Comments on: Significand of the figure

Significand of the figure

Talkie Toaster — Thu, 08 Nov 2012 08:18:38 -0800

From the somewhat arbitrary and dimensionful to the timeless classic, one man's subjective list of notable properties of specific numbers.

Some have stories attached, others appear to be gleaned from trawling The On-Line Encyclopedia of Integer Sequences (previously: I and II). Some other people's takes on the same idea. Previously, from the same author. Previously with broken links.

By: Alterity

Alterity — Thu, 08 Nov 2012 08:27:59 -0800

Cracked.com's Ten Fucked Up But True Numbers Under Eleven That Might Get You Laid in a Zombie Apocalypse

By: Wolfdog

Wolfdog — Thu, 08 Nov 2012 08:34:17 -0800

The reciprocal of 17, 1/17=0.05882352941176470588235..., has a 16-digit repeating decimal, which is the longest possible. I know what he's trying to say, but he hasn't said it.

By: unSane

unSane — Thu, 08 Nov 2012 08:37:16 -0800

What IS he trying to say?

By: notyou

notyou — Thu, 08 Nov 2012 08:41:01 -0800

It would be cooler if the reciprocal had a 17 digit repeating decimal pattern.

By: ubiquity

ubiquity — Thu, 08 Nov 2012 08:42:48 -0800

It can't. Just like 1/7 couldn't have more than a 6 digit repeating pattern.

By: notyou

notyou — Thu, 08 Nov 2012 08:44:02 -0800

Really? That's curious. Why not?

By: DU

DU — Thu, 08 Nov 2012 08:48:28 -0800

Try doing 1/7 division long hand and I think it'll become clear why not.

By: JHarris

JHarris — Thu, 08 Nov 2012 08:52:05 -0800

In the context of the article, it's more obvious that he means that a 16-digit repeating decimal is the longest possible reciprocal repeating decimal, which caught me up when I read Wolfdog's summation. It's possible to have longer repeating decimals that are not of the form 1/(integer). Aren't transcendental numbers effectively repeating decimals of infinite length?

By: benito.strauss

benito.strauss — Thu, 08 Nov 2012 08:53:25 -0800

If you do long division, dividing 7 in to 1.00000000000...., think about the "left over" amount at the bottom, before you bring down the next digit. It obviously has to be less than 7, and (less obviously) it can't be zero. Also, you're always bringing down a '0'. So once you get a '10', you're back in the same exact situation as when you generated the first digit, at which point the process repeats. And there where only six possible values you went through (10, 20, 30, 40, 50, 60). (It's not the most rigorous proof, and it assumes people still do long division.)

By: komara

komara — Thu, 08 Nov 2012 08:54:49 -0800

I just did long division on paper for the first time in ... lord, at least 18 years. I want to say "Thanks, DU" but then it occurs to me I don't really mean it.

By: slogger

slogger — Thu, 08 Nov 2012 08:57:51 -0800

A few rules I used in this list: Everything can be understood by a typical undergraduate college student. Heh. I don't think typical means what he thinks it means.

By: benito.strauss

benito.strauss — Thu, 08 Nov 2012 08:58:24 -0800

To clarify, 1 / N expressed in as a repeating decimal when N is an integer, always has a repeating section length less than N. (The argument above works the same for 1/7, 1/37, or 1/137). But there are many N's where the length of the repeating block is less. E.g. 1/11 = 0.09090909.... Since all number nerds know this fact as second nature, it's just an interesting thing to note when a particular integer requires this maximum.

By: notyou

notyou — Thu, 08 Nov 2012 08:59:48 -0800

It obviously has to be less than 7... Ahh. I see. Thanks. Thanks, also DU for the suggestion to do long division. It was more satisfying than I expected it to be.

By: RobotVoodooPower

RobotVoodooPower — Thu, 08 Nov 2012 09:04:44 -0800

This fellow sure has a lot of er, data published on his web site, including core values for men and movie notes. And it's good to know if I ever need the derived RGB values for the Apple ][ colors I know where to find them. There's something nostalgic about the "here is an arbitrarily organized brain dump of every wacky thing I've ever thought about" web site format in this age of the chronological blog.

By: brenton

brenton — Thu, 08 Nov 2012 09:08:46 -0800

I was sad to see that .866025403784438... is not on there. I have been looking into time dilation and relativity lately and found the number interesting. It's the speed (fraction of the speed of light) that you have to travel which makes time go exactly twice as fast for you, relative to your starting point. Being the esoteric nerd that I am, I posted .8660254037844386521c as my status and forgot about it until a few days later when a friend randomly sent me the following message without context: cos(π/6) I actually didn't know what she was referring to. But the next day she sent me this: sqrt(3)/2 It turns out that she is a math tutor and runs into the decimal .8660254 all the time and it looked familiar, so she took a few days to figure out why. It was a special moment of connection. Here I had posted what I thought was an utterly uncrackable inside joke about the significance of a number and someone else discovered a totally different significance for the number. Math is weird.

By: Egg Shen

Egg Shen — Thu, 08 Nov 2012 09:11:45 -0800

If you'll indulge me, here is my attempt to explain Graham's number.

It begins with mathematicians coloring the edges of n-dimensional hypercubes. Why would they do such a thing? I don't know. Maybe they wanted to gussy them up a bit. Anyway, the mathematicians wanted to know how many dimensions your cube would have to be in order to guarantee that a certain method of coloring its edges would contain at least one example of a particular result. If you want to understand what the method is - or what result they're looking for - you'll have to ask someone else. I'm a long way from understanding that part of it. So how many dimensions does the cube have to be? The mathematicians don't know exactly. But they've narrowed it down. It must be at least 13. But it won't be any larger than Graham's number. [As in, Ronald Graham. He's the guy who invented the "Erdos number".] Then what is Graham's number? Before you can talk about Graham's number, you have to understand Knuth's up-arrow notation. [As in Donald Knuth. He's the guy who created the TeX typesetting system.] Now, regular exponents are often represented by a single up-arrow. A^B = A*A*A...*A with there being B copies of A. 3^2 = 3*3 = 9 3^3 = 3*3*3 = 27 3^4 = 3*3*3*3 = 81 What Knuth did was define a system using additional arrows. A^^B = A^(A^(...^A) with there being B copies of A [With exponents, you always work from right to left.] In other words, A raised to the power of itself in a tower of exponents with B levels. 3^^2 = 3 to the power of 3 = 27 3^^3 = 3 to the power of (3 to the power of 3) = 3 to the power of 27 = 7,625,597,484,987 That's 7.6 trillion - which is nothing to sneeze at. But it's visualizable. If you took 3 Sears Towers and filled them with pennies, that's more or less the number of pennies you'd have. 3^^4 = 3 to the power of (3 to the power of (3 to the power of 3)) = 3 to the power of 7,625,597,484,987 Here we have gone beyond the visualizable. By comparison, the number of Planck volumes - the smallest volume in which the known laws of physics make any sense - contained in the entire known universe is 10 to the power of 185. So with 3 to the power of 7,625,597,484,987, we can safely say that we have entered the realm of Big. Now let's add a third arrow: A^^^B = A^^(A^^(...^^A)) with there being B copies of A [As with exponents, when using arrows, you work from right to left.] Note how, as before, the number of arrows between terms on the right side of the equal-sign is one less than the number of arrows between terms on the left side of the equal-sign. 3^^^2 = 3^^3 = 7,625,597,484,987 3^^^3 = 3^^(3^^3) = 3^^7,625,597,484,987 In other words: an exponent tower of 3's that is stacked 7,625,597,484,987 levels high. To give some idea of this: if it took you a second to calculate each new level of the stack, it would take more than 240,000 years to finish. Now we are in the realm of Stupid Big. Let us call this number X. 3^^^4 = 3^^(3^^(3^^3)) = 3^^X = An exponent tower of 3's that is X levels high. 3^^^5 = An exponent tower of 3's that is 3^^^4 levels high. 3^^^6 = An exponent tower of 3's that is 3^^^5 levels high Continue this series until the number after the three arrows is X. This last number equals 3^^^^X. Let us call this Insanely Big number G1. And remember that it took only four arrows to get there. Now things get interesting. G2 = 3^^...^^3 with there being G1 arrows. G3 = 3^^^...^^^3 with there being G2 arrows. Continue this series until you reach G64. THAT is Graham's number.

By: JHarris

JHarris — Thu, 08 Nov 2012 09:12:24 -0800

Is there an i like constant that is completely inexpressible in electronic form? If so, I move that we name it .

By: benito.strauss

benito.strauss — Thu, 08 Nov 2012 09:12:26 -0800

BTW, for me the fact that iⁱ is a Real number has always endeared it to me. First you you get weird and create an imaginary number, i. Then you push the weirdness by defining what it means to raise one imaginary number to the power of another imaginary number — and it looks nothing like "Oh, 2³ just means multiply 2 by itself 3 times". And what happens when you raise i to itself? You land back on the real number line. And the value you land on isn't 1, or 1/2, or something difficult to evaluate. It's e^-π/2, a small greeting committee composed of the most distinguished exotic real numbers, pi and e. It's like you've got a bunch of weird friends, and one of them decides one day that the strangest thing they can do is live an utterly conventional life.

By: kmz

kmz — Thu, 08 Nov 2012 09:19:15 -0800

Aren't transcendental numbers effectively repeating decimals of infinite length? Not really? Because there's never actually a sequence that repeats. A repeating decimal has to be rational, pretty much by definition. Note that there can still be patterns though in a transcendental number. Something like 0.10010001000010000010000001... Can't remember off the top of my head if that particular one is actually transcendental, but you get the idea.

By: eruonna

eruonna — Thu, 08 Nov 2012 10:14:08 -0800

Jharris, there is Chaitin's constant Ω, the probability that a random Turing machine will halt. It is uncomputable (on a machine of the same type it is the probability for). By a counting argument, there are uncountably many uncomputable real numbers, in fact.

By: invitapriore

invitapriore — Thu, 08 Nov 2012 10:24:29 -0800

The infinitely vast majority of them, even.

By: invitapriore

invitapriore — Thu, 08 Nov 2012 10:25:24 -0800

...a fact which scares me on a Lovecraftian level that I don't really understand.

By: benito.strauss

benito.strauss — Thu, 08 Nov 2012 11:29:36 -0800

kmz, the example you gave is almost the Liouville constant, which has the nice distinction of being the first number that was rigorously proved to be transcendental.

By: stebulus

stebulus — Thu, 08 Nov 2012 12:33:44 -0800

brenton: It turns out that she is a math tutor and runs into the decimal .8660254 all the time and it looked familiar, so she took a few days to figure out why. By the way, for tasks of this sort, the Inverse Symbolic Calculator is a handy tool. Doing a lookup of 0.8660254 finds sqrt(3)/2 pretty quickly. someone else discovered a totally different significance for the number. Here's an argument that it's not really totally different. In that thought experiment with the beam of light bouncing between two mirrors which is often used to derive the formula for time dilation — e.g., as seen in Wikipedia's article on time dilation — there's a right triangle whose hypotenuse is the distance travelled by the light beam going from one mirror to the other according to the moving observer (D in this figure from the Wikipedia article), one of whose legs (L) is the distance travelled by the light beam according to the stationary observer, and the other of whose legs is the distance travelled by the moving observer themselves (D*v/c). You want D/L to be 2, so the angle opposite L is π/6 and v/c = cos(π/6).

By: kengraham

kengraham — Thu, 08 Nov 2012 12:56:34 -0800

This is an awesome FPP, and benito.strauss's "...It's like you've got a bunch of weird friends, and one of them decides one day that the strangest thing they can do is live an utterly conventional life." an awesome comment. It was said of (I think) Srinivasa Ramanujan that every integer was one of his personal friends; benito.strauss keeps more transcendental company that Ramanujan. Heh. I don't think typical means what he thinks it means. I'm in my very first semester teaching math to more-or-less typical undergraduates, and the stuff they can discover for themselves with the correct prodding is pretty amazing. You're right, though, that they'll not know what all of the words in TFA mean.

By: Talkie Toaster

Talkie Toaster — Thu, 08 Nov 2012 12:59:35 -0800

and the stuff they can discover for themselves with the correct prodding is pretty amazing ... while there are PhD students and postdocs out there who can be taken right to the water but still refuse to drink. I'm probably one of them.

By: Jpfed

Jpfed — Thu, 08 Nov 2012 13:42:35 -0800

stebulus - relativity noob here. After reading the articles you linked, it's clear enough how the Lorentz factor emerges in time dilation. But doesn't it show up in a bunch of other contexts? Or do those other equations with the Lorentz factor all involve time as well? (I ask this because I was lied to in my education regarding "relativistic mass" and now I'm super confused.)

By: kengraham

kengraham — Thu, 08 Nov 2012 14:50:52 -0800

... while there are PhD students and postdocs out there who can be taken right to the water but still refuse to drink. I'm probably one of them. I'm definitely one of them, sometimes; everyone I know is, I think. Math has the weird feature that sometimes, when the distance in question is between oneself and the water, Zeno's paradox actually takes effect. (The difference, of course, is that the stuff I mentioned my students discovering is fairly basic, but more complicated than I would have expected them to spot on their own. The frustrating converse also happens, of course.)

By: madcaptenor

madcaptenor — Thu, 08 Nov 2012 15:57:41 -0800

Since all number nerds know this fact as second nature, it's just an interesting thing to note when a particular integer requires this maximum. As a number nerd, I checked the OEIS to see which integers have this property. It's A001913, and the sequence starts 7, 17, 19, 23, 29, 47...; assuming the generalized Riemann hypothesis the proportion¹ of all primes that have this property is Artin's constant (about 0.374). So it's not all that rare of a property. 1. If you know enough to object to my use of the word "proportion" here you probably also know how to fix this.

By: 23

23 — Thu, 08 Nov 2012 17:17:08 -0800

Aren't the years of groundless slander enough without even technical resources calling me a cult leader?

By: benito.strauss

benito.strauss — Thu, 08 Nov 2012 17:29:29 -0800

madcaptenor, does it have to be a prime to have period n-1? I can't think of any off- hand composite with that property, but my number theory sucks and I can't come up with a proof that it can't?

By: Wolfdog

Wolfdog — Thu, 08 Nov 2012 18:32:16 -0800

does it have to be a prime to have period n-1 Yes. The period of the decimal representation of 1/n is the same as the order of the number 10 in the multiplicative group of units mod n. If n is composite, that group has order strictly less than (n-1), and so the order of 10, as a group element, is also strictly less than (n-1).

By: stebulus

stebulus — Thu, 08 Nov 2012 18:38:03 -0800

Jpfed: relativity noob here. After reading the articles you linked, it's clear enough how the Lorentz factor emerges in time dilation. But doesn't it show up in a bunch of other contexts? Or do those other equations with the Lorentz factor all involve time as well? Dunno. I don't know much about relativity either. I'll ask physicsmatt to drop by.

By: physicsmatt

physicsmatt — Thu, 08 Nov 2012 21:12:23 -0800

Hi, stebulus asked me to check in to answer the relativity question. Apparently I really don't want to do work tonight, so let's do this thing. Definitions first: beta is the normalized speed beta = v/c (since v < c, beta < 1). The Lorentz factor, gamma = 1/sqrt(1-beta^2) is therefore > 1, and it will show up in a host of places in relativity, not just in time dilation. There's no way for me to go through all of it here, but let me try a few examples, as the multitude of places in which it shows up is how relativity avoids all the weird paradoxes that would seem to be inherent. First, there's the time contraction. If you travel past me at speed beta, I'll see your clock ticking at a rate slower than mine. The time dilation factor will be gamma. That is, for every t seconds passing for me, I'll see tau=t/gamma pass for you (since gamma > 1, t > tau). In addition, I will see you compressed in the direction of travel. So if your spaceship is length L, if I measure it as it passes, I will see it as length L/gamma. Of course, you see me traveling past you, so you see my clock ticking slower and me getting the length contraction. Both of these effects are NECESSARY for relativity to work. Here's why. Imagine we're in a situation where you will make a trip that I measure as being two units of length long (say, 2 light-years). You travel at beta = 0.866... (gamma = 2), according to me. So I see you finishing the trip in 2/0.866 = 2.31.. years. In that time, I see you aging by 2.31/2=1.15... years (applying the Lorentz time dilation). But what do you see? You see yourself stationary (you're not accelerating, after all), and the Universe moving past you at beta=-0.866 (the minus sign here means "backwards") and gamma =2. So you don't see yourself aging "slowly," and if there was no length contraction, we could never agree on what happened (namely, how old are you at the end of the trip). However, the Universe is moving past you, so you see the distance you must travel shortened by that factor of gamma. So instead of 2 ly, you see the trip as only 2/2 ly = 1 ly long. How long does it take for you to travel that distance? 1 ly/0.866c = 1.15.. years. So both you and I will agree on how old you are when you get done. When it comes to observables that we both can measure, physics has to give us the same answer. (many other paradoxes of relativity are resolved by the fact that not all observers can measure everything simultaneously, but that's not that situation here). Clearly, there's some symmetry underlying this. The thing that's preserved by "boosts" (that is, moving from one frame of reference to another in relativity) is the space-time interval between two events: s^2 = c^2t^2-x^2-y^2-z^2 where t is the time difference between the two events, and x, y, and z are the difference of space coordinates between the two events. So, in our previous example, you see the "events" of departure and arrival as occurring at the same place (where you are), so the interval is only in the "time" component. I see the interval as occurring with a space separation of 2 light-years, which therefore means that the TIME interval I see the trip taking must be longer. This is because of that all important minus sign between the time and space parts of the interval. Without that minus sign, our Universe would be a very different place, and the fact that the 4th dimension (time) has a different sign than the other 3 (this is called "the metric") is probably one of the factors that defines what time is (that is, why is time so unusual compared to space, if they're all just dimensions?) As you can see, relativity mixes up space and time but preserves a particular combination, and so the gamma factor must effect both of them. Similarly, relativity mixes up energy and momentum. Just as the interval is conserved in boosts, it turns out that there's a combination of energy E and momentum p that's conserved: E^2-p^2. This combination turns out to be that mass squared of an object: m^2 = E^2-p^2. Now, sometimes you'll hear that moving particles gain mass. This is wrong. What happens is that the *energy* increases by a factor of gamma. A stationary particle just has mass m as energy, and so a moving particle has energy E = gamma m. If you call the energy = mass, you'll think that the mass went up. But really, that's not true, the combination E^2-p^2 is the same (meaning that in the simple case momentum went to gamma*beta*m, it turns out). If you assume that v << c (beta very small), then we can expand gamma E into m+1/2 v^2 m+...; the 2nd term here is why we living in the slow non-relativistic world normally call "kinetic energy." So that's a taste of the gamma factor in relativity. In general, the Lorentz transformation acts on pairs of variables (like t and x, or E and p), to maintain some invariant (the interval s, or the mass m). Interestingly, the Heisenberg uncertainty principle links t and E as well as x and p. So you know, consider that.

By: physicsmatt

physicsmatt — Thu, 08 Nov 2012 21:19:11 -0800

Clarification, if you're accelerating, life gets more complicated, and you move from special relativity to general relativity. Most everything I said still holds true, but you can get yourself into massive trouble if you blindly apply SR equations to GR without correcting for the fact that acceleration is occurring. For example, the famous "twin paradox," which is "how old are each of us after you return from your trip of 2 ly at beta = 0.866, if we both see each other's clocks ticking slowly?" can only be resolved by noticing that "returning" means you decelerated and then accelerated back, and so the situation between you and me is no longer symmetric (you accelerated, I didn't, so you aged less than I did).

By: neuron

neuron — Thu, 08 Nov 2012 22:45:11 -0800

The Penguin Book of Curious and Interesting Numbers

By: stebulus

stebulus — Fri, 09 Nov 2012 07:58:55 -0800

Thanks, physicsmatt. That thing about space-time interval being the preserved quantity certainly clarifies why the factor appears in both time contraction and length contraction.

By: Jpfed

Jpfed — Fri, 09 Nov 2012 10:41:03 -0800

Thank you so much! That was much clearer than anything I got in any classes I took. In general, the Lorentz transformation acts on pairs of variables (like t and x, or E and p), to maintain some invariant (the interval s, or the mass m). Interestingly, the Heisenberg uncertainty principle links t and E as well as x and p. So you know, consider that. I've heard that there is some difficulty in reconciling QM with GR; does this have something to do with that?

By: Wolfdog

Wolfdog — Fri, 09 Nov 2012 12:07:46 -0800

there is some difficulty in reconciling QM with GR That is tastefully understated.