# Proving 0.999... Is Equal To 1

#### CmdrTaco posted about 4 years ago | from the nerds-like-math-right dept.

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."*

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

## Anonymous Coward | about 4 years ago | (#33892528)

(0.999...)st Post!

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

## arivanov (12034) | about 4 years ago | (#33892616)

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

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

## L4t3r4lu5 (1216702) | about 4 years ago | (#33892818)

Maths is fun!

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

## derrida (918536) | about 4 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) | about 4 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) | about 4 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) | about 4 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) | about 4 years ago | (#33892606)

## Re:I went one further (1, Interesting)

## Anonymous Coward | about 4 years ago | (#33892580)

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

## Re:I went one further (2, Funny)

## atisss (1661313) | about 4 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) | about 4 years ago | (#33892540)

## Re:I'm Surprised (3, Funny)

## Anonymous Coward | about 4 years ago | (#33892630)

## Re:I'm Surprised (0)

## Anonymous Coward | about 4 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) | about 4 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) | about 4 years ago | (#33892550)

## Re:Finally (1, Insightful)

## operagost (62405) | about 4 years ago | (#33892668)

## Re:Finally (5, Funny)

## Vectormatic (1759674) | about 4 years ago | (#33892704)

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

## Re:Finally (3, Insightful)

## SoVeryTired (967875) | about 4 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) | about 4 years ago | (#33892552)

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) | about 4 years ago | (#33892610)

## Re:This is second place (4, Informative)

## MyLongNickName (822545) | about 4 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) | about 4 years ago | (#33892688)

## Re:This is second place (-1)

## Jezza (39441) | about 4 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) | about 4 years ago | (#33892816)

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) | about 4 years ago | (#33892822)

## Re:This is second place (1)

## MyLongNickName (822545) | about 4 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) | about 4 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 | about 4 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) | about 4 years ago | (#33892886)

## Re:This is second place (1)

## NYMeatball (1635689) | about 4 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) | about 4 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) | about 4 years ago | (#33892828)

I wish that would fit in my sig.

## Re:This is second place (1)

## MindStalker (22827) | about 4 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) | about 4 years ago | (#33892866)

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

## betterunixthanunix (980855) | about 4 years ago | (#33892974)

## Re:This is second place (1)

## jasmusic (786052) | about 4 years ago | (#33892970)

## Seriously? (0)

## Anonymous Coward | about 4 years ago | (#33892558)

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

## Sometimes (0)

## Reilaos (1544173) | about 4 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

knowingbetter.## Re:Sometimes (1)

## betterunixthanunix (980855) | about 4 years ago | (#33892732)

## Re:Sometimes (0)

## Anonymous Coward | about 4 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) | about 4 years ago | (#33892782)

## There are other, less tricky, proofs (1)

## pyrognat (233814) | about 4 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) | about 4 years ago | (#33892598)

"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:"## metacafe (1)

## iamhassi (659463) | about 4 years ago | (#33892662)

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

## iamhassi (659463) | about 4 years ago | (#33892900)

0.999.... might be equal to 1, but 0.999 isnotequal 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 | about 4 years ago | (#33892614)

Now I can replace my SLA with 100% uptime.

## Then again... (1, Troll)

## kannibul (534777) | about 4 years ago | (#33892618)

## Re:Then again... (1)

## kannibul (534777) | about 4 years ago | (#33892646)

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

## kannibul (534777) | about 4 years ago | (#33892876)

## Re:Then again... (1)

## Vectormatic (1759674) | about 4 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) | about 4 years ago | (#33892622)

## Re:Um (0)

## Anonymous Coward | about 4 years ago | (#33892832)

Except then you never prove a=1. ;)

## Re:Um (0)

## Anonymous Coward | about 4 years ago | (#33892862)

* Hand over head motion *

Whooooosh!

## Or (3, Insightful)

## Anonymous Coward | about 4 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) | about 4 years ago | (#33892628)

## I learned this in grade 6 (0)

## Anonymous Coward | about 4 years ago | (#33892638)

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

## sum the geometric series (1)

## sevenfactorial (996184) | about 4 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) | about 4 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) | about 4 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 | about 4 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) | about 4 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) | about 4 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) | about 4 years ago | (#33892660)

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

In decimal form:

So,

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

## DiSKiLLeR (17651) | about 4 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) | about 4 years ago | (#33892672)

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 | about 4 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) | about 4 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

exactlyinfinite 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 | about 4 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

iswhat it is, and what it is is 1.## Re:This is just faulty math (1)

## JohnFluxx (413620) | about 4 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 | about 4 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 | about 4 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) | about 4 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 | about 4 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) | about 4 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) | about 4 years ago | (#33892674)

## Recipe for Forum Disaster (0)

## Anonymous Coward | about 4 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) | about 4 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) | about 4 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) | about 4 years ago | (#33892710)

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

## Speare (84249) | about 4 years ago | (#33892910)

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) | about 4 years ago | (#33892716)

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 | about 4 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 ... and there goes the neighborhood

5/x=inf

9/x=inf

inf/x=inf

## Ahh I see... (1)

## fishlet (93611) | about 4 years ago | (#33892734)

So 2+2 really is 5?

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

## Anne_Nonymous (313852) | about 4 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) | about 4 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) | about 4 years ago | (#33892978)

## Ouch... (1)

## whisper_jeff (680366) | about 4 years ago | (#33892746)

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

## -sweeps problem under the rug- (1)

## hort_wort (1401963) | about 4 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 | about 4 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) | about 4 years ago | (#33892812)

I always used

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

2/3 =

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

## 9th grade algebra... (0)

## Anonymous Coward | about 4 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) | about 4 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) | about 4 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) | about 4 years ago | (#33892836)

## The problem (0)

## Anonymous Coward | about 4 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 | about 4 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) | about 4 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 | about 4 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

## That is nothing comparing with 1+2+3+4...=-1/12 (1)

## S3D (745318) | about 4 years ago | (#33892960)

## Mass Effect software glitch (1)

## Drakkenmensch (1255800) | about 4 years ago | (#33892990)