This is a question that has bothered me for almost 6 years now on and off, and I still don't really know enough to tackle it.
To phrase it somewhat formally:
Let P be the series of prime numbers such that P(i) is the i-th prime number.
Let N be a positive integer with a value greater than two
Let X be the series of numbers such that X(0) = 0 and X(i) = (X(i - 1) + P(i)) % N
Does there exist an N for which the set of elements X(0)..X(N-1) contain every number from 0 to N-1 exactly once? Can we prove it one way or another?
I've made computer simulations and let them run overnight and the results seem to suggest that no, there is no such N, but the graphs I got out of doing this are fairly interestingly shaped.
For instance, this is a scatter plot where the X axis is N and the Y axis is the percent of numbers in X(0)..X(N-1) that were reached before a duplicate.
And if we zoom in we can see some definite "structure" to the proportions, which is also interesting.
And on the long tail there seems to be a definite "range", with some outliers
I don't know how to explain why the graph is shaped like it is, or what the structure in it means, or why it seems like a "thick" logarithmic curve.
from Hot Weekly Questions - Mathematics Stack Exchange
My question is about understanding a remark John Conway made in On Numbers and Games (ONAG), where he proposes a method for constructing the real numbers from the rationals. I will have to assume familiarity with Conway's construction of the surreal number field.
The context of this quote is Conway discussing of an unfortunate feature of constructing $\mathbb R$ from Dedekind cuts of $\mathbb Q$. Namely, when defining multiplication $x\cdot y$ of real numbers, you need to break into four cases based on the signs of $x$ and $y$. This means that proving associativity requires breaking into eight cases, which is clunky. One could instead modify Conway's construction of the surreals to construct $\mathbb R$, adding an extra axiom so the extra infinities and infinitesimals are not constructed. This solves the problem clunky associativity proof, since multiplication in the surreal numbers is defined without need for case splitting. However, the full surreal number construction has some drawbacks (being difficult to understand for undergraduates, and giving special status to the dyadic rationals).
To this end, Conway proposes the following compromise:
There is another way out. If we adopt a classical approach as far as the construction of $\mathbb Q$, and the define the reals as Dedekind sections of $\mathbb Q$ with the definitions of addition and multiplication given in this book, then all formal laws have 1-line proofs and there is no case splitting.
I surmise this means that real numbers would be defined as ordered pairs $(L,R)$ of sets of rational numbers, where for all $\ell \in L$ and $r\in R$ we have $\ell < r$, and $L\cup R$ excludes at most one member of $\mathbb Q$. My question is, how does Conway's suggestion actually work, in detail?
That is the end of my question, what follows are my thoughts on the problem.
Following the surreal definition of multiplication, letting $x$ and $y$ be real numbers, and letting $x^L$ and $x^R$ be variables which range over the left and right sections defining $x$, it seems that Conway would define $$ xy=(\{x^Ly+xy^L-x^Ly^L,x^Ry+xy^R-x^Ry^R\},\{x^Ly+xy^R-x^Ly^R,x^Ry+xy^L-x^Ry^L\})\tag1 $$ However, this raises the question, "what does $x^Ly$ mean?". This is the product of a rational with a real number. You cannot use the above definition to define this product, since Conway stipulated the rationals were to be constructed as normal, so they do not have a left and right section. However, it seems the only sensible way to define the multiplication of a rational with $q$ with a real number $x$ would involve case-splitting, as below, defeating the whole purpose. $$ xq= \begin{cases} (\{x^Lq\},\{x^Rq\}) & q>0\\ (\{x^Rq\},\{x^Lq\}) & q<0\\ 0 & q=0 \end{cases} $$
The same problem would occur for the definition of addition: $$ x+y=(\{x+y^L,x^L+y\},\{x+y^R,x^R+y\}) $$ However, there is a nice workaround, where you instead define $$ x+y=(\{x^L+y^L\},\{x^R+y^R\}) $$ This is equivalent to the usual definition of addition in Dedekind's construction of the reals. It works because $x^L<x$ and $y^L<y$ implies $x^L+y^L<x+y$, so that $x^L+y^L$ can be safely put in the left section of $x+y$. Is there a similar fix for the multiplicative definition?
I feel there must be something of value here. The definition in $(1)$ is very clever. Namely, if you define $x$ as a Dedekind-like cut with left members $x^L$ and $x^R$, and the same for $y$, then you can derive the cut for the product $xy$ using the fact that $x^L<x<r^R$, and similarly $y^L<y<y^R$, which implies the inequalities $$ (x-x^L)(y-y^L)>0,\qquad (x-x^L)(y^R-y)>0, \dots, $$ which implies $xy$ is between the elements of the complicated looking expression in $(1)$. No case splitting is needed. I just cannot quite figure out how to apply that same clever logic to define multiplication of cuts of the already constructed rational numbers.
from Hot Weekly Questions - Mathematics Stack Exchange
I'm currently a couple of years out from college with a bachelors in math and biology, along with a masters in finance. My current role is comparable to that of a data analyst, and have experience with SAS, Excel, and a little bit of Python. I was wondering whether working for the NSA would be a possibility for someone whose background is more from the applied side of mathematics and is entertaining the idea applying for an entry level position.
This recurring thread will be for any questions or advice concerning careers and education in mathematics. Please feel free to post a comment below, and sort by new to see comments which may be unanswered.
Please consider including a brief introduction about your background and the context of your question.
Developing Our MathMINDs series of conversations and resources about math that is intended to be a journey of growth with families over several weeks. Each week, MIND's Lead Mathematician and Product Director Brandon Smith and Content Development Manager Nina Wu will be talking about the adjustments families are making to learning at home, and the opportunities this situation provides for changing our relationship with math.
Each week, both the video conversations and resource links will be posted here on the MIND Blog, as well as on our YouTube page.
In week three, Brandon and Nina explain informative feedback, how it's built into the ST Math learning model, and also how to infuse it into any learning activity.
The following videos feature a couple of our ST Math families demonstrating activities that provide informative feedback:
And this one involves using coordinates:
We Want to Hear From You!
We look forward to embarking on this mathematical journey together with families, and we want to hear about your challenges, success, questions and experiences along the way!
This week:
We'd love for you to share your own examples of activities that involve informative feedback! You could describe the activity, or act in out in a video like the ones above.
To reply, you can either comment on the blog below provide a link to your video, head over to our Facebook group, or comment on YouTube. Make sure to provide a link to your own video if you create one!
For more information and resources that expand on this week's discussions, check out the links below.
Hello, I'm going to be starting a mathematics undergraduate program in September and I've never studied any probability or statistics. With my exams finishing next week I'd love to get started on a book that would give me a good understanding. At least enough to be at semester 1 undergrad level if you get. Not looking to rush, just to lay a good foundation. Any questions and suggestions are welcome and appreciated, thank you.
The official subreddit for Mathematics Competitions in the USA, for all who are interested. For those who are knowledgeable, want to learn, or those who just want to know more about competition math.
Computer-assisted proofs aren't entirely new in mathematics, but the form of assistance we're used to hearing about are either naive (e.g. Four colour theorem) or formal (using proof verification software like Coq or Lean).
I usually don't pay that much attention to developments in AI, however reading about GPT-2 really piqued my attention and got me wondering about a possible third form of computer proof which I shall call generative.
The idea would be to train some kind of general-purpose generative model on a large database of mathematical proofs and see if it were able to generate semblances of correct proofs by feeding it statements as prompts. The resulting text would be the result of something profoundly different to mathematical reasoning as we know it, and instead based on only form and context. Basically 100% "machine intuition" if that makes any sense.
A first obvious caveat is the size of the database: I am unsure that even the entirety of arXiv would large enough to get a workable model.
Secondly, I think such a model would be fundamentally ineffective against any statement that is in the least technical. It would be unrealistic to expect for a machine to generate even a few lines of correct calculation using contextual cues alone. This rules out a large class of theorems and lemmas from those which we might expect our putative model to have a reasonable shot at.
Nonetheless, the idea doesn't seem entirely unpromising as a whole. I feel like a sufficiently-trained model should have a reasonable chance at a certain class of non-technical, but non-trivial theorems. Obviously nothing close to cutting-edge (it would be absurd to expect such a model to reason about perfectoid spaces) but still strong enough to potentially teach us something new, like a link between two things that was never noticed, or a simplification of an otherwise complicated proof. Kind of like how AlphaGo taught chess players something new (IIRC some line of some opening had been considered losing in modern play, but then AlphaGo used it and showed that it could provide a decisive advantage 20+ moves in).
Has such a work been attempted, or theorised? To those with experience of knowledge on the subject, to what extend is this flawed or otherwise unrealistic?
So, I have been unlucky enough to have many a problem on my shoulders that have been a nuisance and a setback to the development of my mathematical maturity. I have many classes that I haven't passed and I want to start with a solid foundation so that I can cover the most gaps in the least amount of time. I am thinking that Real Analysis I-IV is the most important class of all. Therefore should I focus on real analysis or other classes first? I will have below the math related classes from each semester.
Semester I: Real Analysis I, An Introduction to Abstract Algebra and Set Theory, Analytic Geometry
Semester II: Real Analysis II, Discrete Mathematics, Linear Algebra I
Semester III: Real Analysis III, Ordinary Differential Equations I, Probability Theory I
Semester IV (current): Real Analysis IV, Ordinary Differential Equations II, Probability Theory II, Abstract Algebra I, Number Theory
So should I focus on this semester and cover ODEs and PT so that I can catch two birds with one stone and also focus on Abstract Algebra I and Number Theory? Then should I go over all of Real Analysis and Linear Algebra I and lastly focus on An Introduction to Set Theory and Abstract Algebra/Analytic Geometry/Discrete Mathematics? If you can recommend any resources like books and notes it would be greatly appreciated!
Here is a question in model theory I need help. Suppose $T$ be a set of universal sentences and $T\models\forall x\exists y\:P(x,y)$. Prove that there exists terms $t_1(x),\cdots,t_n(x)$ such that $$ T\models\forall x \bigvee^n_{k=1}P(x,t_k) $$ Intuitively, it is obvious true but I need a rigorous proof. Does $T$ universal mean that there is a predicate $\phi=\forall x\forall y \:Q(x,y)$ that $\phi\implies \forall x\exists y\:P(x,y)$? Then it follows by the compactness theorem.
from Hot Weekly Questions - Mathematics Stack Exchange
Can a subset of $\mathbb{R}^n$ be homotopy equivalent to $S^n$? I am pretty certain the answer is no and I suspect it might be provable using homology groups, but I do not see how. Note that, because $S^n$ is the one-point compactification of $\mathbb{R}^n$, this question is equivalent to asking whether $S^n$ is homotopy equivalent to one of its proper subsets.
Edit: By freakish's comment, the answer is no when the subset is a manifold. However, I am now less certain whether the answer is no for any subset by the example Tyrone showed.
Edit 2: As Maxime Ramzi pointed out, the answer is also no if $X$ is compact by Alexander duality.
from Hot Weekly Questions - Mathematics Stack Exchange
Equation of the locus in complex numbers – Loci problems in complex numbers. Hello friends, today I’ll show how to derive the equation of the locus of any complex number. Have a look!! The much-needed posts to understand the loci problems are: Modulus and argument of the complex numbers Multiplication and division of complex numbers...
Heyy, so I'm majoring in Computer Science in my college. Now as a career, I'm more into ML/DL but good at Development (which has amazing career opportunities here) So I love my branch but I love Maths too and as time passes the time I get to explore mathematis reduces. I've taken a Real Analysis course in my sophomore year and a few in my freshmen year but that's that and now my college is out of mathematics courses.
So, my question is should keep studying Calculus/Analysis/Geometry as they're interesting and borederline addicting and will that help in the long run as a CS guy?
$ke^{-kx^2}\leq ke^{-kx^2}\arctan(x)$ for $[\tan(1),\infty)$, but by integrating we know that $\int_{0}^{\infty}ke^{-kx^2}\to \infty$ and so out sequence of original integrals diveres too.
Is this correct?
from Hot Weekly Questions - Mathematics Stack Exchange
I am graduating with a bachelor's degree in applied mathematics, and I’m interested in perusing a master’s degree in applied mathematics. I came here to ask about some of the advantages and disadvantages of going the nonthesis route, as opposed to going the thesis route.
Why are the Partial Differential Equations are so named? i.e, elliptical, hyperbolic, and parabolic. I do know the condition at which a general second order partial differential equation becomes these, but I don't understand they are so named?
Does it have anything to do with the ellipse, hyperbolas and parabolas?
from Hot Weekly Questions - Mathematics Stack Exchange
In Rudin's Principles of Mathematical Analysis 1.1, he first shows that there is no rational number $p$ with $p^2=2$. Then he creates two sets: $A$ is the set of all positive rationals $p$ such that $p^2<2$, and $B$ consists of all positive rationals $p$ such that $p^2>2$. He shows that $A$ contains no largest number and $B$ contains no smallest.
And then in 1.2, Rudin remarks that what he has done above is to show that the rational number system has certain gaps. His remarks confused me.
My questions are:
If he had shown that no rational number $p$ with $p^2=2$, this already gave the conclusion that rational number system has "gaps" or "holes". Why did he need to set up the second argument about the two sets $A$ and $B$?
How does the second argument that "$A$ contains no largest number and $B$ contains no smallest" showed gaps in rational number system? My intuition does not work here. Or it is nothing to do with intuition?
from Hot Weekly Questions - Mathematics Stack Exchange
I have a problem in the proof of the following result: Given $f \in L^1(\mathbb{R}^n)$, we have that $$|\hat{f}(\xi)| \rightarrow 0, \;\;\; as |\xi| \rightarrow \infty.$$
This result is known as the Riemann-Lebesgue Lemma (Proposition 2.2.17 from Grafakos's book Classical Fourier Analysis, third edition). In the proof of this proposition, one considers the function $$g := \prod_{j=1}^n \chi_{[a_j,b_j]},$$ that I suppose is the characteristic function of the cube $\prod_{j=1}^n[a_j,b_j] \subset \mathbb{R}^n$, and whose Fourier transform is $$\hat{g}(\xi) = \prod_{j=1}^n \frac{e^{-2\pi i \xi_ja_j} - e^{-2\pi i \xi_jb_j} }{2\pi i \xi_j},$$ in the meaning that if some $\xi_j = 0$, the correspondent factor is equal $b_j-a_j$. Now, if $\xi =(\xi_1, ..., \xi_n) \neq 0$, choose $j_0$ such that $|\xi_{j_0}| \geq |\xi|/\sqrt{n}$. So that $$\left| \prod_{j=1}^n \frac{e^{-2\pi i \xi_ja_j} - e^{-2\pi i \xi_jb_j} }{2\pi i \xi_j} \right| \leq \frac{2\sqrt{n}}{2\pi|\xi|}\sup_{1\leq j_0\leq n}\prod_{j\neq j_0}(b_j-a_j).$$ This inequality is what I'm not being able to prove. Once it's proved, I have the desired result.
from Hot Weekly Questions - Mathematics Stack Exchange
Title is supposed to say "Can a person with little mathematical training go into pure math?" Sorry.
Hello all. This has been on my mind for quite sometime now.
Only till 10th grade did I formally study math and I wasn't that great at it also (managed to get B+/A- at best). I opted for the humanities stream for my +2 and took up an Honours course in Philosophy which I'm getting my degree in.
The thing is, I'm really into logic, language and the likes of Russell, Riemann, Gödel, Wittgenstein etc., and wanted to know if, in addition to the philosophy of mathematics and logic, I could do pure math as well (maybe not get a degree in it per se, but at least be able to do so that I, like I can "do" [for a lack of a better term] Philosophy, can do pure math as well).
Computer-assisted proofs aren't entirely new in mathematics, but the form of assistance we're used to hearing about are either naive (e.g. Four colour theorem) or formal (using proof verification software like Coq or Lean).
I usually don't pay that much attention to developments in AI, however reading about GPT-2 really piqued my attention and got me wondering about a possible third form of computer proof which I shall call generative.
The idea would be to train some kind of general-purpose generative model on a large database of mathematical proofs and see if it were able to generate semblances of correct proofs by feeding it statements as prompts. The resulting text would be the result of something profoundly different to mathematical reasoning as we know it, and instead based on only form and context. Basically 100% "machine intuition" if that makes any sense.
A first obvious caveat is the size of the database: I am unsure that even the entirety of arXiv would large enough to get a workable model.
Secondly, I think such a model would be fundamentally ineffective against any statement that is in the least technical. It would be unrealistic to expect for a machine to generate even a few lines of correct calculation using contextual cues alone. This rules out a large class of theorems and lemmas from those which we might expect our putative model to have a reasonable shot at.
Nonetheless, the idea doesn't seem entirely unpromising as a whole. I feel like a sufficiently-trained model should have a reasonable chance at a certain class of non-technical, but non-trivial theorems. Obviously nothing close to cutting-edge (it would be absurd to expect such a model to reason about perfectoid spaces) but still strong enough to potentially teach us something new, like a link between two things that was never noticed, or a simplification of an otherwise complicated proof. Kind of like how AlphaGo taught chess players something new (IIRC some line of some opening had been considered losing in modern play, but then AlphaGo used it and showed that it could provide a decisive advantage 20+ moves in).
Has such a work been attempted, or theorised? To those with experience of knowledge on the subject, to what extend is this flawed or otherwise unrealistic?
I have just finished my final calculus class and am looking forward to moving on with my mathematics education. However, I can’t help but feel like I have already forgotten a lot of stuff from earlier calc classes. For example, partial fractions from cal B crossed my mind for some odd reason..and I was just thinking about how I forgot some of the finer details that come with it. Additionally, certain trig substitutions and other details stick out as things that I should remember more clearly. There was a time when I was sharp with both partial fractions and integrals with trig substitution.
I will no doubt look back to review stuff this summer, but what do you guys do to remember stuff? I am always trying to become a better a learner, and it is quite frustrating when I forget stuff that I should remember. Basically, I want to keep things fresh in my head and limit the amount of stuff I forget. I truly enjoy maths and I want to teach it, so I obviously want to be well-rounded and retain what I am learning.
Since any open set A is basically the interior of the set A itself, and the boundary of a set is basically the end points of the set being that it could be in the set A or the complement of A, then is A-BdA = IntA and is A-IntA= BdA
from Hot Weekly Questions - Mathematics Stack Exchange
I would like to compute the determinant of a matrix with the following structure: \begin{equation} \begin{pmatrix} D_1 & l_1 & l_1 &\cdots & l_1 \\ l_2 & D_2 & l_2 &\cdots & l_2 \\ l_3 & \cdots & D_3 &\cdots & l_3 \\ l_4 & \cdots & l_4 & D_4 & l_4 \\ l_5 & \cdots & \cdots & l_5 & D_5 \\ \end{pmatrix} \end{equation} That is, it is constant on each line apart from the diagonal. $l_i, D_i \in \mathbb R^+$.
Is there a way to make use of such symmetric structure to simplify the calculation of the determinant?
from Hot Weekly Questions - Mathematics Stack Exchange
I’m familiar with programs like Budapest Semesters in Mathematics and REU’s, but are there other prestigious and high-quality undergraduate programs that significantly help a math student grow so that they can be a capable grad student?
When I consider my list of things I’m applying to, it’s quite small. The only other things that people aiming for grad school seem to do are industry jobs with tech/finance firms. Is there more out there?
Suppose that $F$ is defined via the recurrence relation $$F(1)=1, \qquad F(n)=\sum_{k=1}^{n-1}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr)$$ What is $F(N)$? I don't have any idea how to solve this problem. Only one thing that I've noticed is that: $$ 0=\sum_{k=1}^{n}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr).$$
Edit:
From the comment section:
I was trying to rewrite $\sin\left(\frac{\pi}{2^n}\right)$ by the formula for a double argument and I've ended up with $$\sin\left(\frac{\pi}{2^n}\right)=\frac{\sin\left(\frac{\pi}{2}\right)}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)}=\frac{1}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)},n\gt 1,$$ but it doesn't seem to help. $F$ is used in another formula. It should be true for most of the functions $$g(x)=\sum_{n=1}^\infty \left(\sin\left(x2^{n-1}\right)\sum_{k=1}^n\left(F(k)g\left(\frac{\pi}{2^{n-k+1}}\right)\right)\right),\;x\in\left(0,\frac{\pi}{2}\right)$$ First four values of $F$ are:
These terms look like if they created some pattern, but the fifth term which is \begin{align*}F(5)=-\sin\left(\frac{\pi}{32}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{16}\right)-3\sin^2\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)+\sin^4\left(\frac{\pi}{4}\right)\end{align*} meses the pattern up.
from Hot Weekly Questions - Mathematics Stack Exchange
For some time, it seems like pure math has been developing without science, as in the application of a lot of math in the "real world" has been trailing behind theoretical math. I know a lot of mathematics has eventually been applied to computer science, cryptography, the like. At the same time, so many things can't, like, as far as I can understand, the Banach-Tarski paradox? The question is, will all math inevitably be applied? Is there a limit to what can conceivably be used in the real world?
The centre of gravity of a solid of revolution. Hello friends, today I’ll talk about the centre of gravity of a solid of revolution. Have a look!! The centre of gravity of a solid of revolution Now let’s suppose is the equation of a curve. Also, let’s say, is the position of the centre...
Any collection of possible outcomes, including the sample space $\Omega$ and its complement, the empty set $\emptyset$, may qualify as an event. Strictly speaking, however, some sets have to be excluded. In particular, when dealing with probabilistic models involving an uncountably infinite sample space; there are certain unusual subsets for which one cannot associate meaningful probabilities.
Question 1
What is meant by "meaningful" probabilities?
Question 2
Can you provide an example in which we cannot assign meaningful probabilities to the events of the sample space?
from Hot Weekly Questions - Mathematics Stack Exchange
Let $H$ be a Hilbert space, let $A\in B(H)$ satify $\|A\|\le 1$. If $A$ is positive, i.e. $A$ is a self-adjoint operator and for all $x\in H$, $\langle A(x),x\rangle\ge 0$, proof that $${\|x-A(x)\|}^2\le {\|x\|}^2-{\|A(x)\|}^2, \forall x\in H.$$
This is a exercise. For $A$ is positive \begin{align*} {\|x-A(x)\|}^2 &={\|x\|}^2+{\|A(x)\|}^2-\langle A(x),x\rangle-\langle x,A(x)\rangle\\ &={\|x\|}^2-{\|A(x)\|}^2+2{\|A(x)\|}^2-2\langle x,A(x)\rangle \end{align*} so we should prove that $${\|A(x)\|}^2\le \langle x,A(x)\rangle$$ or $${\|A(x)\|}^2=\langle A(x),A(x)\rangle = \langle x,A(A(x))\rangle \le \langle x,A(x)\rangle$$ but i don't know how to use the condition "$\|A\|\le 1$" and "$\langle A(x),x\rangle\ge 0$".
from Hot Weekly Questions - Mathematics Stack Exchange
A pet peeve of mine is measuring things to far too many decimal places. For example, notice that the thickness of these trash bags is 0.0009 inches (0.9 mil) but is 22.8 microns in metric. There are two mistakes: While the conversion factor is correct, there’s no way that the thickness is known within only […]
Pat Ashforth and Steve Plummer explore maths through knitting and crochet at woollythoughts.com.
It may not be immediately obvious what this represents. You might get a clue if you ask yourself why you see yellow stripes in alternate columns but nowhere else. Why do some columns have so few colours?
The blanket is called Counting Pane. It has 100 squares in 10 rows. The first row represents the numbers 1 to 10, the second is 11 to 20, and so on. The colours across the top represent factors (or divisors, or whatever you want to call them). Blue stands for 1, yellow stands for 2, red is 3, green is 4, pink is 5, brown is 6, light pink is 7,light blue is 8, orange is 9, and purple is 10.
Before you read on, ask yourself some questions about what jumps out at you. Here are some you might like to think about:
Why are there so many plain blue squares?
Why are there so many light-blue/yellow/green squares?
Why is the end column so crowded?
There were questions and problems, at the designing stage. The first was that there were more factors than we had thought about and we couldn’t possibly include them all. No factors above 10 are included because there simply wouldn’t be room for more. Also, and more importantly, the colours had to be easy to identify. It had to be immediately obvious which was which.
Another problem was working out what size the squares must be to fit in the different numbers of stripes. We went for twenty-four ridges of knitting in each square. This worked for squares with 1, 2, 3, 4, or 6 colours and I only had to cheat a little where we needed 5 or 7 colours.
Making this blanket was a shock. I had assumed I knew most of what one could ever know about the numbers up to one hundred but there was far more there than I had anticipated
What struck me first was the large number of colours in the tens column. I must have known that as I was knitting it but it only jumped out when it took its place with the other columns. I have often been known to talk about ‘nice numbers’. To me, nice numbers are those that have lots of factors because they are easy to juggle with. I had never classed multiples of ten as nice numbers. They were ‘just tens’. Everyone knows you add a nought if you want to multiply by ten or take one off if you want to divide. That didn’t make them interesting enough to be ‘nice numbers’. Now they were the first thing to shout at me. That column was far more colourful than any of the others. There were more factors than anywhere else.
The blue/yellow/green squares also seemed to stand out. It is obvious, when you stop to think about it, that any number that will divide by 4 also divides by 2. Everywhere there was a green stripe representing 4, there would be a yellow stripe. Similarly, every square containing light blue, for 8, must also have green and yellow. A brown 6 always went along with a yellow 2 and a red 3.
Oranges, representing multiples of 9, made an interesting sloping line. We could have expected this because adding 9 is the same as adding 10 and taking away 1 so the squares are one away from lining up. We hadn’t expected the lines of red that seemed to criss-cross in all directions. Bright pink only appeared in two columns but, as it represented 5, perhaps that was not so surprising.
The other big surprise, that changed my perception of certain numbers, was the number of plain blue squares. All the plain blue squares are primes but not all of the primes are plain blue. 2, 3, 5, and 7 are all prime numbers but they have a coloured stripe. However, what was very apparent was that the blue squares were clustered in certain columns. Whenever I think about prime numbers now I think about them in terms of which column they are in. We have made three versions of Counting Pane. The first was bought by the Science Museum (London). It is slightly different from the others because we tried to resolve the problem of 1 looking like a prime number when technically it isn’t. For versions two and three we decided this was an anomaly we would have to live with and introduce as a discussion point when appropriate. Versions 2 and 3 have other slight cosmetic differences (including at the tops of the columns).
At the most basic level this afghan offers a very graphic way for children to understand the difference between odd and even and to realise that this is really the same as knowing whether a number divides by 2. Yellow only appears in half of the columns. These columns contain only even numbers.
Numbers which divide by 5 are also spotted quickly, by young children, as raspberry is noticeably only in the 5’s and 10’s columns. Putting these two ideas together there is only one column with yellow and raspberry so this must be the column of 10’s. There are many questions that can be asked. I could go on and on. The more you look the more you will see.
More than twenty years ago we provided resources for using this design in schools. Some of them are lost but others are cobbled together in a file to download and use. It includes a colouring-in sheet www.woollythoughts.com/information/countingpane.html
(If you are old enough to remember the days when bedspreads were called counterpanes, the title will make more sense.)
Nikki Rohlfing – The Phoenix Number
Nikki Rohlfing is a heavy metal roadie turned maths teacher. On twitter he is@heavymetalmaths.
What is your favourite number? Surely everyone has one, and Alex Bellos even did a survey to see which number is the most favourite in the world (or at least from the 30,025 submissions he got).
Spoiler: it was 7.
If you ask me though, at a glance, I’d say 6, it’s just, y’know, so perfect…..b’dum. Also hexagons are awesome, and the more personal reason of wearing the number 6 shirt when scoring my first football goal for school. Being half German, I can also provide a somewhat low-brow reason as well for why I like 6….
However, if giving it some thought, then despite a range of impressive numbers out there, I will not go with 6, but rather a 6 digit number instead; The Phoenix Number: $142857$.
So what’s so special about $142857$? Well, here’s a start:
It explodes! But then, just like the phoenix, it comes back again. Although because you then get a number with more than 6 digits, there’s a little additional work to get back to a permutation. For example,
What you do now is, counting from the units, get a 6-digit number, and whatever digits are before that create a new number. These you then add together. So in the case above, you would have $428455$ as the 6 most right digits, and it leaves 116 in front. So you compute $428455 + 116 = \color{orange}{4}\color{green}{2}\color{blue}{8}\color{darkblue}{5}\color{purple}{7}\color{red}{1}$.
What if there are more than 6 digits in front of the far most right 6 digits I hear you ask? Well, you just repeat the above process until only 6 digits are left, and you will get a permutation of the Phoenix Number (if not itself). Magic!
And if that’s not enough of a wow, here’s a couple of small bonuses:
Splitting the number up into two 3 digit numbers: Then $857^2-142^2=\color{purple}{7}\color{red}{1}\color{orange}{4}\color{green}{2}\color{blue}{8}\color{darkblue}{5}$.
(Is it just me, or does the difference of two squares somehow pop up everywhere?)
Also:
\[14+28+57=99\\ 142+857=999\]
Even 4 digits works, if you allow a bit of a cycle and therefore also ensuring each number appears an equal amount of times for fairness:
$1428+5714+2857=9999$ and this carries on for 5 digits, 6 digits etc.
Speaking of cycle, $142857$ is in fact an example of a cyclic number. There are quite a few out there, the next has 16 digits, so they get quite long quite fast! As you’d expect, the OEIS lists them: https://oeis.org/A004042.
So what makes it cyclic? Well, it’s all to do with the fact that $142857$ is the recurring sequence of the decimal of $\frac{1}{7}$, which also gives away at what happened when the phoenix number was multiplied by 7, or in fact what is likely to happen if multiplied by any multiple of 7.
All cyclic numbers have this feature of being the sequence in a recurring decimal. For example, the 16 digit cyclic number mentioned, is the repeating decimal part of $\frac{1}{17}$. In fact, to find more cyclic numbers from fractions, the denominator has to always be a prime number, but that does not mean that all primes necessarily generate a cyclic number. According to a few sources I found just now, it’s around 37.395% of all primes that will generate a cyclic number. These primes are known as ‘full repentant primes’: https://oeis.org/A001913.
In fact, there’s no doubt quite a bit of serious maths going on linked to this, as the study of primes always has, but probably way beyond me and my more simple pleasure at finding such an awesome number!
So, which bit of maths made you say “Aha!” the loudest? Vote:
Note: There is a poll embedded within this post, please visit the site to participate in this post's poll.
The poll closes at 9am BST on Wednesday the 29th, when the next match starts.
If you’ve been inspired to share your own bit of maths, look at the announcement post for how to send it in. The Big Lockdown Math-Off will keep running until we run out of pitches or we’re allowed outside again, whichever comes first.
Consider the set $$S=\{1,2,3,4,5,6,7,8,9,12,13,14,15,16,17,18,19,23,24,\ldots,123456789\},$$ which consists of all positive integers whose digits strictly increase from left to right. This set is finite. What is the median of the set?
This problem is harder than I thought at first. I first simply though the solution was $\frac{123456789+1}{2}=61728395.$ Turns out I'm wrong! Where am I going wrong? I checked my work and $61728395-1+1=61728395.$ Also,
$123456789-61728395+1=61728395.$ These are equal, so the distance should also be equal. Where am I going wrong?
from Hot Weekly Questions - Mathematics Stack Exchange
Having a bit of a problem calculating the volume of a take-away box:
I originally wanted to use integration to measure it by rotating around the x-axiz, but realised that when folded the top becomes a square, and the whole thing becomes rather irregular. Since it differs in circumference I won't be able to measure it like I planned.
Is there any method or formula that can be used to measure a shape like this, or do I just have to approximate a cylinder and approximate a box and add those two together?
from Hot Weekly Questions - Mathematics Stack Exchange
I was hoping to show this by Lagrange's theorem. As the general linear group over $\mathbb{F}_p$ (the field with $p$ elements), $\mathrm{GL}_n(\mathbb F_p)$ has order $(p^n - 1)(p^n - p)\cdots(p^n - p^{n-1})$, we just need to find a subgroup that has an order of $n!$.
We know that $S_n$ (symmetric group) has order $n!$ and since it is isomorphic to the set of permutation matrices with matrix multiplication as its operation, we already have or proof constructed.
My problem here is that it works for every $p \in \mathbb{N}$, it is not required for $p$ to be prime. Where am I wrong? Am I wrong?
from Hot Weekly Questions - Mathematics Stack Exchange
When I browsed Zhihu(a Chinese Q&A community), I met this question. That is
Let $\{a_n\}$ be recursive s.t. $$a_1=2,\ a_{n+1}=\ln |a_n|(n\in \Bbb N).$$ Show that $\{a_n\}$ is unbounded.
I want to investigate a subsequence $\{a_{t_n}\}$ of $\{a_n\}$, where $t_n$ is greatest integer satisfying $$a_{t_n}=\min_{1\leqslant k\leqslant n}a_k.$$ Thus $a_{t_n}\to -A(<0),n\to \infty$.
However, it helps little with the origin question. So how can I solve it ?
from Hot Weekly Questions - Mathematics Stack Exchange
Let $V$ be a finite dimensional vector space over an infinite field $k$. The ring of polynomial functions on $V$ is the subalgebra of the $k$-algebra of all functions $V\to k$ generated by the dual space $V^*$, and is denoted by $k[V]$.
Let $(e_1,\dots,e_n)$ be an ordered basis of $V$ and let $(f_1,\dots,f_n)$ be its dual basis, then an element of $k[V]$ is a polynomial in $f_1,\dots,f_n$. We can then define a (formal) derivative as follows: First, fix $i\in\{1,\dots,n\}$ and define $$ \partial_{e_i}(f_1^{r_1}\cdots f_{i-1}^{r_{i-1}}f_i^{r_i}f_{i+1}^{r_{i+1}}\cdots f_n^{r_n}) = r_i f_1^{r_1}\cdots f_{i-1}^{r_{i-1}}f_i^{r_i-1}f_{i+1}^{r_{i+1}}\cdots f_n^{r_n}, $$ for all $r_1,\dots,r_n\in \mathbb{Z}_{\geq 0}$. Extending by linearity we obtain a well defined derivation $\partial_{e_i}:k[V]\to k[V]$. Then for $v\in V$, write $$ v = \sum_{i=1}^n a_i e_i, \qquad a_1,\dots,a_n\in k $$ and define $$ \partial_v(f) = \sum_{i=1}^n a_i \partial_{e_i}(f), \qquad \forall f\in k[V]. $$
When we take $V=k^n$ and $(e_1,\dots,e_n)$ as the canonical ordered basis, the $i$-th vector in the dual basis is the coordinate function $x_i:k^n\to k$ given by $x_i(a_1,\dots,a_n) = a_i$, and $k[V]$ is precisely the polynomial ring $k[x_1,\dots,x_n]$ and the derivation $\partial_v$ coincides with the known formal directional derivative on that polynomial ring.
The main issue with this definition is that it depends on the chosen basis $(e_1,\dots,e_n)$. I would like to know if there is a basis-free definition of the derivative $\partial_v$ for a ring of polynomial functions $k[V]$ on a finite dimensional vector space $V$ over an infinite field $k$.
from Hot Weekly Questions - Mathematics Stack Exchange
How big is one coronavirus? How big is a micron? How well do these masks protect against the virus? What can you observe about the size of particles mentioned? Why are we wearing masks?
I've been doing some digging into Surreal Numbers, and I've seen this "inductive" proof of transitivity. Basically, we assume, that there exists the simplest triplet of surreal numbers, x, y, z, such that x<=y, y<=z, but x is not smaller or equal to z. Then, applying the axiomatic definition of what it means to be a number and the definition of <=, we conclude that there exists a simpler triplet satisfying the same property, and hence, contradiction, QED, GG EZ.
My issue is, that we seem to assume that an infinite chain of simpler surreal numbers is impossible, but I'm not convinced that it's true, since some surreal numbers are "created" on an infinite "day", having therefore an infinite sequence of surreal numbers that are simpler.
I am aware that just because a number has infinite "day" of "creation" it doesn't mean that it's possible to construct such a sequence of decreasing simplicity, but I'm not convinced otherwise either.
Here is the document in which I found the proof I'm talking about, maybe I'm missing something. The proof is outlined on page 3, in Theorem 1.1
Also, if anyone knows a better, more mathematically clean description of "day of creation" of a number, please let me know
TL; DR: I'm not convinced to the proof of transitivity of <= over surreal numbers
For fields $K$ and $L$, I am interested in proving that "$GL_n(K)$ and $GL_m(L)$ are isomorphic (as groups) if and only if $m=n$ and $K\simeq L$".
I don't know how generally this is true, but:
assume $K=L$. In that case, if $char(K) \neq 2$ then there is a usual proof that $n=m$ by looking at the group of involutions of each, that yields $2^m = 2^n$ and this the result.
if $char(K) = 2$, I do not find any proof of this fact.
is there a stronger result without supposing $K = L$? Can we at least conclude that $char(K) = char(L)$?
from Hot Weekly Questions - Mathematics Stack Exchange
Let's name each set of more than three points "connectable", when it is possible to connect all of the points that belong to set with n line segments (where n is number of points) in such a way that they create a n-gon that don't intersect itself. Of course, points are placed in two dimensional space.
Here are two examples of connectable sets(Sorry for bad quality of the pictures):
Next example shows points connected in improper way, the created polygon have only six sides, while the set consists of seven points(The orange point lies on the straight line segment).
This set, and the previous one are not connectable.
My question is: Can we easily determine whether given set is connectable or not?
For example the set below looks like if it wasn't connectable, but I didn't managed to prove this.
Thanks for all the help.
from Hot Weekly Questions - Mathematics Stack Exchange
Apparently the book is riddled with small errors / typos. But I can’t find a list of errata or corrections anywhere. Is there such a list?
Ran into the book by mistake and quite like it’s style/*. But not sure I want to commit if I have to clean up after the author constantly / risk wasting time in false bogs.
*(though problems without solutions, especially in grad level math books, is an educational war crime imo. Needless obstacles to use and self-enrichment. Online solutions are fine like Harold Simmon’s intro to category theory. But I digress...)
Are there more than four isosceles, right-angled, INSCRIBED triangles in an ellipse? By inscribed, I mean all three vertexes should lie ON the ellipse.
I am attaching a simplified picture that shows my count of four such triangle. Two of them form the only inscribed square of an ellipse in the middle. The other two would exist at each end.
Are there any proof or studies showing that these are the only four, or are there more?
*Edit: As Deepak pointed, when I said four, I was not careful. If you draw the diagonal line in the inscribed square the other way, you can also have two others.
To give you a better idea, I'm specifically wondering whether there are isosceles right angled inscribed triangles at different locations entirely, as is shown in my supplementary diagram that has a question mark.
Note that in that diagram, I'm using a FAKE isosceles triangle (the two sides are not actually of same lengths), as I cannot find a true isosceles triangle. But I'm wondering whether a real one exists that looks similar to what is drawn.
from Hot Weekly Questions - Mathematics Stack Exchange
I would really like some feedback on this. Please feel free to share your thoughts. I will be very glad for every comment (even negative ones). Please, if you downvote, can you at least explain why. Thanks.
___
This theory helps proving a+b=c, especially if equations are hard to prove (plus some more). We also have unsolved mathematics problems that include relationship between a,b and c and they are hard to prove of disprove. Theory is here:
We have:
a,b,c
a,b,c are any possible numbers
How do we know if sum of any of two numbers (and which ones) is 3rd number?
We use this equation:
(c²-b²- a²):(b*a)=
If:
___
- result is 2, than a+b=c and b+a=c
Proof:
If a+b=c; than c²=a²+2ab+b², so
Equation is: (c²-b²- a²):(b*a)=
= (a²+2ab+b²-b²-a²):(b*a)=2ab:(b*a)=2
___
- result is -2, than a+b (or b+a) is not c. But, if we take bigger number of a and b and we swap it with c, than we will get correct result (for example:a is bigger than b, than:a+b=c is not true; but c+b=a is true (and b+c=a true also)).
Proof:
If a-b=c or(not and) b-a=c; than c²=a²-2ab+b², so
Equation is: (c²-b²- a²):(b*a)=
= (a²-2ab+b²-b²-a²):(b*a)=-2ab:(b*a)=-2
____
- result is different than 2 or -2. Then all of next statements are false:
A+b=c
A+c=b
B+a=c
B+c=a
C+a=b
C+b=a
Proof:
Result of equation is different than in previous cases, meaning that our conclusion is correct.
______________________
Example #1:
a=2
b=4
c=6
a+b=c
2+4=6 : Correct or False?
(c²-b²- a²):(b*a)=2
(6²-4²-2²):(4*2)=
=(36-16-4):8=
=16:8=
=2
Because our result is 2, we know that 2+4=6 is correct.
___
Example #2
a=2
b=5
c=3
a+b=c
2+5=3 : Correct or False?
(c²-b²- a²):(b*a)=
(3²-5²-2²):(5*2)=
=(9-25-4):10=
=(-20):10=
=-2
Because our result is -2, we know that 2+5=3 is not correct. But if we take bigger number of a and b and we swap it with c, than we will get correct result. In this case we swap b and c (this means that equation:2+3=5 is correct).
___
Example #3:
a=2
b=7
c=3
a+b=c
2+7=3 : True of False?
(c²-b²- a²):(b*a)=
(3²-7²-2²):(7*2)=
=(9-49-4):14=
=(-44):14=
=-3,143
Because our result is not 2 and also not -2 we know that all of these statements are wrong:
A+b=c
A+c=b
B+a=c
B+c=a
C+a=b
C+b=a
____
For easier understanding I have used simple examples to show this theory. As mentioned above, it becomes useful when equation is harder to prove or solve.
___
Thank you for reading. Have a lovely day. Thoughts?
Peter Rowlett is a maths lecturer at Sheffield Hallam University. You can find him on Twitter at @peterrowlett.
About Quarto
Quarto is a fun, commercially-available game. It’s played on a \(4 \times 4 \) grid. The game pieces are all unique, and have four attributes taking one of two values: colour, height, shape and whether it has a dimple in the top or not.
Since each combination of attributes is used once each, and each can take two values, the game uses \( 2^4=16 \) pieces. This is the same as the number of spaces on the board, which is \( 4 \times 4 = 16 \).
The aim of the game is to be the player who places the fourth piece in a row that matches in at least one attribute (i.e., the fourth tall piece, the fourth dimpled piece, etc.). The twist? You don’t choose your own piece — your opponent chooses the piece you will play, and you choose theirs.
Quarto game size
Let’s say it’s essential for a game of Quarto to have the same number of pieces as spaces on the board. This is so the game can end in a draw with all pieces used and all spaces filled.
This works for the standard Quarto you buy, because there are \( 2^4=16 \) pieces on a board with \( 4 \times 4 = 16 \) spaces.
Using this constraint, we can’t play on an arbitrary-sized board, even if we vary the number of attributes we use. For example, we can’t play on a \( 3 \times 3 \) board, because this would have \( 9 \) spaces but three binary attributes would only generate \( 2^3 = 8 \) pieces.
Stop reading and have a think about what size boards work for valid games? How many pieces would these use?
Rather than thinking about numbers of pieces directly, it helps to think about the number of attributes we have. If we have \(m\) binary attributes (each taking two values), this generates \(2^m\) pieces.
Say we play on an \( n \times n \) board, using pieces which have \( m \) binary attributes. Then we require
\[ n^2 = 2^m \text{.} \]
This can be rearranged to give
\[ m = 2 \log_2(n) \text{,} \]
meaning, for a viable game (i.e. \(m\) and \(n\) as positive integers), we need \(m\) to be even and \(n\) to be a power of \(2\).
For example, we can play Quarto on a \( 8 \times 8 \) board if we use pieces with \(6\) binary attributes, because \( 8^2 = 2^6 \); and we can play Quarto on a \( 128 \times 128 \) board if we use \( 14 \) attributes, because \( 128^2 = 2^{14} \).
Notation for Quarto pieces
How would we arrange a game using more attributes? We could invent new attributes for our game, say making pieces spotty or not, have handles or not, and … You can see that as we increase the number of attributes, we are going to run out of ways to make pieces quite quickly.
One way around this is to invent a notation. Since attributes can take two values, label them \(0\) or \(1\). For example, label the pieces in standard \(4 \times 4 \) Quarto as follows: white (\(0\)), black (\(1\)); short (\(0\)), tall (\(1\)); round (\(0\)), square (\(1\)); dimpled (\(0\)), flat (\(1\)).
Using this labelling in this order, we can represent Quarto pieces as binary strings. For example, \(0100\) would be white, tall, round and dimpled; \(1001\) would be black, short, round and flat-topped. The winning condition is met if four pieces in a line share one digit in common. For example, \(0100\) and \(1001\) are both round, so both have third digit \(0\).
You now have the notation you need to play Quarto with larger numbers of attributes, or in settings where you don’t have a shared game board. I recently enjoyed a game of \( 4 \times 4 \) Quarto over Slack; some years ago I spent a summer playing \( 8 \times 8 \) Quarto via Twitter with a student.
More flexibility
To generalise further, we could think about attributes which have more different values, e.g. three different colours, seven different heights, etc. Say we have \(m\) different attributes each taking one of \(k\) different values, then we have \(k^m\) pieces and require that
\[ k^m = n^2 \text{.} \]
This means we can play using our \( 3 \times 3 \) game board after all, by using two attributes taking each of three values (\(k=3\),\(m=2\)), for:
\[ 3^2 = 3^2 \text{.} \]
Higher-dimensional thinking
But why stop there? Just like we can make 3D Noughts and Crosses by stacking three boards on top of each other to form a cube, and 4D Noughts and Crosses by stacking three cubes on top of each other (in the fourth dimension) to make a hypercube, we can play Quarto in different dimensions.
A Quarto board with side length \(n\) on a board in \(d\) dimensions, would have \(n^d\) spaces.
\[ n^d = k^m \iff m = d\log_k(n) \]
for a valid game, with \(n\), \(d\), \(m\) and \(k\) all positive integers.
So now you can play Quarto on all sorts of boards! Setting \( n = k^p \) to give \( m = dp \) allows viable game configurations to be generated for positive integer \(p\).
For example, playing pieces which have 15 ternary attributes on a \( 27 \times 27 \times 27 \times 27 \times 27 \) 5-cube would use \( 3^{15}=27^5=14,\! 348,\! 907 \) pieces.
For a simpler example, if you have a standard Quarto set, you can play using the standard binary pieces on a \( 2 \times 2 \times 2 \times 2 \) hypercube.
You might wonder how to play on a \( 2 \times 2 \times 2 \times 2 \) hypercube. Here I took inspiration from David Butler’s 4D rules for Noughts and Crosses. Because there are just so many ways to win with lines in 4-dimensions, it seems to make sense to make a plane be a winning move. Visualising planes in 4D is mind-bending, so I present a 4D Quarto worksheet to try to help you through. Take a look, have a play and let me know how you get on!
Zeno Rogue – A truly self-referential formula
Zeno Rogue works mostly in theoretical computer science and non-Euclidean geometry. They tweet at @zenorogue, and made the non-Euclidean game HyperRogue.
In Match 2, Sam Shah has shown us the formula known as the Tupper’s self-referential formula. This is a formula which, where we draw its graph at a specific point, can be read as the formula itself.
But is it really self-referential? I would argue that it is not. It is possible to encode all 17 pixel tall pictures made of white and black squares into integer values (by replacing all black squares as 1 and white squares as 0, and considering the result, read column by column, as the binary representation of a number \(k\)). Tupper’s formula works like a decoder: given the number encoding (by looking close to \(k\)), we get back the original picture.
You could say, so what? It can encode anything, so it can be used to draw itself, so it is self-referential! The problem here is that this formula is not self-referential by itself — it can only draw itself when supplied with external information (the correct window of arguments). Being able to produce a copy of self, without external information, is a much more difficult task! Tupper himself did not call his formula self-referential.
Let’s try to encode the data in the formula itself. Let \(\frac{1}{2} < p(x,y)\) be Tupper’s formula, and \(k\) be the value where we get the self-referential window. We could try something as follows:
If we draw the whole graph of this, we can read Tupper’s formula. But this is not exactly the formula above!
Of course, we can change the value of \(k\), to get another image as a result. We could choose \(k_1\) which reads the formula above. But it still refers to \(p\) which is not defined!
Of course we can replace \(p\) by the definition of Tupper’s formula. This will be a bit better, but the formula still does not tell us what \(k_1\) we shall use!
We can try to replace \(k_1\) in the formula with our current value of \(k_1\), and get a new \(k\) (let’s call it \(k_2\)) which produces the formula with \(k_1\). Still not self-referential — as we can easily imagine, \(k_2\) will be a different number than \(k_1\) (it will be greater)!
We can try to continue this process: get a \(k_3\) which displays \(k_2\), then get an \(k_4\) which display \(k_3\), and so on. But that does not help: the \(k\)-number we use will be always greater than the \(k\)-number we get as a result. It appears that we cannot get a truly self-referential formula using this method. But is it really the end? Maybe it is possible by using some other method? If possible, would not it be much more impressive?…
It is a bit difficult to find a truly self-referential formula if we think about numbers and pure mathematics. So let’s move to the programming world, where everything will be easier. In Python we can write a program print(k) which prints whatever is given as k.
This program is very similar to Tupper’s formula: it can output anything, given the correct k. It is “self-referential” in the same sense as Tupper’s formula is: if k is "print(k)", it will print itself. We could attempt to write a program like print("print(k)"), or print("print(\"print(k)\")") — these attempts are similar to the our attempts at making a truly self-referential formula, and fail for similar reasons.
But if we think a bit, we can actually solve this. The trick is that we do not really have to make the value \(k\) include a description of itself! We can have \(k\) which lists the whole program except for the value of \(k\); we can let our program fill the ‘hole’ itself. This works as follows:
k=
def insert(text, q):
# insert a quotation of q after the second character of text
# return text
s = replace(k, k)
print(s)
except that after “=” we should insert the quotation of the program above. Look, a truly self-referential program! We have left some details as an exercise, but if you do not want to have some programming fun, look here. Such programs are called quines.
Now let’s go back to the world of pure mathematics. Can we use the same trick? It is difficult, but we can! We already have print(s): this is the Tupper’s formula. We have to write a mathematical function ‘replace’ which checks that the two coordinates \(k\) and \(s\) are in the following relation: Tupper’s images of \(k\) and \(s\) are similar, except that \(s\) inserts some extra columns after the column \(i\) which contain a drawing of the number \(k\). This can be done, as follows:
\[ s = k{\rm\ mod\ }2^{17i} + \lfloor k / 2^{17i} \rfloor k \cdot 2^{17i} \cdot 2^{170 \lceil \log_10 k \rceil} + \sum_{j} 2^{17i+170j} d(\lfloor k/10^j\rfloor{\rm\ mod\ }10) 2^{17i} \]
where \(d(0)\) to \(d(9)\) are 10×17 Tupper representations of the digits 0 to 9.
The whole self-referential formula will be:
\[ k=\ \wedge s = k{\rm\ mod\ }2^{17i} + \lfloor k / 2^{17i} \rfloor k \cdot 2^{17i} \cdot 2^{170 \lceil \log_10 k \rceil} + \sum_{j} 2^{17i+170j} d(\lfloor k/10^j\rfloor{\rm\ mod\ }10) 2^{17i} \wedge \frac{1}{2} < p(x,y+s) \wedge y>0 \wedge y < S \]
but where we first replace \(i\) and \(d\) with their correct values, \(S\) by the size, and then we replace \(p\) with Tupper’s formula. After this, we insert the Tupper representation of the obtained formula after \(k=\). And this way, we get a truly self-referential formula!
This puzzle actually has very deep consequences. For example, we can construct a formula \(\phi\) which says that:
“preceded by its own quotation cannot be proven” preceded by its own quotation cannot be proven
Note that \(\phi\) works similar to our quines above: it actually talks about itself! \(\phi\) says that \(\phi\) cannot be proven. It could be false or true. If it was false, this would mean that \(\phi\) could be proven, and hence \(\phi\) would have to be true. So \(\phi\) is a formula which is true but could not be proven. So not everything that is true in mathematics could be proven. (Note: it is very easy to misunderstand Gödel’s theorem — claiming that we understand it after reading this short, popular explanation is definitely not recommended.)
Another consequence of self-reference is the life itself. Live beings are also able to create copies of themselves. If you know a bit about biology, you should recognize the parts of our quine in our cells. The number \(k\) is our genetic code. We have a biological machinery \(p\) which can use the genetic code to create proteins; this machinery can construct a copy of itself, given the correct genetic code. However, this machinery does not construct a copy of the genetic code itself; the genetic code is simply copied using another special biological machinery \(r\). I have read some popular articles about the origin of life, claiming that chemical molecules such as RNA and proteins could appear spontaneously on the young Earth quite easily, and that’s how life was created. But is it really life? I could believe that something like a biological Tupper’s formula could easily appear spontaneously, but to create life, we actually need all three components \(k\), \(p\) and \(r\) at the same time: \(k\) is worthless without \(p\) to copy it, and \(p\) is worthless without \(k\) to describe it. It seems that the spontaneous creation of life is as likely as randomly writing out the self-referential formula above! (But if the universe is infinite, it must happen somewhere!)
So, which bit of maths made you say “Aha!” the loudest? Vote:
Note: There is a poll embedded within this post, please visit the site to participate in this post's poll.
The poll closes at 9am BST on PUT THE DATE IN HERE, when the next match starts.
If you’ve been inspired to share your own bit of maths, look at the announcement post for how to send it in. The Big Lockdown Math-Off will keep running until we run out of pitches or we’re allowed outside again, whichever comes first.
I don't know how to explain why the graph is shaped like it is, or what the structure in it means, or why it seems like a "thick" logarithmic curve.
