. Near the end of the review, the author quotes the following theorems, indicating that proofs can be found in his precalculus textbook:

1. Two nonvertical lines are parallel if and only if they have the same slope.
2. Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1m_2=-1$; that is, their slopes are negative reciprocals:$$m_2=-\frac{1}{m_1}$$

That got me to thinking about how I would prove those things, and I ended up spending the rest of my study time working on them. I have a complete proof of Theorem 1. It is not that exciting, but it did give me the chance to use some of the logic that I am reviewing in Analysis with an Introduction to Proof.

Proving Theorem 2 seems like the more interesting problem. What does it mean for lines to be perpendicular? If it means that they meet at a right angle, what is that? How do you prove an angle is a right angle? My best idea so far is an argument involving right triangles, but in order for it to work, I need to establish to my satisfaction not just that the Pythagorean theorem holds for all right triangles, but also that only right triangles have sides that relate in the way described in the Pythagorean theorem. It’s pretty easy to convince oneself of the latter, given the former, and I may decide that’s enough. We’ll see.

Anyway, here’s what I did for Theorem 1:

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

For two lines to be parallel means that either they do not intersect or they are the same line.

Thus, we need to prove two statements:

  1. If two nonvertical lines do not intersect or they are the same line, then they have the same slope.
  2. If two nonvertical lines have the same slope, then either they do not intersect or they are the same line.

To prove statement 2, assume that two nonvertical lines have the same slope, $m$. Their equations can be written as $y=mx+b$ and $y=mx+c$, where $b$ and $c$ are real numbers.

Now assume that the lines intersect, and let $(x_1, y_1)$ be a point that lies on both lines. It follows that $y_1=mx_1+b$ and $y_1=mx_1+c$, and thence that

$$mx_1+b=mx_1+c$$$$b=c$$

When $b=c$, the two lines have the same equation, and are thus the same line.

Thus, when two nonvertical lines have the same slope it follows that if the two lines intersect, then they are the same line.

Hence, if two nonvertical lines have the same slope, then either they do not intersect or they are the same line, which is statement 2.

(The logic in play here is: $[p\implies (q\implies r)]\equiv[p\implies(\lnot q\lor r)]$)

To prove statement 1 by contraposition, assume that two nonvertical lines have different slopes, $m$ and $n$. Their equations can be written as $y=mx+b$ and $y=nx+c$ where $m\neq n$ and $b$ and $c$ are real numbers.

Since their equations are different, they are not the same line.

Furthermore, consider the point $$\left(\frac{c-b}{m-n},\frac{mc-nb}{m-n}\right)$$ which must exist, since $m\neq n$.

Notice that $$m\left(\frac{c-b}{m-n}\right)+b=\frac{(mc-mb)+(mb-nb)}{m-n}=\frac{mc-nb}{m-n}$$ and $$n\left(\frac{c-b}{m-n}\right)+c=\frac{(nc-nb)+(mc-nc)}{m-n}=\frac{mc-nb}{m-n}\text{.}$$

Thus, the point lies on both lines and the lines intersect.

Therefore, if two nonvertical lines have different slopes, then the two lines intersect and they are not the same line.

By contraposition, it follows that if two nonvertical lines do not intersect or they are the same line, then they have the same slope, which is statement 1.

(The logic in play here is: $[\lnot p\implies (q\land\lnot r)]\equiv[\lnot(q\land\lnot r)\implies p]\equiv[(\lnot q\lor r)\implies p]$)

Stay tuned for related proofs about perpendicular lines, probably tomorrow.

Algebra Completion Day

Olly / / Math Posts

Today I finished the 160 exercises in the review of algebra that I’ve been working through. This is an important achievement for me. Over the years, I’ve made repeated stabs at reviewing math, but I’ve never been able to tolerate it long enough to finish anything. To finish such a marathon problem set is positive proof that this time is different.

There isn’t much to talk about mathematically. One of the final exercises was to prove that $|ab|=|a||b|$, which I was prompted to do using the fact that $|a^b|=|a|^b$.1 Yet surely the latter statement is proven using the former. Isn’t it? In the end, I decided that doing the proof that way would be silly, and opted for a somewhat clunky casewise argument instead.

Anyway, happy Algebra Completion Day, everyone!

  1. To do (see comments). ↩︎

Exponent Negative

Olly / / Math Posts

Today I did more than two hours of algebra practice. It was fun, except for the sections devoted to simplifying messes of exponents. There are just too many opportunities for arithmetic errors in problems like that.

I did run into a bit of confusion in another section regarding one of the answers provided in the review materials, but I got help at the Math Help Forum, which is a good resource for those self-studying.

Stewart’s Calculus

Olly / / Math Posts

Today was better than yesterday, though I was still not at full capacity. I spent about an hour on algebra review, then thought for a while about the decimal representations of rational numbers. I also read the Wikipedia article on long division, which says that the debate about its place in the curriculum actually dates back to the 1980s. There are some interesting examples there of the ways long division problems are written in different countries, as well.

For anyone interested, here is the review of algebra I am using. It’s made available as a supplement by the publishers of Stewart’s calculus textbooks. There are also reviews of analytic geometry and conic sections, which I plan to work through as well, along with the review of trigonometry provided as an appendix to my textbook. (My textbook has reviews of algebra, analytic geometry, and conics, too, but they are not as complete as those offered on the website.)

I have a lot of affection for Stewart’s calculus texts. They were used in both my high school and my college courses, and represent good times to me. A person I met online once asked me to name a book that had changed my life, and Stewart’s Single Variable Calculus was my choice. Had I not enjoyed calculus so much in high school, I would not have been attracted to economics as a college major, would not have been encouraged to minor in math, would not have taken Discrete Math my first semester, and might never have had my love of math kindled such that I am still carrying a torch for it. His discovery of Ramanujan might have been “the one romantic incident” in Hardy’s life, but my encounter with math was mine, and Stewart’s calculus was one of the sign posts on the way.

About Division

Olly / / Math Posts

Today’s study time was devoted to more algebra review, including some problems that involved factoring using polynomial division. I have a vague memory of being taught polynomial division around the turn of the millennium, but I had heard the name more recently. There is debate these days about whether, in a society where almost everyone carries a calculator almost everywhere, it is still worth the instructional time to teach students numerical long division. (Eliminating it wouldn’t be unprecedented. I was never taught to compute square roots by hand, as older generations were.) At any rate, one of the arguments in favor of keeping numerical long division in the curriculum is that, without it, students won’t be able to do polynomial division. I’m not sure that would be much of a loss, honestly. It is a fairly niche method of factoring, and I managed to study a lot more math than most people without running into any other application of it. I think the greater drawback to eliminating long division would be that, without that tool to convert one to the other, the connection between fractions and decimals would become something mysterious only the calculator understands.

Speaking of conversion between fractions and decimals, in the portion of The Joy of X that I read last night, Steven Strogatz mentions the fact that the decimal representations of rational numbers always terminate or repeat. Of course, I wondered how one could prove that. I haven’t had time to work on it yet, but I think I can see the vague outline of how it might be done (probably using an argument related to the long division algorithm). The thing I don’t have any ideas about yet is how to prove that a number represented by a repeating decimal must be rational. I may find time to work on both of these tomorrow, but I may decide on more algebra practice instead. I am growing impatient to move on to a new review topic.