# Clock for geeks

Here's a clock for geeks. It's \$20 on Etsy.

### 50

1. dougrogers says:

Repost :-)

2. Ceronomus says:

Best clock EVER!

3. noen says:

I bet we can get a good argument going over “does 6.999… really equal 7.0? and is it really a different number”.

4. arkizzle says:

Curses!

Noen, I just wrote: In before arguments about “0.9999.. = 1”

MeFail

5. retchdog says:

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.

6. arkizzle says:

..or

Auto-Fail

7. brooklyntwang says:

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.

8. arkizzle says:

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?

9. Takuan says:

this enrages me since I am too stupid to understand it!

10. coreymaley says:

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).

11. spazzm says:

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.

12. Avram / Moderator says:

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…).

13. ReboundDesigns says:

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.

14. mdh says:

re: 6.99999 vs. 7

it’s a math clock, not an atomic clock. One significant digit please.

15. arkizzle says:
16. eagleapex says:

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
Sweet.

17. UncleBaggy says:

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.

18. arkizzle says:

Tee hee hee!

It’s beginning! Mu Ha haha aha aha ha!

*titter*

19. eagleapex says:

20. Banksynergy says:

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

21. Anonymous says:

how does 2^(-1)(mod7) == 4 ?

It seems to me that it equals 1/2, and that the mod7 is irrelevant.

22. trimeta says:

@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.

23. TheChickenAndTheRice says:

Coming this summer to high school math rooms across the country…

24. mdh says:

Banksynergy – the former are called PhD’s. ;)

25. skyjack23 says:

ths clck s knd f lm. T bd t sn’t bttr lkng. t lcks ll f th lgnc gks nrmlly ssct wth mthmtcs.

Fl.

26. Banksynergy says:

27. callmebazza says:

5 – Don’t you mean Golden Ratio? (not Mean?)

28. Bathblogger says:

lk th grl wth th pn lgs bttr.
bthblggngdtblgsptdtcm

1. Antinous / Moderator says:

Bathblogger,

I’ve suspended you for repeated blogwhoring. Write me if you’d like to explain.

29. Anonymous says:

Anonymous,

“…that the mod7 is irrelevant.”

30. dougrogers says:

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?

31. Anonymous says:

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.

32. js7a says:

Whenever this .999… thing comes up, I like to ask this: does 1.000…0001 = 1?

33. coreymaley says:

@JS7A

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.

34. barnaby says:

To quote a former bar patron of great renown:

SEMANTICS!

35. Anonymous says:

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.

36. drmacro says:

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.

37. Anonymous says:

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 :(

Kudryavtsev rocks!!!

38. sabik says:

@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.

39. retchdog says:

#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).

40. michaeltheotherone says:

#35 js7a / #36 coreymaley : js7a’s notation is nonsensical, but we can ask:

does 2 – 0.999… == 1?

41. Anonymous says:

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)

but

h0.FFF~ = d0.999~ (Wait, I thought F > 9…)

Worse yet…

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.

42. Anonymous says:

Here’s an easier to understand explanation of modular multiplicative inverses:
http://en.wikipedia.org/wiki/Modular_multiplicative_inverse
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.

trickledown

43. trimeta says:

@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.

44. Anonymous says:

I meant an integer multiple, not an integer factor.

45. retchdog says:

@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…

46. 1 = .999…

Proof:

1 = 1/3 + 1/3 + 1/3

1/3 = .333…

Therefore,

1 = .333… + .333… + .333…

1 = .999…

Q.E.D.