Mark Frauenfelder at 2:30 pm Thu, Jan 29, 2009
ADVERTISE AT BOING BOING!
Here's a clock for geeks. It's $20 on Etsy.
Explanation of clock numerals:
1 Legendre's constant
2 "An infinite number of mathematicians..."
3 A Unicode character as a HTML entity
4 Modular arithmetic
5 The Golden Mean reworked a little
6 Three Factorial (3*2*1)
7 6.99999.. Though a different number than 7, still equals 7
8 Graphical representation of Binary code
9 An example of a base-4 number
10 A Binomial Coefficient
11 An example of Hexadecimal encoding
12 The cube root is the inverse of 12^3
1 = .999…
1 = 1/3 + 1/3 + 1/3
1/3 = .333…
1 = .333… + .333… + .333…
Maybe the little black line marking 6.9 is actually imperceptibly ever so slightly moved in the anti-clockwise direction, so that it does mark 6.9 rather than 7?
#38: The people who are bickering about the 7, don’t actually understand the 2 and thus accept it on faith of authority (probably because it explicitly incorporates a formal expression).
#35 js7a / #36 coreymaley : js7a’s notation is nonsensical, but we can ask:
does 2 – 0.999… == 1?
And the answer is yes.
Best clock EVER!
@anonymous #22, it’s because 2*4=1 (mod 7).
It’s all in how you define division. One definition of a/b is “what number, multiplied by b, gives a?”. Here, we have 1/2; in other words: “what number, multiplied by 2, gives 1?” When working with integers modulo 7, the answer is 4 (as above).
When working with ordinary integers, there is no such number (the answer doesn’t exist). With integers modulo 8, there is also no such number.
When working with fractions or reals, the number is a half.
The practical importance of this is mostly in cryptography; several well-known ciphers (including RSA) rely on related facts about modular arithmetic.
for a more specific, albeit not as funny, explanation of 2.
I bet we can get a good argument going over “does 6.999… really equal 7.0? and is it really a different number”.
Noen, I just wrote: In before arguments about “0.9999.. = 1″
Obscure; shallow; and ugly. If I wanted to see that everyday, I’d hang a mirror.
Also, 6.999… is not a “different number than 7″, it is 7.
More precisely, 4 is the multiplicative inverse of 2 in Z^x_7 : 4 mod 7 * 2 mod 7 â‰¡ 1 mod 7, so 4 mod 7 â‰¡ 2^-1 mod 7. The additive inverse in Z_7 would be -2 mod 7 â‰¡ 4 mod 7.
Whenever this .999… thing comes up, I like to ask this: does 1.000…0001 = 1?
if you believe 6.999… is equal to 7, then it is the same number. Two equivalent numbers are the same number. If they are not equivalent, then they are not the same number. But I bet we all agree (except for Mark, maybe) that you can’t have two quantities that are equal but not the same number. No need to debate the 0.999… = 1 thing.
“No need to debate the 0.999… = 1 thing.”
Ah ha ha hahah hah ha…
Are you new to this, or have you just suppressed the horror?
Unfortunately, 1.000…0001 is a nonsensical notation. The “…” ellipsis doesn’t stand for some really large number, like what you’re suggesting. It’s infinite. So it doesn’t make sense to have an infinity of 0′s, and then something after that. It’s not the name of any number, and so it’s not a number at all.
this enrages me since I am too stupid to understand it!
I think the best way to express the 6.99… and 7 thing is to say that they’re different names for the same number, much like 0.5 and 1/2, or 111 (base 2) and 7 (base 10).
A lot of confusion can be avoided by carefully distinguishing the expressions we use to refer to the numbers (the numerals, along with decimals and slashes), and the numbers themselves (whatever kind of abstract object those are).
6.999… is equal to 7, but it is a different symbolic representation.
So same value, different representation.
Sort of how 0.0, -0 and 0 all have the same value, but are different symbolic representations.
Since one definition of ‘number’ is ‘symbolic representation’, then yes – different numbers.
The whole thing hinges on whether you believe that a symbol and the value it represents are two different things or not.
Ceci n’est pas une pipe.
A friend of mine has a similar, but not identical, clock. If I remember correctly, it also uses various geekish symbols such as scientific constants and mathematical symbols, but doesn’t constrain itself to the traditional twelve clock positions. So it’s got π instead of three, but slightly past the three position (because it’s 3.14…).
To quote a former bar patron of great renown:
Here’s another geek clock on etsy, I got this one for my Nobel Laureate physicist father for Christmas. He got a big kick out of it. Then again, physicists are not known for having the best taste in home decor.
re: 6.99999 vs. 7
it’s a math clock, not an atomic clock. One significant digit please.
Easy proof that 0.999~ does not = 1. Let’s compare repeating digits in decimal and hexadecimal notation. Hex numbers are preceded by h; decimal by d
Assertion: h0.FFF~ = 1 (By similar rule that d0.999~ = 1)
hF > d9 (Hope you agree)
h0.FFF~ = d0.999~ (Wait, I thought F > 9…)
hE < hF (Hope you agree)
h0.EEE~ < h0.FFF~ (Hope you agree)
hE > d9 (You see where I’m going here…)
h0.EEE~ < d0.999~ (???)
Zeno's Paradox approach: You need to get to one. I give you 9/10. Now you need an additional 1/10. I give you 1/100. And so forth. Doing this an infinite number of times is not going to help.
And, no, I’m not talking about the number moving or approaching anything, I’m talking about how to determine where that point is by evaluating its representation. I get the feeling that some refutations confuse these two points.
The problem is that saying “an infinite number of nines” is equivalent of saying “and then a miracle occurs”. Yes, there is a place on the number line that represents 1/3, you just can’t accurately express where that place is with base ten digits, no matter how long you try. Claiming that it works out if you go to infinity doesn’t help.
In short, a repeating decimal is not a number any more that 1/0. It may be used as a shorthand for a concept, but you can’t do real math with it because it’s inherently inaccurate. Or, if you insist, show your work.
I think it’s really odd that people happily accept the infinite sum used on this clock for 2, but not the one for 7. They’re completely analogous situations.
Takuan, take your pick, they disagree.
Hey I got Boinged! If you want to buy one they are currently sold out. But as soon as I can I will repost more quantities of math clocks
this is just the kind of NERDiculator I’ve been looking for.
FWIW, 6.9… is not the same number as 7, but it does have the same value. To say 6.9… = 7 is correct, but 6.9… == 7 is not.
Tee hee hee!
It’s beginning! Mu Ha haha aha aha ha!
I posted more clocks link
More of my math clocks always posted here: http://eagleapex.etsy.com
I’ll keep them in stock.
I wish there were separate, commonly accepted terms for math/science-y geeks and gamer/1337speak-y geeks.
I understand many of you probably fall into both categories, but surely I’m not the only person who got excited seeing the title and then sighed heavily when I saw all that equation jazz. :P
how does 2^(-1)(mod7) == 4 ?
It seems to me that it equals 1/2, and that the mod7 is irrelevant.
@UncleBaggy: Would you likewise say that 0.5 and 1/2 have the same value but are not the same number? In my opinion, 0.5 == 1/2, not just 0.5 = 1/2. I suppose you could say that they’re different representations of the same underlying number, but (aside from making awesome clocks) the actual representation is pretty unimportant.
Coming this summer to high school math rooms across the country…
Banksynergy – the former are called PhD’s. ;)
ths clck s knd f lm. T bd t sn’t bttr lkng. t lcks ll f th lgnc gks nrmlly ssct wth mthmtcs.
Just to make sure we have a full measure of pedantry: three is not an “HTML entity” but an XML “numeric character reference”. The referenced Wikipedia page has it right.
MDH – TouchÃ©.
@45: If repeating decimals such as 0.999… are not allowed in mathematics, how can pi, or e, or sqrt(2) possibly be? If anything they are more complicated…
5 – Don’t you mean Golden Ratio? (not Mean?)
That’s the notation for the modular inverse. So 4 is the number n such that 2n is congruent to 1 modulo 7.
lk th grl wth th pn lgs bttr.
I’ve suspended you for repeated blogwhoring. Write me if you’d like to explain.
“…that the mod7 is irrelevant.”
would be your problem. perhaps you should click on the link…
6.9… is not a number. It another notation for an infinite sum: 6+0.9+0.09+0.009+… that yields 7 as a result in the limit. You can look it up in the 3-Volume Kudryavtsev’s Foundations of Math Analysis, can’t remember the right volume and page, sorry…
So, it is similar to the 2 O’Clock formula: (1+1/2+1/4+1/8… = 2 in the limit), because the geek who designed it did not know the math well enough :(
Here’s an easier to understand explanation of modular multiplicative inverses:
The ^-1 stands for inverse in this context, not negative power of 1, isn’t that right? In this context, it’s saying 2^-1 times something will be congruent to 1 in mod 7.
So we’re trying to figure out in the equation where 2 * x is congruent to 1 in mod 7, what’s the x? In order to be congruent to 1 in mod 7, a number has to be an integer factor of 8 because 8-7=1. Finally, 2 * x = 8 gets you x = 4.
@Anonymous#45: Though it is true that hE > d9, what you’re neglecting is that while h0.FFF… starts out closer to 1 than d0.9… does, it approaches it more slowly. For example, h0.F > d0.9, but h0.0F < d0.09 (really!). Thus, they both get to exactly 1 when they hit infinity. To demonstrate this, I’ll work the decimal, hexidecimal, and binary cases formally, using limits. The similarity to the formula for 2 in the original clock should become apparent here.
First step: expansion of the …
b0.111… = b0.1 + b0.01 + b0.001 + …
d0.999… = d0.9 + d0.09 + d0.009 + …
h0.FFF… = h0.F + h0.0F + h0.00F + …
Second step: conversion into (decimal) fractions
b0.111… = 1/2 + 1/(2^2) + 1/(2^3) + …
d0.999… = 9/10 + 9/(10^2) + 9/(10^3) + …
h0.FFF… = 15/16 + 15/(16^2) + 15/(16^3) + …
Third step: changing into summation notation
b0.111… = sum(i=1, infinity, 1/(2^i))
d0.999… = sum(j=1, infinity, 9/(10^j))
h0.FFF… = sum(k=1, infinity, 15/(16^k))
Fourth step: rearranging terms to take advantage of the formula sum(a=0, infinity, a^r) = 1/(1 – a)
b0.111… = sum(i=0, infinity, (1/2)^i) – 1
d0.999… = 9*(sum(i=0, infinity, (1/10)^i) – 1)
h0.FFF… = 15*(sum(i=0, infinity, (1/16)^i) – 1)
Fifth step: substituting in with the aformentioned formula
b0.111… = 1/(1 – 1/2) – 1
d0.999… = 9/(1 – 1/10) – 9
h0.FFF… = 15/(1 – 1/16) – 15
Sixth step: continuing to simplify our expressions
b0.111… = 1/(1/2) – 1
d0.999… = 9/(9/10) – 9
h0.FFF… = 15/(15/16) – 15
Seventh step: canceling out the fractions
b0.111… = 2 – 1
d0.999… = 10 – 9
h0.FFF… = 16 – 15
Eight step: performing the subtraction
b0.111… = 1
d0.999… = 1
h0.FFF… = 1
Now, you might object to the formula sum(a=0, infinity, a^r) = 1/(1 – a), but it is true by the rules of mathematics that describe infinite sums. (So long as |a| < 1, but that’s beyond the scope of this discussion.) If you don’t like them, tough; just because Zeno didn’t know about something doesn’t mean we must likewise remain in ignorance.
I meant an integer multiple, not an integer factor.
Mail (will not be published) (required)