The Daily WTF: Curious Perversions in Information Technology

2010-09-29 Reply Admin

smxlong:
hatterson:
One:
Do we get to have a discussion about whether or not 1 is a prime number?

I would rather a discussion about whether .999... == 1

They are equal.

If 0.999... != 1, then by the density of the real numbers, there must be some number X where 0.999... < X < 1. Suppose such a number exists. Please show me how to write this number.

(The above is not a proof, but hopefully it's convincing enough)

Whoooo. I guess if that "proof" is true, then, for integers, 0=1 after all. There goes number theory. TWTF.

2010-09-29 Reply Admin

cicobuff:
smxlong:
hatterson:
One:
Do we get to have a discussion about whether or not 1 is a prime number?

I would rather a discussion about whether .999... == 1

They are equal.

If 0.999... != 1, then by the density of the real numbers, there must be some number X where 0.999... < X < 1. Suppose such a number exists. Please show me how to write this number.

(The above is not a proof, but hopefully it's convincing enough)

Whoooo. I guess if that "proof" is true, then, for integers, 0=1 after all. There goes number theory. TWTF.

Since the integers are discrete, bringing them in to ridicule an argument that explicitly mentioned denseness only exposes yours.

2010-09-29 Reply Admin

I think the point would be to define what we all mean... rather than trying to cite any smart sounding rule that comes to mind.

Rules created to make sense of chaos like "Any 2 Real numbers always have a number between them" are just that, made up rules some dudes decided was needed. It's not 'real' and without any hard reason to back it up makes no sense in relation to the discussion i think (other than its there to make things simpler and workable).

To mean 0.9999999 etc to infinity in actuality means not 1 but zero point 9 to infinity so as to never be 1. There is no 'real' reason to have it equal 1 other than to simplify math and so people can feel better about having to deal with it and can use basic formulas on it using multiplication etc.

Simple normal math isn't designed to work with infinite numbers in the first place and that's the real issue.

0.333... = 1/3 multiply both sides by 3 and you get 0.999... = 3/3 collapse the fraction 0.999... = 1

This is incorrect as 0.333 inf does NOT = 1/3 as 0.333 inf is not a perfect third of 1 (its damn close... but not exact), that's the whole point... they are not equal.

0.999 inf does NOT = 3/3 either because if 0.333 inf is not a perfect third 3 of them cannot be a perfect set of 3 thirds (its 'close enough' perhaps but not technically equal)

Given x = .999... Then: 10x = 9.999... .....{snip}.....

This is also wrong as 10 times an infinite number isn't something that has set rules in basic math. e.g. How do you multiply a fraction correctly that's infinite? you would never stop calculating and so it simply cannot be done at all.

In short the rule that 0.999 infinite equals 1 is just a farce to simplify a problem in mathematics and is in fact wrong in actuality.

===============

How did i do? please rate my attempt at helping the losing argument side on a scale of 'bah' to 'wtf!', all proceeds go to help the 'i need more wild turkey: american honey' fund.

2010-09-29 Reply Admin

Ok, I think that was at least 0.8 TC (timecubes). Probably even more than 0.9. (No, not more than one. No one can do more than TC without their head asploding.)

2010-09-29 Reply Admin

anon:
(No, not more than one. No one can do more than TC without their head asploding.)

More than 1 TC, of course. Dammit.

2010-09-29 Reply Admin

Ouch!:
cicobuff:
smxlong:
hatterson:
One:
Do we get to have a discussion about whether or not 1 is a prime number?

I would rather a discussion about whether .999... == 1

They are equal.

If 0.999... != 1, then by the density of the real numbers, there must be some number X where 0.999... < X < 1. Suppose such a number exists. Please show me how to write this number.

(The above is not a proof, but hopefully it's convincing enough)

Whoooo. I guess if that "proof" is true, then, for integers, 0=1 after all. There goes number theory. TWTF.
Since the integers are discrete, bringing them in to ridicule an argument that explicitly mentioned denseness only exposes yours.

hatterson:
Anonymous:
hatterson:
One:
Do we get to have a discussion about whether or not 1 is a prime number?

I guess I am too dense to know of a number theory that states that 2 adjacent numbers have an equality relationship.

If you had bothered to read and understand what people are posting you would have seen the fragment below.

I would rather a discussion about whether .999... == 1
Well, technically there is nothing to discuss because they most certainly are equal and it is a simple proof to confirm this beyond doubt. But damn, what a mindfuck eh? I remember when I first had to tackle this concept back in maths class, many years ago. Took a long time to get my head around. But once it clicked it was like a lightbulb going off in my head, I literally blurted out "fuck me!" in the middle of my maths class, directly to the lecturer. He was a bit of an old bastard and I thought he was going to go ballistic, but he just looked at me with a knowing smile and gently nodded. I had joined the club.

Oh I'm well aware that they're the same number (just different representations) I just thought it would be fun to troll a thread on that rather than on if 1 is prime.

.999... == 1 usually provides more entertaining responses to read since the is 1 prime "debate" just ends with "it's not because we said it's not."

Edit: And looking at the last page or so, I greatly succeeded.

Ouch indeed, so I guess that make you an airhead as compared to being dense?

As usual, TWTF is in the comments.

2010-09-29 Reply Admin

shimon:
Sup dawg, I heard you like references, so we've put a reference in your reference so you can get self while you have self!

+1, best use of a meme on this site in the past month. Definitely deserves to be a FEATURED COMMENT!

Transverbero...

2010-09-29 Reply Admin

Hmmm, quoting system is obviously not working as intended.

The above post should also include.

Since which number theory states that 2 adjacent numbers have an equality relationship. Because if that is true, all numbers would have been equal with a proof by induction.

Callin · 2010-09-29 Reply Admin

public static boolean isMagical(char c)
{
    return isLetter(c) && isDigit(c);
}

2010-09-29 Reply Admin

Is it even worth pointing out that 215 isn't a prime number?

2010-09-30 Reply Admin

cicobuff:
Since which number theory states that 2 adjacent numbers have an equality relationship. Because if that is true, all numbers would have been equal with a proof by induction.

Good thing there are no adjacent real numbers.

DaveK · 2010-09-30 Reply Admin

Callin:

public static boolean isMagical(char c)
{
    return isLetter(c) && isDigit(c);
}

I tested this function on all the letters of the alphabet, and it returned true for 'C', 'D', 'I', 'L', 'M', 'V' and 'X'!

2010-09-30 Reply Admin

cicobuff:
Hmmm, quoting system is obviously not working as intended.
The above post should also include.

Since which number theory states that 2 adjacent numbers have an equality relationship. Because if that is true, all numbers would have been equal with a proof by induction.

There are no “adjacent” real numbers. That is, for two real numbers x, y with x < y, there is another real number z such that x < z < y. If there is no such number for x ≤ y, then x = y.

2010-09-30 Reply Admin

I am not sure the first one is WTF: Suppose we have something like

assingWork(chooserFunct) {
  while work=getWork() { chooserFunct().doIt(work) }
}

then we can call it like that:

assignWork(callCenter.getFreeOperator());
assignWork(market.getShortestQueue());
assignWork(oneManArmy.getSelf());

DaveK · 2010-09-30 Reply Admin

Robert:
How did i do?

It wasn't a triumph. (I'm making a note here: EPIC FAIL.) It's hard to understate my satisfaction.

TDWTF postings We tear them all down, because we can. For the good of all of us Except the ones who get flamed

Now it's no use pointing out every mistake You just keep on posting 'til you run out of flames And the CSoDs all get done, and the Error'ds come and come From the posters who are still awake...

[ ... ]

2010-09-30 Reply Admin

frits:
Arnold Vriezekolk writes, "one of the guys on our team likes to be thorough. Very, very thorough."
/* This program will only run if the laws of mathematics hold */
if(1 == 0) 
{
    fprintf("Oh crap - we are not running in the correct Universe\n");
    exit(17);
} 
Right. Because in your universe fprintf() doesn't need a file pointer.

In that universe, one file pointer parameter equals no file number parameters.

2010-09-30 Reply Admin

Anonymous:
Hey there Ancient Mathematician, you might be just the person I need to help me with an ancient Roman math problem. The following sum is written in Roman numerals. Can you solve it and provide your answer also in Roman numerals? (IV + I) - V = ?

Easy: (IV + I) - V = I - I

2010-09-30 Reply Admin

hatterson:
One:
Do we get to have a discussion about whether or not 1 is a prime number?

I would rather a discussion about whether .999... == 1

It depends on the number system you are using. In standard analysis they are equal, but in non-standard analysis -- which is used for some things like the analysis of fractals -- the number system is defined in a different way and they are not equal.

2010-09-30 Reply Admin

digitig:
hatterson:
One:
Do we get to have a discussion about whether or not 1 is a prime number?

I would rather a discussion about whether .999... == 1
It depends on the number system you are using. In standard analysis they are equal, but in non-standard analysis -- which is used for some things like the analysis of fractals -- the number system is defined in a different way and they are not equal.

Also, they are clearly not equal in base 11 or greater.

2010-09-30 Reply Admin

Steve The Cynic:
Anonymous:
Hey there Ancient Mathematician, you might be just the person I need to help me with an ancient Roman math problem. The following sum is written in Roman numerals. Can you solve it and provide your answer also in Roman numerals?
(IV + I) - V = ?
Ah, but you forget that the things we call Roman numerals were for writing down numbers, not for performing arithmetic. One of the principal advantages of the modern versions of Arabic numerals (the 0..9 system that we use today) is that they use the same notation for writing down numbers and for doing arithmetic. The Romans used a form of abacus (note that the wire-and-bead calculators that we call abacuses are in fact NOT abacuses) when performing non-trivial arithmetic.

All that, of course, does not change the fact that there is no way to use Roman number-recording-notation to write zero.

Um... actually.. they did have a representation for nothing; but there was no need to represent a zero figure since it did not figure in their numeric symbols (Roman numerals are not positional). If a value was what we call zero (no apples, no eggs, no life, whatever), then it was regarded as nulla/nullae - which has actually been represented in some old roman texts as N (rare).

There is really no need for the number zero either - cnsider the value MM, which is 2000, or MMI which is 2001.

2010-09-30 Reply Admin

boog:
Fair enough, it's valid syntax in both C# and Java.
But you said it is Java.

Severity One:
Unless 1 == 0, that's Java.
And you clarified your argument, for those of us who are "running in the correct Universe":
Severity One:
...so in short: 'That's Java'.
I've never heard of someone taking a snippet of code that exists in the context of a C# application and outright calling it "Java" (or vice versa). So this implies that either A) you know more about the actual habitat of this code snippet than we do, or B) you're making assumptions. I think the GP was arguing B.
I'm not saying you're wrong, I'm just suggesting that you revise your statement to avoid these silly arguments.

what Severity One should have said:
Unless 1 == 0, that's valid Java syntax.
Isn't that more accurate?

Yep, exactly what boog said. You really don't have to be an asshole just for the sake of it, we all knew what you meant and we all knew what I meant. Is it problems at home Sev One? Why don't you get it off your chest instead of lashing out at strangers? I'm here if you want to talk about it.

2010-09-30 Reply Admin

hatterson:
Anonymous:
hatterson:
One:
Do we get to have a discussion about whether or not 1 is a prime number?

I would rather a discussion about whether .999... == 1
Well, technically there is nothing to discuss because they most certainly are equal and it is a simple proof to confirm this beyond doubt. But damn, what a mindfuck eh? I remember when I first had to tackle this concept back in maths class, many years ago. Took a long time to get my head around. But once it clicked it was like a lightbulb going off in my head, I literally blurted out "fuck me!" in the middle of my maths class, directly to the lecturer. He was a bit of an old bastard and I thought he was going to go ballistic, but he just looked at me with a knowing smile and gently nodded. I had joined the club.

Oh I'm well aware that they're the same number (just different representations) I just thought it would be fun to troll a thread on that rather than on if 1 is prime.

.999... == 1 usually provides more entertaining responses to read since the is 1 prime "debate" just ends with "it's not because we said it's not."

Edit: And looking at the last page or so, I greatly succeeded.

Mission accomplished my friend, I haven't seen 220+ comments on an article for quite some time!

2010-09-30 Reply Admin

0.999 infinite has ambivalent meanings due to the unability of the decimal system to represent thirds of one correctly.

1/3 != 0.333 infinite

It's just the closest number to 1/3 we know in the decimal system - that's the reason for the widespread use of 1/3 = 0.333 infinite. In the ternary numeral system it would be simply 1/3 = 0.1. Basically, that means both sides are correct, because both meanings can be applied. It's like the M in roman numbers - it stood for 1,000 as well as infinity. By the way, the same problem arises if a double is supposed to represent 1. I don't think I'll have to elaborate this one in this place.

On another note, like somebody else already pointed out, there is a number between 0.999 infinite and 1 - it's 1.999 infinite divided by 2.

Always be open-minded, and check the validity of the arguments of other people before answering. Sometimes they're right, sometimes both are right.

(I love "real world" comparisons: "Okay, this stick shall now represent a gun. Hey, why can't I shoot with the stick? Why doesn't it behave like a gun?" "Okay, 0.333 infinite shall now represent 1/3. Hey, why can't I calculate with 0.333 infinite so 3 * 1/3 = 3 * 0.333 infinite = 1? Why doesn't 0.333 infinite behave like 1/3?")

2010-09-30 Reply Admin

One:
Do we get to have a discussion about whether or not 1 is a prime number?

It's all down to the definition. "Prime" used to be defined in such a way that 1 was prime (see Goldbach's original formulation of the Goldbach conjecture, for instance). Now "prime" is usually defined in such a way that 1 is not prime (and mathematicians who don't know the history of the subject get cross if you suggest that "prime" /could/ be defined in such a way as to include 1). I've never managed to find out when the change of definition happened, though. Some time between the 18th century and the late 20th.

2010-09-30 Reply Admin

Sammy:
0.999 infinite has ambivalent meanings due to the unability of the decimal system to represent thirds of one correctly.
1/3 != 0.333 infinite

It's just the closest number to 1/3 we know in the decimal system

It's either 1/3, or it's not actually defined, and so isn't a number, and so can't be "the closest number" to anything. There's no way you can sensibly* define 0.333 recurring as a number that != 1/3.

Obviously if you want to remove this constraint, it's quite easy. But not very helpful.

2010-09-30 Reply Admin

Anonymous:
Ancient Mathematician:
WthyrBendragon:
One:
Do we get to have a discussion about whether or not 1 is a prime number?

Well, it is only integer factorable to 1 and itself so, yes, 1 is prime.

However, random it is not.
Well, if you were old enough, you'd know that 1 is not even a number, so it can't be a prime. And if you're young enough, you should know that 1 is a unit, so it can't be a prime. Only if you're stuck some time between -200 and 1900 is 1 a prime.
Hey there Ancient Mathematician, you might be just the person I need to help me with an ancient Roman math problem. The following sum is written in Roman numerals. Can you solve it and provide your answer also in Roman numerals?

(IV + I) - V = ?

V - V

2010-09-30 Reply Admin

Whether 1 is a prime number or not is essentially an arbitrary definition.

If you say it is prime, there will be some theorems and proofs that say "for all primes except 1".

If you say it isn't prime, there will be some other theorems and proofs that say "for all primes and 1".

The primality or not of 1 is defined so it makes sense in the majority of cases in actual use.

GalacticCowboy · 2010-09-30 Reply Admin

"This is one of our developer's misguided attempts to prevent a null pointer exception," writes Yamee.
    if (_tblItem.getTable().getModel() != null 
        && _tblItem.getTable() != null 
        && _tblItem != null)

So close... and yet so very, very far...

The Nerve:

Fixed?

if (&& _tblItem != null != null 
    && _tblItem.getTable() != null 
    _tblItem.getTable().getModel())

I think you forgot the comments on this one. Or else your parser quit on you... Looking forward to your future submissions to this site.

2010-09-30 Reply Admin

IV + I) - V = (I-I)

2010-09-30 Reply Admin

I've seen GameFAQs topics with a more intelligent 0.999~ = 1 discussion that this.

2010-09-30 Reply Admin

THE NUMBER ONE:
Do we get to have a discussion about whether or not I am a number?
FTFY

I am not a number, I am a FREE MAN!

2010-09-30 Reply Admin

DCRoss:
THE NUMBER ONE:
Do we get to have a discussion about whether or not I am a number?
No, you are a free man.

Iron Maiden!

2010-09-30 Reply Admin

Sammy:
On another note, like somebody else already pointed out, there is a number between 0.999 infinite and 1 - it's 1.999 infinite divided by 2.

Finding yet a third way of writing the same thing doesn't prove anything, I'm afraid. If 0.999 infinite is a number, and is not equal to zero, then there must be a number between them (an infinite number of numbers, in fact) that can be represented by a finite decimal.

The only sensible (there's that word again) definition of 0.999 infinite is as an infinite geometric series. And they're fairly well understood. And the limit of this particular series is 1.

dkf · 2010-09-30 Reply Admin

Robert:
Simple normal math isn't designed to work with infinite numbers in the first place and that's the real issue.

The real issue is that you're a dumbass in the old math department.

To put things simply, there is no mathematical operation that you can perform on 0.999… that you can't also do on 1 to get the same result. (Assuming you're sticking with the real domain; in the integer domain the 0.999… is not syntactically valid.) If you think about it, that's got to be the case (what is 1-0.999… anyway? Must be 0.000… That in turn means that no measure function can distinguish them) which means that, at least with the standard representation of the continuum, they must be the same thing.

The cause of this weirdness is the fact that real numbers are not discrete, and that has some very non-obvious consequences. The natural way for people to think about things is in terms of discrete entities, but that sort of thinking just doesn't work for real numbers. (There are mathematical structures that can distinguish these things, but they're even odder than the reals and I've yet to see any convincing argument that they're useful for anything at all.)

Severity One · 2010-09-30 Reply Admin

Anonymous:
Yep, exactly what boog said. You really don't have to be an asshole just for the sake of it, we all knew what you meant and we all knew what I meant. Is it problems at home Sev One? Why don't you get it off your chest instead of lashing out at strangers? I'm here if you want to talk about it.

Strangers, like people called "Anonymous (unregistered)"?

Look, whoever made the first comment was claiming that this was C#, because it had generics, notwithstanding that Java has had generics for the last five or six years.

It was pointed out that is, in fact, Java, and he said, fair cop, I was unaware of that.

But then he came back with the pedantic rubbish about definitions, "patently false", and who should have written what.

I don't care if someone makes a mistake; I make them myself all the time. But I can't stand people trying to cover their arse and making excuses.

Or coming up with silly arguments about "problems at home". There's a name for such a debating "technique" (for want of a better word), but quite frankly, I can't be arsed to look it up.

Ilya Ehrenburg · 2010-09-30 Reply Admin

dkf:
Robert:
Simple normal math isn't designed to work with infinite numbers in the first place and that's the real issue.
The real issue is that you're a dumbass in the old math department.
To put things simply, there is no mathematical operation that you can perform on 0.999… that you can't also do on 1 to get the same result. (Assuming you're sticking with the real domain; in the integer domain the 0.999… is not syntactically valid.) If you think about it, that's got to be the case (what is 1-0.999… anyway? Must be 0.000… That in turn means that no measure function can distinguish them) which means that, at least with the standard representation of the continuum, they must be the same thing.

It's even simpler. Notations like 0.(142857) or 0.999 infinite or such have no intrinsic meaning, so their meaning has to be defined.

The meaning has been defined so that these representations denote the limit of the sequence of the finite decimal fractions (e.g. 0.1, 0.14, 0.142, ... , 0.1428571428, 0.14285714285, ...) of increasing length formed from it[1] (you all know the rule by which they are formed, I believe).

Since the limit of (1 - 10^(-n)) as n tends towards infinity is 1, 0.(9) = 1 by the definition of the notation. Similarly 0.(3) = 1/3 by the notation's definition.

dkf:
(There are mathematical structures that can distinguish these things, but they're even odder than the reals and I've yet to see any convincing argument that they're useful for anything at all.)

I haven't seen a convincing argument for non-standard reals either. I would appreciate one.

[1] Technically, the definition is a little different, but I won't go into that without a blackboard or LaTeX available.

Ilya Ehrenburg · 2010-09-30 Reply Admin

anon:
Whether 1 is a prime number or not is essentially an arbitrary definition.

One can argue about the "arbitrary", but yes, it's undeniably a matter of definition.

If you say it is prime, there will be some theorems and proofs that say "for all primes except 1".
If you say it isn't prime, there will be some other theorems and proofs that say "for all primes and 1".

The primality or not of 1 is defined so it makes sense in the majority of cases in actual use.

And since in the overwhelming majority of cases not including 1 amongst the primes (or, for that matter, 0, if you use the more modern definition of primes instead of the traditional definition taught in school, which characterizes what are nowadays called irreducible numbers), makes things far more elegant and succinct, that's the definition adopted by the overwhelming majority of mathematicians.

2010-09-30 Reply Admin

Severity One:
Anonymous:
Yep, exactly what boog said. You really don't have to be an asshole just for the sake of it, we all knew what you meant and we all knew what I meant. Is it problems at home Sev One? Why don't you get it off your chest instead of lashing out at strangers? I'm here if you want to talk about it.
<Snipped aggression that obviously stems from far more than a trivial anonymous comment>

Let it out Sev One, let it all out. I can tell you're having a bad day at the very least, probably a bad year by the sounds of things. If it makes you feel better to argue your corner then I'm not going to stop you. It's perfectly clear from boog's response that my comment was valid and correct, but this isn't about correctness anymore. This is about you getting the help you need to sort out whatever it is that's causing all this unfounded aggression. As I said before, I'm here if you want to talk. Hey, I don't even mind if you want to continue defending your logically flawed point of view - whatever it takes to make you feel a bit better about life.

2010-09-30 Reply Admin

LB:
ShatteredArm:
smxlong:
hatterson:
about whether .999... == 1
by the density of the real numbers, there must be some number X where 0.999... < X < 1.
Why does there have to be a number between the two?
Maybe there is a proof, but this is how I look at it:

What is the largest number that is less than 1? I think you'd have to conclude that it is 0.999... wouldn't you?

And if a < b, then a != b.
Indeed there are number systems in which there is such a thing as "the largest number that is less than 1", but the only number system in widespread use in which 0.999... is a valid number is the real number system, and that one involves a continuum (the number line) rather than a series of discreet points.

Of course, if you want to posit a number system of discreet points in which there is a point for 0.999... distinct from the point for 1, you can. (There are, after all, an infinite number of possible number systems.) But you would need to define both the notation and rules for this number system yourself and identify that it's the number system you're using, since it's not one in regular use by mathematicians.

Wow I'm sure glad those points are able to keep a secret. I think you mean "discrete"? You won't win any math points around here if you get that one wrong.

2010-09-30 Reply Admin

0.999.... vs. 1 - it really depends on context, those two things could be clearly different in some cases - lets take function "floor" which maps Real numbers to Integers by returning the highest Integer number less or equal its parameter.

floor(0)=0
floor(0.9)=0
floor(0.99)=0
floor(0.9999999999999999999999999999999999999999)=0
floor(0.9...)=0
floor(1)=1

and there are not only functions, which are not continuous in some points, there are also functions, which are not continuos in any point :)

2010-09-30 Reply Admin

gilhad:
floor(0.9...)=0

Sorry. You're assuming that limit(floor(x)) = floor(limit(x)) there. It doesn't.

gilhad:
and there are not only functions, which are not continuous in some points, there are also functions, which are not continuos in any point :)

My favourite "strange" function is:

f(x) = 0 when x irrational f(x) = 1/q when x rational and p/q in its lowest fractional form

Continuous at all irrational points, discontinuous at all rational. Proving this is left as an exercise for the reader.

2010-09-30 Reply Admin

Ozzie:
Is it even worth pointing out that 215 isn't a prime number?

Not sure. Everyone else seemed to think so when they pointed it out. ;)

2010-09-30 Reply Admin

gilhad:

floor(0.9...)=0
floor(1)=1

Time to call your residential building insurance. You floor is broken.

CAPTCHA: secundum (the CAPTCHA is backing me up!)

2010-09-30 Reply Admin

00Davo:
Steve The Cynic:
Anonymous:
Hey there Ancient Mathematician, you might be just the person I need to help me with an ancient Roman math problem. The following sum is written in Roman numerals. Can you solve it and provide your answer also in Roman numerals?
(IV + I) - V = ?
Ah, but you forget that the things we call Roman numerals were for writing down numbers, not for performing arithmetic. One of the principal advantages of the modern versions of Arabic numerals (the 0..9 system that we use today) is that they use the same notation for writing down numbers and for doing arithmetic. The Romans used a form of abacus (note that the wire-and-bead calculators that we call abacuses are in fact NOT abacuses) when performing non-trivial arithmetic.

All that, of course, does not change the fact that there is no way to use Roman number-recording-notation to write zero.

Could one not write zero in Roman numerals like this: " "?

They would have written it as NIHIL, NULLUS or NIL to prevent confusion.

It's generally thought that the Romans did not contribute anything of significance to mathematics, yet European mathematicians were using Roman numerals and base 60 (from the Babylonians) until well into the first Renaissance. Had at some time they formalized a zero, say N/n for "nil", and a sexagesimal point say E/e for "et", we might today still be using it.

E.g. the ratio of the circumference of a circle divided by its diameter is approximated as:

iii e viii xxix xliv n xlvii

It actually not too hard to add, subtract or multiply in Roman numerals. There are only a few symbols, so the multiplication table is really small. Division is a pain though.

AquaDuck · 2010-09-30 Reply Admin

gilhad:
0.999.... vs. 1 - it really depends on context, those two things could be clearly different in some cases - lets take function "floor" which maps Real numbers to Integers by returning the highest Integer number less or equal its parameter.
floor(0)=0
floor(0.9)=0
floor(0.99)=0
floor(0.9999999999999999999999999999999999999999)=0
floor(0.9...)=0
floor(1)=1
and there are not only functions, which are not continuous in some points, there are also functions, which are not continuos in any point :)

Except floor(x) isn't actually mapping reals to integers but floats to integers. Floating-point numbers aren't reals, they're just an approximation that's close enough most of the time.

2010-09-30 Reply Admin

gilhad:
0.999.... vs. 1 - it really depends on context, those two things could be clearly different in some cases - lets take function "floor" which maps Real numbers to Integers by returning the highest Integer number less or equal its parameter.
floor(0)=0
floor(0.9)=0
floor(0.99)=0
floor(0.9999999999999999999999999999999999999999)=0
floor(0.9...)=0
floor(1)=1
and there are not only functions, which are not continuous in some points, there are also functions, which are not continuos in any point :)

The only "context" it depends on is the number system. I think we've established that we're talking about real numbers. Note that digital computers cannot accurately calculate real numbers, which may cause some confusion with your example since most people around here probably see floor() in computer programs (where it typically maps from IEEE FLOAT, not real numbers).

The floor() function operating on Reals does not map 0.999... to 0. You can't just drop everything from the decimal point onward. Per your definition, floor() returns the greatest integer less than or equal to the Real number you give it. You can't use your assumptions about its output to prove something about its input; you have to calculate its output based on what's known about the input. As has been discussed at length, 0.999... is just a funny decimal notation for 1, so the greatest integer less than or equal to 0.999... is 1. The correct evaluation of floor(0.999...) is 1.

2010-09-30 Reply Admin

Several people have suggested that 0.333... != 1/3 but is an approximation. This is provably incorrect. (I'm not going into the details. The importance to me of you understanding this is exactly enough for me to post a suggestion that you educate yourself; not enough that I will try to educate you.)

At least one called 0.333... an "infinite number". It isn't; it's finite and has a well defined value. The math operations we know and love (like multiplcation and subtraction) work on all real numbers, including those whose decimal representation is infinite. They even work on irrational numbers, or you couldn't do any geometry involving circles.

A lot of people get uncomfortable with infinite representations, or with math that extends to talk about limits or other infinite concepts. This doesn't mean that math doesn't work with those concepts; it means you don't.

2010-09-30 Reply Admin

smxlong:
If 0.999... != 1, then by the density of the real numbers, there must be some number X where 0.999... < X < 1. Suppose such a number exists. Please show me how to write this number.

I realize the discussion has gone far beyond this particular post, but I thought I'd throw one more brain-teaser out there for anyone who is still continuing this idiotic debate.

Assuming that 0.999... != 1 (which is assuming wrong, but bear with me), or rather that 0.999... is supposedly the largest number less than 1, I ask you this: what is the largest number that is less than 0.999... and how do you write this number?

hatterson · 2010-09-30 Reply Admin

Matthew:
gilhad:
floor(0.9...)=0
Sorry. You're assuming that limit(floor(x)) = floor(limit(x)) there. It doesn't.
gilhad:
and there are not only functions, which are not continuous in some points, there are also functions, which are not continuos in any point :)
My favourite "strange" function is:
f(x) = 0 when x irrational f(x) = 1/q when x rational and p/q in its lowest fractional form

Continuous at all irrational points, discontinuous at all rational. Proving this is left as an exercise for the reader.

Heh, I remember proving that...it was actually on a test. It was the first time any of us had seen the function or it's proof.

The prof usually liked to throw things like that on the test because he felt tests weren't an accurate assessment unless we had to actively think/deduce as opposed to just recall.

hatterson · 2010-09-30 Reply Admin

boog:
smxlong:
If 0.999... != 1, then by the density of the real numbers, there must be some number X where 0.999... < X < 1. Suppose such a number exists. Please show me how to write this number.
I realize the discussion has gone far beyond this particular post, but I thought I'd throw one more brain-teaser out there for anyone who is still continuing this idiotic debate.
Assuming that 0.999... != 1 (which is assuming wrong, but bear with me), or rather that 0.999... is supposedly the largest number less than 1, I ask you this: what is the largest number that is less than 0.999... and how do you write this number?

Given that the standard rules of mathematics don't apply under this assumption I would be perfectly happy with saying that .999...8 is the largest number less than .999...

In this crazy world .999...8 would be .000...1 less than .999... which is .000...1 less than 1

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