Today I worked on exercises from both the review of conics and Analysis with an Introduction to Proof. I also thought some more about the questions from my post About Division. I’ve come up with a new argument that repeating decimals must represent rational numbers. It doesn’t require any theorems about geometric series, but it is going to be a bit of a pain to formalize. I may decide just to give an example of the maneuver it is based on and wave my hands for the generalization. Today I was going to share my proof that the decimal representations of rational numbers must repeat, but it is taking longer than I expected to write that up, so I will have to wait until tomorrow. Stay tuned, faithful readers!
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Jiggety Jig
I’m back home after a wonderful trip. I only fit in a little math during my week away, but I did make some progress on the questions posed in my post About Division. I have developed a good argument that the decimal representations of rational numbers must repeat (or terminate, which is simply ending in a repeating zero). An argument that a repeating decimal must represent a rational number is proving more elusive. I could use facts about the sums of geometric series, but I feel as if there should be another way. More on these things later in the week; today I have a lot of neglected chores to do.
A Week Off
Hooray! Today, with a battery of hyperbolas, I finished the last of the repetitive graphing exercises from the review of conics.
I was busy with a lot of other tasks, as well, because I’m leaving tomorrow on a trip with my family. I hope to do some math while I am gone, but I probably won’t be updating the blog, as I am not taking my laptop. Expect me back on Monday, 12 February.
Ellipses
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
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
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.
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 $BC=AD$.
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.
Hit and Miss
First of all, if you are subscribed to post notifications, please note that those seem to be hit and miss. Not everyone is getting a notification for every post. I may have to find another notification tool. If I do, I will try to migrate the email addresses of those already signed up.
I got a late start today and was distracted by a lot of small tasks, so that I wasn’t able to sit down to study until evening. I did manage about an hour and a half of analytic geometry review, though. It’s going well. The concepts are coming back to me fairly easily. One thing I find curious is that circles are covered in the review chapter on basic analytic geometry, not the one on conics. That seems odd.
Part 2 of Revenge of the Squares will probably go up tomorrow. I don’t have the energy for it today.
Revenge of the Squares, Part 1
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.
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.
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.
- To do. ↩︎
Perpendicular Lines
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.
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 $x_F=x_G=\frac{c-b}{m-n}-1$.
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.
- To do. ↩︎
Parallel Lines
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:
- If two nonvertical lines do not intersect or they are the same line, then they have the same slope.
- 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.