Tag: number theory

Trying

Apr 29, 2024
I didn’t feel very well today, either. I read some of Infinite Powers and worked on my logarithm problem. I am not ready to present a proof, but I have satisfied myself that if $a$ and $b$ are natural numbers greater than $1$, then they must be powers of the same base in order for $\log_a b$ to be rational. I’m not sure how to begin thinking about other types of numbers, and it’s possible I will lay that question aside for now.¹

I am still not making much progress on my calculus review. It’s an important part of this project, because I hope it will prepare me to earn some money tutoring. The truth is, though, that a lot of the time I am just not hale of mind. I’m doing my best.
1. To do. ↩︎
Math Posts

number theory, to-do
Day of Unrest

Apr 28, 2024

I had a lot of anxiety today, which didn’t lend itself to study. I did a little work on the logarithm question I’ve been talking about, but nothing else.

(For those wondering, the question is what the relationship between $a$ and $b$ must be in order for $\log_a b$ to be rational. It arose from a exercise in Analysis with an Introduction to Proof that asked me to prove that $\log_2 7$ is irrational.)

Updates

number theory
Ow

Apr 22, 2024

My back pain continued today. I did manage to get some things done, but was too tired to do any math besides reading more of Infinite Powers.

One thing Strogatz relates in that book is that Greek mathematicians emphasized geometry partly because they had no way of expressing irrational numbers, such as $\pi$, except in geometric terms: as lengths, areas, and their ratios. If you think about it, that has not changed much. We may have discovered more contexts in which $\pi$ or $\phi$ crop up, but we can still only express them exactly as solutions to particular problems. Any effort to write them down in isolation is only an approximation.

Math Posts

geometry, number theory
Shiny New Week

Apr 1, 2024
Today I spent two hours reading a section of Analysis with an Introduction to Proof and doing the associated exercises. I’m feeling better than I was for most of last week, though the mental weather is still not entirely clear.

Part of the section I read explained the inadequacy of examples, even quite a few examples, in proving general statements. As an illustration, it offered the statement that $n^2+n+17$ is prime for every natural number $n$. This statement is true for values of $n$ up to 15, but is not true for 16 or 17. The latter, in particular, could be predicted without trying each value of $n$, since $17^2$, $17$, and $17$ are clearly all divisible by a common factor (meaning that their sum must also be divisible by that factor).

Later, an exercise asked me to find a counterexample to the statement that $3^n+2$ is prime for every natural number $n$. This had me wondering whether there was a way to predict a counterexample, as there had been in the earlier case.¹ I could not think of one, however, so I answered the question by trying each $n$. It turns out that I didn’t have to go far, as $n=5$ breaks the pattern. I still wonder if there is a better way, though.

Thank you to all of you who have commented on my posts recently. It really helps me stay committed to my project. I will try to reply soon.
1. To do. ↩︎
Math Posts

number theory, to-do
Not Doing Much

Mar 24, 2024
My mental health was better today, but I haven’t yet bounced back physically. I took a long nap after church, and I was only up for some light math reading after. The section of The Mathematical Tourist that I read was about tests for primality, which is appropriate given my current interest in primes. It made me wonder if SageMath tests numbers directly or looks them up in a table. There might be a forum where I could ask.¹
1. To do. ↩︎
Math Posts

number theory, to-do
Persistent Puzzlement, Part 2

Mar 22, 2024

Today I continued to play with the problem from my post Puzzling Primes. Among other things, I installed the computer algebra system SageMath, which I can use to test quickly whether numbers are prime. With that tool, I might be able to solve the problem by brute force. I can’t imagine that this would be chosen as a problem of the week if there were not a better way, though.

So far I have been able to prove (a) that a superprime may contain at most two occurrences of the digits 1 or 7, (b) that all the other digits must be 3 or 9, except the first digit, which may also be 2 or 5, and (c) that if the first digit is 2 or 5, there will be no occurrences of 1 or 7.

All this means that, if there were some factor limiting the number of trailing 3s and 9s a prime could have, I would at least be able to prove a limit on the length of superprimes, and therefore that there must be a largest one. So far I’ve had no luck with that avenue of inquiry, though.

I will share the actual proofs of these claims in the future, but I’m still feeling unwell today. In the meantime, here is a chart of the ways that potential superprimes can be constructed. Red represents potential starting digits while blue represents digits that may be added.

Math Posts

number theory
Repairs are Ongoing

Mar 6, 2024
The depressive episode I started yesterday persists, but has been much less severe today. This afternoon I finished the last few chapters of The Joy of X without too much struggle. Now I will have to find a new book for my light math reading. I have several options in my library already, but I would welcome recommendations, as well.

The final chapter of The Joy of X was about infinity and included a treatment of Cantor’s diagonal argument that the set of real numbers is uncountable. (I’m sorry, non-specialist readers, but I’m not up to explaining what that is right now, though it is fairly accessible.) The Joy of X did not account in any way for the fact that some numbers have more than one decimal representation (e.g. that $0.1\overline{0}=0.0\overline{9}$). I’m sure a more formal presentation of the argument would have. I intend to think for a while about how I would do it before looking up the accepted method. Not today, though.¹
1. To do. ↩︎
Math Posts

number theory, to-do
More Digit Sums

Feb 22, 2024

Today I read more about trigonometry, both in the appendix of my calculus textbook and in OpenStax’s Algebra and Trigonometry. (The calc book does not discuss the Unit Circle, the Law of Sines, the Law of Cosines, or for some mysterious reason, the half-angle trig identities.) I also started on the calc book’s review exercises. Depending on how I feel after finishing those, I may do some of the exercises from Algebra and Trigonometry. It has a lot of applied problems that look as if they might be interesting.

I thought that today I would share the other fruit of my investigation of digit sums. I actually proved this proposition before the one from yesterday, but I didn’t have the energy to write it up. I originally found it in my notebook stated for base 10 only, but it occurred to me that it applied to other bases, as well. (Other than that, it’s not all that interesting.)

Proposition: In base $n>1$, if $m$ is a natural number then the digit sum of $m$ is less than or equal to $m$.

If $m$ is a natural number, then $m=n^md_m+n^{m-1}d_{m-1}+…+nd_1+d_0$ where $d_m, d_{m-1},…,d_1, d_0$ are the digits, in order, of $m$ expressed in base $n$.

The digit sum of $m$, expressed in base $n$, is $d_m+d_{m-1}+…+d_1+d_0$.

Since $n>1$, $d_p<n^pd_p$ for all $p\in\mathbb{N}$.

It follows that $d_m+d_{m-1}+…+d_1+d_0\leq n^md_m+n^{m-1}d_{m-1}+…+nd_1+d_0\text{.}$

Therefore, the digits sum of $m$ is less then or equal to $m$.

Math Posts

number theory, proof
Elegance and Regret

Feb 21, 2024

I didn’t do any trigonometry today, but instead read more of Steven Strogatz’s The Joy of X and worked on the propositions about digit sums that I found in my notebook.

The most interesting tidbit from The Joy of X was Strogatz’s comment regarding a particular proof of the Pythagorean Theorem that, “The proof does far more than convince; it illuminates. That’s what makes it elegant.” That caught my attention and had me wondering whether all elegant proofs must be illuminating. I haven’t come to a conclusion yet.

To be illuminating is certainly not the only requirement for an elegant proof. The main proposition I worked on today was “the sum of the digits of any multiple of 9 is also a multiple of 9.” I came up with the outline of a proof by induction, the meat of which was a description of what happens to the digit sum when you add 9 to any natural number. It was full of cases and loops and was decidedly not elegant. I do think it gave a good sense of what was happening, though, and it could have been reworked to prove the general theorem that, in base $n$, the digit sum of any multiple of $n-1$ is also a multiple of $n-1$.

I was sure there must be a better proof, however, and I ended up searching for guidance on the Internet. I regret that, as I would rather have discovered the answer I found for myself. I did get to generalize it, at least, since it applied only to two-digit numbers. I’m not sure I would say that the result is illuminating. I don’t think it reveals why the theorem is true, except in as much as “why” is that 9 is one less than the base of the number system being used. It is simple, though. Does that make it elegant?

Proposition: For all $n\in\mathbb{N}$, the digit sum of $9n$ is a multiple of 9.

When $n$ is a natural number, $9n = 10^md_m+10^{m-1}d_{m-1}+…+10d_1+d_0$, where $d_m, d_{m-1},…,d_1, d_0$ are the digits of $9n$, in order.

$$10^md_m+10^{m-1}d_{m-1}+…+10d_1+d_0=$$ $$\left[\left(10^m-1\right)d_m+\left(10^{m-1}-1\right)d_{m-1}+…+(10-1)d_1\right]+\left[d_m+d_{m-1}+…+d_1+d_0\right]$$

Thus, $$d_m+d_{m-1}+…+d_1+d_0=9n-\left[\left(10^m-1\right)d_m+\left(10^{m-1}-1\right)d_{m-1}+…+(10-1)d_1\right]\text{.}$$

Note that the left-hand side of the equation is the digit sum of $9n$.

Note also that, for any natural number $x$, $$10^x-1=\sum_{i=0}^{x-1}10^i(9)=9\sum_{i=0}^{x-1}10^i\text{.}$$

[For instance, $10^3-1=1000-1=999=9(111)$.]

Hence, because each term of the expression is a multiple of 9, $$9n-\left[\left(10^m-1\right)d_m+\left(10^{m-1}-1\right)d_{m-1}+…+(10-1)d_1\right]=9p$$ for some integer $p$.

It follows that the digit sum of $9n$ is a multiple of 9 for all $n\in\mathbb{N}$.

This proof could also be rewritten to cover all bases. In addition, the same argument could be used to prove that, in base 10, the digit sum of any multiple of three is also a multiple of three.

(And it occurs to me as I write this that you could also use it to prove the analogous fact, in base $n$, for any factor of $n-1$. That’s very cool and makes me feel like I have come up with something worthwhile on my own today.)

Math Posts

arithmetic, number theory, proof
Conic Completion Day

Feb 19, 2024

Today I worked through poor concentration to finish the exercises from the review of conics. There wasn’t much left to do, in fact, because the five final exercises all called for tools from calculus that I have not reviewed yet.

Next up will be trigonometry. I didn’t start reading the appendix on that topic today, but I looked over it. I also did a little review of the Unit Circle (which, oddly, isn’t pictured in the review of trigonometry). I remember Mike, my high school math teacher for two out of three years, was very strong on the Unit Circle. He used to say that if he ever met us again as adults, his first question would be, “What’s $\sin\left(\frac{\pi}{6}\right)$?” I did remember the answer to that before my review today, so perhaps threatening years-late pop quizzes is an effective teaching technique.

I reckon it’s about time I present my hand-waving argument that repeating decimals must represent rational numbers. I proved the converse in my post Rational-a-Rama with what I think most of my audience found a boring amount of rigor. Here I’m going to err far on the other side.

Consider the repeating decimal $x=0.\overline{142857}$.

Since $x$ repeats every six digits, multiplying $x$ by $10^6$ will yield another number with the same decimal part: $10^6x=142857.\overline{142857}$

Observe that $10^6x-x=142857.\overline{142857}-0.\overline{142857}=142857$.

Thus $(10^6-1)x=142857$ and $x=\frac{142857}{10^6-1}=\frac{142857}{999999}\text{.}$

Both $142857$ and $999999$ are integers, so it follows that $x=0.\overline{142857}$ is rational.

(In fact, $x=0.\overline{142857}=\frac{142857}{999999}=\frac{1}{7}$.)

A similar argument can be made about any repeating decimal, so any repeating decimal represents a rational number.

(Note that some repeating decimals do not begin repeating right after the decimal point. An example is $y=0.58\overline{3}$ (which is $\frac{7}{12}$). Multiplying $y$ by ten and subtracting $y$ from the result yields $10y-y=5.83\overline{3}-0.58\overline{3}=5.25$. This is not an integer, but it can be made one by multiplying by 100. This leaves the equation $100(10y-y)=525$, which can be used to show that $y$ is rational in the same manner as above.)

Math Posts

number theory, proof, trigonometry