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Proving 0.999... Is Equal To 1

CmdrTaco posted more than 3 years ago | from the nerds-like-math-right dept.

Education 1260

eldavojohn writes "Some of the juiciest parts of mathematics are the really simple statements that cause one to immediately pause and exclaim 'that can't be right!' But a recent 28 page paper in The Montana Mathematics Enthusiast (PDF) spends a great deal of time fielding questions by researchers who have explored this in depth and this seemingly impossibility is further explored in a brief history by Dev Gualtieri who presents the digit manipulation proof: Let a = 0.999... then we can multiply both sides by ten yielding 10a = 9.999... then subtracting a (which is 0.999...) from both sides we get 10a — a = 9.999... — 0.999... which reduces to 9a = 9 and thus a = 1. Mathematicians as far back as Euler have used various means to prove 0.999... = 1."

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(0.999...)st Post! (5, Funny)

Anonymous Coward | more than 3 years ago | (#33892528)

(0.999...)st Post!

Re:(0.999...)st Post! (3, Funny)

arivanov (12034) | more than 3 years ago | (#33892616)

2+2=5 for sufficiently big values of 2.

Re:(0.999...)st Post! (0)

L4t3r4lu5 (1216702) | more than 3 years ago | (#33892818)

In the case above, the same sufficiently large values of 2 also mean that 2+2=6 to 1s.f.

Maths is fun!

Re:(0.999...)st Post! (5, Funny)

derrida (918536) | more than 3 years ago | (#33892950)

2+2=5 for sufficiently big values of 2.

or for sufficiently small values of 5.

Re:(0.999...)st Post! (4, Funny)

Dan East (318230) | more than 3 years ago | (#33892856)

Geez these first posters. Like spammers, always looking for a new attack vector. I'm sure he's been sitting on this particular exploit for a long time, just waiting for his opportunity to strike. You've won today, but we're all onto your trick when you try to (0.999...)st post the next story...

I went one further (4, Funny)

MyLongNickName (822545) | more than 3 years ago | (#33892532)

I was able to prove that with even one less "9" after the decimal point, it STILL equaled 1. I plan on doing this for a few more iteration until I can prove that . = 1

Re:I went one further (3, Insightful)

MyLongNickName (822545) | more than 3 years ago | (#33892562)

And seriously... is this really front page material? The simplest proof is to say "express 1/9" as a decimal. Now multiply both sides by 9. I remember this in elementary school algebra.

Re:I went one further (4, Insightful)

betterunixthanunix (980855) | more than 3 years ago | (#33892606)

Small numbers usually win; express 1/3 as a decimal, and multiply by 3. The problem with that, though, is that people have trouble accepting that there was nothing wrong with what they did -- a lot of people have this implicit assumption that if a few simple steps bring them to a result that doesn't look like it makes sense, then they did something wrong. If you get them a more complicated proof (assuming they can follow it), they are more willing to accept the result.

Re:I went one further (1, Interesting)

Anonymous Coward | more than 3 years ago | (#33892580)

It must be great to have an infinte amount of time.

Re:I went one further (2, Funny)

atisss (1661313) | more than 3 years ago | (#33892956)

All you have to do now is to prove that / = 1, then you could just type http:1 [1] in your browser to visit /.

I'm Surprised (0)

SpasticWeasel (897004) | more than 3 years ago | (#33892540)

They have mathematics in Montana?

Re:I'm Surprised (3, Funny)

Anonymous Coward | more than 3 years ago | (#33892630)

They damn well better, how else will I measure out all this dental floss.

Re:I'm Surprised (0)

Anonymous Coward | more than 3 years ago | (#33892652)

They have mathematics in Montana?

Dude, that's ALL they have in Montana.

That and empty nuclear silos.

Re:I'm Surprised (1)

Chrisq (894406) | more than 3 years ago | (#33892800)

They have mathematics in Montana?

The surprise is that someone can read wikipedia [wikipedia.org] in Montana where they have had this information complete with the same proof for years.

Finally (2, Funny)

kannibul (534777) | more than 3 years ago | (#33892550)

Someone disproved math. Kids around the world celebrating. Accountants are lighting themselves on fire. Corporate greed accellerates. 'Office Space' now seen as a prophecy.

Re:Finally (1, Insightful)

operagost (62405) | more than 3 years ago | (#33892668)

Government spending like there's no tomorrow... wait, that already happened.

Re:Finally (5, Funny)

Vectormatic (1759674) | more than 3 years ago | (#33892704)

just as long as no-one proves 0 = 1 we computerpeople are safe...

Re:Finally (3, Insightful)

SoVeryTired (967875) | more than 3 years ago | (#33892878)

If pressed, many logicians will admit that the modern foundation of mathematics (ZFC) is probably inconsistent.
See this article:
http://www.math.princeton.edu/~nelson/papers/warn.pdf [princeton.edu]

The author discusses an informal survey he took among loogicians on page three.

If someone ever discovers a paradox, we can simply scale back to some other system and keep most of what we know, but still...

This is second place (4, Insightful)

betterunixthanunix (980855) | more than 3 years ago | (#33892552)

0.999... = 1 is second place to the Monty Hall Problem on the list of things that people have difficulty understanding and accepting the proof of. It is second place because the only department where I do not see graduate students giving me a confused look is the math department; with the Monty Hall problem, I will sometimes get a confused look even from people in the math department.

The other reason I put it in second place is that most people have difficult understanding the problem at all, whereas very few people have trouble understand what the Monty Hall problem is asking.

Re:This is second place (2, Insightful)

bluefoxlucid (723572) | more than 3 years ago | (#33892610)

I'm more interested in the .5 repetand, 5/9. Besides, 8/9 is 0.8 repetand; 9/9 would be 0.9 repetand but 9/9 = 1.

Re:This is second place (4, Informative)

MyLongNickName (822545) | more than 3 years ago | (#33892612)

It is easy to explain.

1. 1/9 = 0.111111111111111111111111111111.....
2. Multiply each side by 9
3. 9/9 = 0.999999999999999999999999999999......
4. Simplify fraction
5. 1 = 0.999999999999999999999999999999......

Monty Hall trips up even serious math enthusiasts.

Re:This is second place (1)

betterunixthanunix (980855) | more than 3 years ago | (#33892688)

Except that I was counting non-mathematicians as well. A lot of people have difficulty grasping what is going on with 0.999... and what it means for that number to equal 1 (the idea that a number could have two representations in the same base goes over a lot of people's heads).

Re:This is second place (-1)

Jezza (39441) | more than 3 years ago | (#33892962)

The point is that no matter how many "9"s you write you're not expressing the number exactly - hence subtracting "a" (which is inexactly expressed) has no real meaning. As soon as we introduce a number of decimals then the whole thing falls down. It's a representation problem. That most people can understand.

Clearly 1 isn't the same as 9.999... but the difference is too small to represent with a decimal. This makes sense as everyone can accept they are ALMOST the same (for practical purposes).

Re:This is second place (4, Interesting)

kannibul (534777) | more than 3 years ago | (#33892816)

This could be done with any fraction represented as a repeating decimal.
The trip-up is that it's repeating...since we have no concept for infinity, and, that there's no method of resolving a fraction w/ repeating decimal...it's not an accurate representation of the fraction - that's the flaw.
Therefore, Fractions are Good. Decimals are Evil!
Good thing our banks, credit card companies, and governments don't use repeating fractions.

Re:This is second place (2, Interesting)

mattj452 (838570) | more than 3 years ago | (#33892822)

Unfortunatelly, your proof is not valid. You are trying to prove something which you postulate in your first step. How do you know 1/9 equals 0.1111111.... ?

Re:This is second place (1)

MyLongNickName (822545) | more than 3 years ago | (#33892890)

Pretty damn easily. Go do your long division, and you will clearly see that the one will repeat forever.

Re:This is second place (5, Informative)

Colonel Korn (1258968) | more than 3 years ago | (#33892958)

Unfortunatelly, your proof is not valid. You are trying to prove something which you postulate in your first step.
How do you know 1/9 equals 0.1111111.... ?

He begged the question! For anyone confused about the term "beg the question," this is exactly what it means: assuming the proposition to be proved in the premise.

But that begs the question: is the classical meaning already dead, replaced with the much more easily understood modern usage demonstrated here?

Re:This is second place (0)

Anonymous Coward | more than 3 years ago | (#33892846)

That is based on the assumption that conventional mathematic rules apply to such numbers ;)

Re:This is second place (1)

lilo_booter (649045) | more than 3 years ago | (#33892886)

1/9 = 0.1111111111111111.... assuming infinite precision, multiply both sides by 9 and you have 9/9 = 1, hence 1 = 1 and we're no further forward.

Re:This is second place (1)

NYMeatball (1635689) | more than 3 years ago | (#33892698)

Its far lower than second place, in my mind.

The monty hall problem has significant and reasonable applications - the understanding and application of the theory behind solving the problem can be, at worst, useful on a game show, and at best applied to hundreds of other similar situations.

Proving, understanding, debating that 0.999... = 1? Okay, great, now what?

I can't see how (dis)agreeing with this theory is going to impact me in any way, shape or form. MAYBE i can make jokes about static const ONE = 0.9999999 on TheDailyWTF now, but that's about it....

Re:This is second place (4, Insightful)

ObsessiveMathsFreak (773371) | more than 3 years ago | (#33892728)

The Monty Hall problem and its delinquent cousin the Tuesday Boy problem are genuinely difficult because the answer is highly dependent on the way that the question is posed.

0.9999...=1 is not genuinely difficult because at the end of the day it's a very informal statement about adding an infinite number of decimals, and the only real controversy about the statement exists among 4chan trolls and Wikipedia users. Most who don't understand don't care and most who do understand also don't care.

The only people with a problem are the people who don't understand but still care, but then that's the problem with most topics these days.

Blimey (1)

tygerstripes (832644) | more than 3 years ago | (#33892828)

The only people with a problem are the people who don't understand but still care, but then that's the problem with most topics these days.

I wish that would fit in my sig.

Re:This is second place (1)

MindStalker (22827) | more than 3 years ago | (#33892776)

And easy way to explain the Monty Hall problem is.
Lets say there are 100 doors, you pick 1 door. Then someone with knowledge of right/wrong doors opens 98 other wrong doors. There is only 2 doors left the one you picked and probably what is the correct door. Would you switch?

Wait, second in what way? (1)

hackwrench (573697) | more than 3 years ago | (#33892866)

?At first I was thinking second in difficulty, and then I read your "The other reason I put it in second place is that most people have difficult understanding the problem at all, whereas very few people have trouble understand what the Monty Hall problem is asking."

Re:Wait, second in what way? (1)

betterunixthanunix (980855) | more than 3 years ago | (#33892974)

Difficulty understanding the question is not the same as difficulty understanding the answer. People will get tripped up by the answer to the Monty Hall problem, but they will usually understand what is being asked -- so in terms of problems whose answer confuses people, Monty Hall is a winner. People won't be confused by the answer to 0.999... = 1 if they cannot even understand what you are asking.

Re:This is second place (1)

jasmusic (786052) | more than 3 years ago | (#33892970)

Any child can see that 0.99... will never be 1 because it is always separated by 0.0...1. Yes people, the Emperor has no clothes, and you ought to laugh at these "mathematicians" who use fallacies to prove fallacies. You're doing "operations" on algorithms that never complete, using a base-10 digit system that is only capable of expressing constant values.

Seriously? (0)

Anonymous Coward | more than 3 years ago | (#33892558)

Uh wtf, I did this in the 8th grade.

Sometimes (0)

Reilaos (1544173) | more than 3 years ago | (#33892588)

I can't help but feel like proofs that 1 = .999... are along the same lines as that joke proof that tries to prove all numbers are the same. I know these proofs are much more legitimate, but my intuition, in the back of my mind, screams "bullshit!" despite me knowing better.

Re:Sometimes (1)

betterunixthanunix (980855) | more than 3 years ago | (#33892732)

You are not alone. I have shown this proof to a lot of people, and I have even proved it multiple ways to those people, and I am still confronted with "this cannot be right."

Re:Sometimes (0)

Anonymous Coward | more than 3 years ago | (#33892976)

How do they respond to the question "how big a number would you have to add to it to get one?"?

Re:Sometimes (2, Funny)

elrous0 (869638) | more than 3 years ago | (#33892782)

That's a real crowd-pleaser at parties. Personally I like the "writing an executable java program without a main method" trick to impress the ladies myself--that is, if I ever get to meet an actual lady who would even get that trick.

There are other, less tricky, proofs (1)

pyrognat (233814) | more than 3 years ago | (#33892904)

For instance, one could take the perspective of analysis. In the real numbers, given a number such as 1 and some other number x, if |1-x| e for any positive real number e then 1 = x (think about this for awhile, if you haven't before, and you will probably believe it). The point here is that the sequence of numbers .9, .99, .999, .9999 etc gets arbitrarily close to the number 1. So the limit of this sequence .999... = 1. I think that this proof is much more intuitive and less "tricky" (ie. does not rely on algebraic manipulation/slight of hand).

they have wikipedia in Montana? (1)

iamhassi (659463) | more than 3 years ago | (#33892598)

I have wikipedia too... [wikipedia.org] : "When a number in decimal notation is multiplied by 10, the digits do not change but the decimal separator moves one place to the right. Thus 10 × 0.999... equals 9.999..., which is 9 greater than the original number. To see this, consider that in subtracting 0.999... from 9.999..., each of the digits after the decimal separator the result is 9 9, which is 0. The final step uses algebra:"

Re:they have wikipedia in Montana? (2, Insightful)

iamhassi (659463) | more than 3 years ago | (#33892900)

0.999.... might be equal to 1, but 0.999 is not equal to 1:
x=0.999
10x = 9.99
10x-x = 9.99 - 0.999
9x = 8.991
x = 0.999

This is true whether there's three 9s or a hundred 9s, so I can see the confusion.

Time to Update my SLA (5, Funny)

Anonymous Coward | more than 3 years ago | (#33892614)

Now I can replace my SLA with 100% uptime.

Then again... (1, Troll)

kannibul (534777) | more than 3 years ago | (#33892618)

Wouldn't 10a (subtract) .999 be exactly 8.991...which breaks the whole "breakthrough"? Given that 'a' is a known value of .999... Math...it's so simple, only a mathemtician can't do it.

Re:Then again... (1)

kannibul (534777) | more than 3 years ago | (#33892646)

And apparently, I can't spell in the mornings.

Re:Then again... (2, Funny)

kannibul (534777) | more than 3 years ago | (#33892876)

I wasn't trolling :(

Re:Then again... (1)

Vectormatic (1759674) | more than 3 years ago | (#33892884)

if the .. in 0.999.. means 9s ad infinitum (which i assume it does), then your reasoning doesnt work

Um (0, Troll)

PsyciatricHelp (951182) | more than 3 years ago | (#33892622)

"both sides we get 10a — a = 9.999... — 0.999... which reduces to 9a = 9 and thus a = 1." But a=.999999......... this is a bit contradictory. I think we need another variable. a!=1 if a already = .999.......

Re:Um (0)

Anonymous Coward | more than 3 years ago | (#33892832)

Except then you never prove a=1. ;)

Re:Um (0)

Anonymous Coward | more than 3 years ago | (#33892862)

* Hand over head motion *

Whooooosh!

Or (3, Insightful)

Anonymous Coward | more than 3 years ago | (#33892624)

1/3 = 0.3333...
2/3 = 0.6666...

0.3333.... + 0.6666.... = 0.9999....

1/3 + 2/3 = 1 = 0.9999.....

What? (1, Troll)

qoncept (599709) | more than 3 years ago | (#33892628)

.999 * 10 = 9.99

I learned this in grade 6 (0)

Anonymous Coward | more than 3 years ago | (#33892638)

Seriously, this kind of bullshit elementary algebra is slashdot news?

sum the geometric series (1)

sevenfactorial (996184) | more than 3 years ago | (#33892640)

The proof I do in my classes uses the formula for summing a geometric series.. .999.. = .9*10^0 + .9*10^-1 + .9*10^-2 + .....

= .9/(1-(1/10)) = .9/.9 = 1

Oh yeah? Well... (2, Funny)

sootman (158191) | more than 3 years ago | (#33892642)

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a + b = b
2b = b
2 = 1

Re:Oh yeah? Well... (3, Informative)

MyLongNickName (822545) | more than 3 years ago | (#33892780)

this probably isn't necessary for most of the Slashdot crowd, but...

(a+b)(a-b) = b(a-b) --> a + b = b

Required division by (a-b) on both sides. Since a = b, this is division by zero.

Re:Oh yeah? Well... (0)

Anonymous Coward | more than 3 years ago | (#33892810)

(a+b)(a-b) = b(a-b)
a + b = b

you can't divide both sides by (a - b) because (a - b) = 0

Re:Oh yeah? Well... (1)

Rob the Bold (788862) | more than 3 years ago | (#33892814)

a = b a^2 = ab a^2 - b^2 = ab - b^2 (a+b)(a-b) = b(a-b) a + b = b 2b = b 2 = 1

Oops, div by zero error. But still funny.

Re:Oh yeah? Well... (1)

Chrisq (894406) | more than 3 years ago | (#33892882)

a = b a^2 = ab a^2 - b^2 = ab - b^2 (a+b)(a-b) = b(a-b) a + b = b 2b = b 2 = 1

For those who didn't spot it the funny bit is (a+b)(a-b) = (a-b) cannot be simply cancelled because "a-b" is zero, so you are saying (a+b)*0 = b*0.

1/3 + 1/3 + 1/3 (1)

wookaru (1521381) | more than 3 years ago | (#33892660)

Though im sure its far from mathematically sound, Ive used this method to convince myself and others of the general "truthiness" of the .99999 = 1 debate in the past:

1/3 + 1/3 + 1/3 = 1
In decimal form:
.3333 + .3333 + .3333 = .9999

So, .9999 = 1

This is so old... (3, Interesting)

DiSKiLLeR (17651) | more than 3 years ago | (#33892666)

This is so old...

Even Blizzard issues a press release about it years ago because people kept arguing about it on the Blizzard forums.

http://www.mbdguild.com/index.php?topic=14915.0 [mbdguild.com]

This is just faulty math (0)

zzsmirkzz (974536) | more than 3 years ago | (#33892672)

Why is it faulty, because the proof is obviously wrong. 9a can't be equal to 9 and 8.99999..991 at the same time.

First off, by multiplying by ten, they lost one 9 at the end of the series. thus if using 6 decimal points (a = 0.999999, 10a = 9.999990. 10a - a = 9.999990 - 0.999999, 9a = 8.999991).

if you want to argue infinite repeating decimals, than yes, 0.9999... is approaching 1. It's limit as we approach an infinite number of decimal points would essentially make it equal to 1. But you cannot reach infinity so this is a moot point.

Re:This is just faulty math (2, Funny)

Anonymous Coward | more than 3 years ago | (#33892848)

But you cannot reach infinity so this is a moot point.

I think you just dismissed most of mathematics.

Re:This is just faulty math (5, Insightful)

Sockatume (732728) | more than 3 years ago | (#33892854)

Actually, if you define 0.999... as having an infinite number of decimal points, then it is true. And that's how that ellipsis is defined! It means exactly infinite repeating decimals.

You've demonstrated the first hurdle that this problem raises in people's brains: they start thinking about adding "one more" decimal point to the expression, meaning they're thinking of a large but finite number of decimal points. And the second hurdle: people find it hard to believe that you can do mathematics with "infinity" as a meaningful quantity.

Re:This is just faulty math (0)

Anonymous Coward | more than 3 years ago | (#33892902)

You make the mistake of thinking of 0.999... as a number that "gets bigger." You can't take the "limit" of a number that never changes. 0.999... is not "approaching" anything. It simply is what it is, and what it is is 1.

Re:This is just faulty math (1)

JohnFluxx (413620) | more than 3 years ago | (#33892906)

In real numbers, there is no such thing as the number 8.99999..991

There is in hyperreals, but not reals. Where are you reading that from?

Re:This is just faulty math (0)

Anonymous Coward | more than 3 years ago | (#33892918)

Why is it faulty, because the proof is obviously wrong. 9a can't be equal to 9 and 8.99999..991 at the same time.

You can't have 8.99999..991. The whole point of the ... is that it means it goes on forever. You can't have something after the forever because if you do, it means you didn't go on forever.

You're committing another version of the fallacy where people claim that 0.999... != 1 because 1 - 0.999... = 0.00...01. Again, that doesn't work because you can't have something after the forever.

Re:This is just faulty math (0)

Anonymous Coward | more than 3 years ago | (#33892928)

First off, by multiplying by ten, they lost one 9 at the end of the series.

What? They are using an infinite number of 9's after the decimal. If you take away one of those infinite number of 9's, there's still an infinite number of 9's.

Re:This is just faulty math (2, Insightful)

goffster (1104287) | more than 3 years ago | (#33892930)

The series is infinite, you don't lose one.

Just because you can not show the number as a whole does not mean
you can not perform operations using it.

i.e. Think of pi.

Re:This is just faulty math (0)

Anonymous Coward | more than 3 years ago | (#33892942)

There is no "end of the series" of nines in 8.9999... . 9a equals 8.9999..., which is very different from 8.9999...991.

Re:This is just faulty math (1)

RobVB (1566105) | more than 3 years ago | (#33892952)

You cannot reach infinity, therefore you can't reach the end of the series, therefore you can't reach the 1 at the end of the series. Which means you can't use the 1 at the end of the series to disprove the proof.

The whole point of "infinity" is that there IS no end. You can't say "but suppose there is" and use that to prove something.

9.999... -- 0.999... = 9 ? (1)

Demablogia (1149365) | more than 3 years ago | (#33892674)

9.999... -- 0.999... = 9 ? why ? If I suppose than 9.999... -- 0.999... limits on 9 because the operation consists of a infinite number of finite operations . In this case , 0.99999 limits on 1 , and this has sense

Recipe for Forum Disaster (0)

Anonymous Coward | more than 3 years ago | (#33892680)

For some reason, discussion of 0.999... = 1 seems to bring out the worst in people on forums. That, the Monty Hall problem [wikipedia.org] , and the question of whether or not a plane on a treadmill can take off seem to be the evil trio of simple, solvable problems that lead to flame wars and arguments as intense as those normally seen in more subjective realms (religion, politics, abortion, tipping, and favorite text editor).

Humans are just biased towards natural numbers (2, Insightful)

elrous0 (869638) | more than 3 years ago | (#33892682)

Humans are used to natural numbers because they're simple. But do natural numbers even exist in the real world? For the vast majority of practical purposes, 0.99999 can be thought of as one. But "one" itself is usually just a construct in the real world. There is no such thing as the perfect one of anything. The more precise we get, the more "one" becomes more of a mathematical ideal than a reality. So we spend our entire lives rounding off, because that's practical. We teach kids to count 1, 2, 3, 4... We can't very well teach them to count 0.000001, 0.00001, 0.0001, 0.001... (or any of the infinite variations of "counting" without resorting to natural numbers).

Proving that 0.99999 = 1 is an interesting intellectual exercise. But in the real world, we do it every minute of every day.

In other words--eh, close enough.

Vindication (1)

Myopic (18616) | more than 3 years ago | (#33892696)

So after all these years, has Intel been vindicated [wikipedia.org] ?

When you add/subtract/multiply/divide infinite set (-1)

Fallen Kell (165468) | more than 3 years ago | (#33892710)

Too bad someone didn't learn the rules of dealing with infinite sets; there are orders of magnitude. Look at the following: 2x > x So what happens when "x" is set to infinity? We know that "2x" is always greater than "x", but since infinity is an irrational number, different things happen. The same is true with an infinite repeating decimal. It is an irrational number. As such, if you set x = 0.999...99, it will always be less than 1, even if you multiply it out because the multiplication carries the same "infinite magnitude" issues as shown above.

Re:When you add/subtract/multiply/divide infinite (2, Informative)

Speare (84249) | more than 3 years ago | (#33892910)

The same is true with an infinite repeating decimal. It is an irrational number.

Far from true. A rational number is a number you could get by expressing as a ratio (real number divided by real number). Any infinite repeating decimal is easily shown as a ratio (and often of simple integers to boot), i.e., a rational number. 0.22222... is 2/9. 0.456456456456456... is 456/999. And so on.

Cat and Mouse (0)

L4t3r4lu5 (1216702) | more than 3 years ago | (#33892716)

A cat has been chasing a mouse for months. He's never been able to quite get hold of it! Every time, the mouse either jumps into a hole, hides under a sofa, or skips off somewhere else just out of reach, and the mouse is fed up. The mouse says to the cat "Ok, I'm fed up. I don't want to be chased anymore. I'm tired and old, and the chase is too much." The cat is obviously elated! "I'll make you a deal, Mr Cat" says the mouse. "I will stop running and let you catch me, but only if you move half of the distance towards me with each step!"
The cat can't believe his luck! "Of course! Of course, I will only move half the distance with each step, as long as you do not run from me anymore!"
To which the mouse sits down gently, safe in the knowledge that he will never be caught by the cat. After all, no matter how close the cat gets, he can only get half the distance closer with each step...

TL;DR: 0.999... != 1. It's just really, really close. I don't care what number say; If you can show me an infinite accuracy measuring device, I'll show you a 0.999... unit of length structure is not 1 unit of length.

Re:Cat and Mouse (0)

Anonymous Coward | more than 3 years ago | (#33892898)

the concept of .999~ =! 1 has a big problem

1-.999~=x

Now let's say you want to say that as an infinitessimal, Ok then, let that equal x

1/x=?

Infinite? Well that was easy... but wouldn't that by proxy mean
5/x=inf
9/x=inf
inf/x=inf ... and there goes the neighborhood

Ahh I see... (1)

fishlet (93611) | more than 3 years ago | (#33892734)

So 2+2 really is 5?

Cribbed, Since My Memory for Jokes Sucks (5, Funny)

Anne_Nonymous (313852) | more than 3 years ago | (#33892738)

In the high school gym, all the girls in the class were lined up against one wall, and all the boys against the opposite wall. Then, every ten seconds, they walked toward each other until they were half the previous distance apart. A mathematician, a physicist, and an engineer were asked, "When will the girls and boys meet?"

The mathematician said: "Never."
The physicist said: "In an infinite amount of time."
The engineer said: "Well... in about two minutes, they'll be close enough for all practical purposes."

And if (3, Funny)

Dunbal (464142) | more than 3 years ago | (#33892740)

0.99999... is equal to 1, then 0.999999...8 is equal to 0.99999... and 0.9999999...7 is equal to 0.999999...6 etc etc etc until 1 = 0! Holy shit!

Or we could just admit that using a tool incorrectly produces idiotic results.

Re:And if (1)

Demablogia (1149365) | more than 3 years ago | (#33892978)

Human stupidity = infinity This axiom explains everything :-)

Ouch... (1)

whisper_jeff (680366) | more than 3 years ago | (#33892746)

That just hurt my brain and made sense at the same time...

Is it any wonder that The Big Bang Theory is one of my favourite shows?...

-sweeps problem under the rug- (1)

hort_wort (1401963) | more than 3 years ago | (#33892762)

Soooo before my coffee, it looks like this is just them moving the problem area infinitely far away. If you just start with 0.99 and do the same thing, you can see that the numerator =/= the denominator. This is kind of like taking a derivative, throwing away the differential parts because they're "so small anyway", then reintegrating to get your answer.
*blinks* Need coffee and donuts.....

Not really. It's the LIMIT that's equal to 1.0 (0)

Anonymous Coward | more than 3 years ago | (#33892768)

You are being very imprecise. The LIMIT expressed by the infinite series 0.999... is equal to 1.

But math is hard, and I don't expect you girls on /. to understand it.

Re:Not really. It's the LIMIT that's equal to 1.0 (1)

Quill_28 (553921) | more than 3 years ago | (#33892812)

I always used

1/3 = .33333....
2/3 = .66666....

1/3 + 2/3 = 3/3 .333333.. + .666666666.... = .999999.....

9th grade algebra... (0)

Anonymous Coward | more than 3 years ago | (#33892790)

I once saw this problem on a 9th grade algebra problem set. It was with a bunch of easier decimal to fraction conversions. I spent three hours on this one, constantly getting the same answer! I finally found the answer on a PHILOSOPHY discussion board!

Ummmm (0)

John Napkintosh (140126) | more than 3 years ago | (#33892826)

If we let a=0.999, then wouldn't 10a = 9.990, not 9.999?

Makes a pretty big difference.

Re:Ummmm (1)

John Napkintosh (140126) | more than 3 years ago | (#33892858)

Oh, yeah. I guess you can tell that my math scores might have been better than my reading comprehension scores.

Disregard.

Mine is better (1)

KiwiCanuck (1075767) | more than 3 years ago | (#33892836)

16/64 cross out the 6s and you get 1/4. Or Pi is exactly 3!

The problem (0)

Anonymous Coward | more than 3 years ago | (#33892860)

OK, for those of you repeating the same old boring 'proof' that 0.9999=1 by 1/9 *9=1 it comes down to this:

What do you MEAN by 0.999... ? Mathematically, what is this object? If it's the sum n=1 to infinity of 9*10^(-n) within real analysis then yes. If you're defining it to be 'the decimal representation of 1/9' then yes by tautology. But sadly that's not ALL it can be - it can be a member of the hyperreals, valued 1-w where w is the smallest number (read up on hyperreals before you respond please - http://en.wikipedia.org/wiki/Hyperreal_number [wikipedia.org] would be a good place to start). In that case it is most definitely NOT 1. TFA has a good list of things that this symbol could mean, and a discussion of the implications of each. I'd advise you to read it and learn what the issues are before effectively making the DEFINITION that 0.999...=1 in order to prove that statement.

The actual reason (1, Insightful)

Anonymous Coward | more than 3 years ago | (#33892872)

Decimal numbers are just names for points on the real number line (relative to a chosen point we call "0"). Thus one reason 0.999... is equal to 1 is that if they were referring to two different point on the number line, there would have to be a point (acutally infinitely many points) between them. Since every point on the real line can be written as a decimal (this is called the completeness property of the reals), and there is clearly no decimal greater than 0.999... and less than 1, then 0.999... and 1 must be the same point on the real line: the same number.

I proved this decades ago (1)

cdn-programmer (468978) | more than 3 years ago | (#33892880)

On a dare I proved this decades ago. Its really easy and took less than 10 minutes.

The issue is really one of notation. 1E0 also equals 1. It is not that 0.99999... is close to 1.0 It is actually equal to 1.0 and just another way to write 1.0.

Why is this is slashdot?

I don't agree... (0)

Anonymous Coward | more than 3 years ago | (#33892894)

multiply by ten and you must have a zero at the end (on the corresponding decimal place of the last 9 of 0.99999, no matter how far down the line that last nine is)

therefore you're actually left with 10a - a = 9.9999999999...0 - 0.99999999999...9
9a = 8.999999999...1
a = 0.9999999

Mass Effect software glitch (1)

Drakkenmensch (1255800) | more than 3 years ago | (#33892990)

One time back when the Quarians still had a planet, the Reapers tricked the Geth into thinking that 1.3382 was really worth 1.3381. Hilarity ensued.
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