## Friday, June 02, 2006

## About Me

**Name:**Karthik Nagaraj

Have you heard of Google+? I was part of the team that built Google+. I went on be the Technical Lead and Manager for Profiles and Pages in Google's Social world and now I lead and manage a bunch of engineers in the Social for Business Experiences space.

## Time since last post

## SCORE SO FAR

- Total puzzles posted: 26
- Puzzles solved so far: 24
- Moderator Involved: 6
- Bobach: 5
- Kala Ramachandra: 5
- Balaji: 4
- Vikram: 3
- Kavya: 2
- Sanjay: 2
- Mayura: 2
- Ishan: 1
- Steve: 1
- Dinesh: 1
- Victor Matheus: 1
- Amit Sarda: 1
- Ram: 1
- Vidya Vaidyanathan: 1
- Aditi Limaye: 1
- Ajit: 1
- Tejas: 1
- Abracadabra: 1

## Puzzles yet to be completed

## Puzzles with most comments

## Previous Posts

- Birthday matching
- Crazy guy on the airplane
- 000000000000000...............
- What's so strange about me?
- Playing quarters with the devil
- Name the hidden card
- Am I red, or Am I blue?
- Hands of a clock
- What's our average salary?
- Survivor of the pirates

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## 28 Comments:

well i dont know if this is correct or not, but here is my shot at it.

let 0.999999... = x

multiplying both sides by 10 we get

9.99999.... = 10x

which can again be written as

9 + x = 10x

solving for x we get

x=1.

but by first principle, it is said that

0.99999 is the closest number possible to 1 but not equal to one, because no matter how many 99999 u have, we can still find a number which has a fraction more than it.

Welcome Amit Sarda,

Good attempt. So what is your answer? Is it that 0.9999999999..... = 1? or 0.999999999...... < 1

I kind of understood your point and you have arrived at 0.9999.... = 1 and then again you say something else at the end. Can you please confirm?

It should be 1.

0.9 = 1 - 0.1 = 1 - 1/10^1

So, your desired number can be represented by 1 - 1/10^n

lim_{n -> Inf} (1 - 1/10^n) = 1

Ram,

The equation is a series and not just 0.9. It is 0.99999....

Hence 1 - 0.1 = 0.9 and not 0.999999.....

Bobach writes:

If N = 0.9999... then this is a repeating decimal of period 1 (that is, the number of digits in the repeating pattern is 1).

If we multiply by 10, we learn that 10N = 9.999... Subtract those 2 equations and we get 10N-N = 9, so N = 1.

Therefore, defying all my intuition, 0.999... is equal to 1!

Karthik,

I guess I should have explained better. What I meant was that your number 0.99999.. with infinite number of 9s is the same thing as the limit point of the infinite sequence 1 - 1/10^n.

So,

0.9 = 1 - 1/10 (n=1)

0.99 = 1 - 1/10^2 (n=2)

0.999 = 1 - 1/10^3 (n=3)

0.9999 = 1 - 1/10^4 (n=4)

...

0.999(with infinite 9s) = 1 - 1/10^n, evaluated as n -> Inf.

This evaluates to 1.

Bobach,

Yes thats true! You have gotten the same result as Amit Sarde. Thats good! But my concern is - is it really that simple to prove 0.999.... = 1, and if it is ture why even bother to have a number like 0.9999.... which is infinite.

Ram,

Your explanation seems more reasonable and more logical and provides a basis of calculus and limits. Let me think more about this...

MathForum seems to have some references here...

http://mathforum.org/library/drmath/view/57035.html

Also,

9 seems to have some peculiar behaviour

say,

1/9 = 0.11111111111...

2/9 = 0.22222222222...

3/9 = 0.33333333333...

4/9 = 0.44444444444...

on the same lines

can 9/9 be 0.9999999999... or 1

Mayura,

Your logic good until 9/9 -> Why should 9/9 be 0.999999..... and then again how this suddenly become 1 - doesn't it defy your own reasoning?

Bobach writes:

I think Amit was on the right path. I agree with others that the limit as n -> inf for any proper representation of your number is 1. However, a limit is not the same as equality (I think of a hyperbola).

No matter how many decimal places one examines, .9999... never equals 1. There is always a difference between .999999... and 1 until you get to the "last n" of infinity - which is a logical contradiction.

So, my previous answer was wrong, .9999... is, indeed, less than 1.

0.9999... < 1 is what it feels like it should be. So how do we confirm ?

1 subtracted from 0.9999... is neither 0 nor a positive number. It is negative. To me that suggests that 1 > 0.9999 ... and 1 != 0.9999... or 1 < 0.9999...

Another way to look at it is to plot both the numbers on a number line.No matter how hard it is to actually pin a point on this line as 0.9999..., we can reasonably assume that it lies between 0.9 and 1 i.e, it lies before 1.

Third, I guess every number in any number system is unique, so definitely the equality 1 = 0.9999... doesn't hold.

Just thought I'll have my say :) comments/corrections/suggestions welcome.

- Bhavan

I don't really grasp the concept of infinity that well.

So amit_sarda comes up with a brilliant proof (no calculus and all.. just plain elementary 6th Std. stuff) that actually proves 0.999... = 1,

But my gut says NO!

Can we actually have an infinite supply of 9's !? (for you to actually shift one 9 out on multiplying by 10 and still be left with the same number of 9's to subtract from the original infinite 9's)

So that means you haven't started out with infinite 9's in the first place!! Because you just reached a number bigger than what you so far thought was infinity !!

I feel the prob. lies with all of us actually 'abusing' infinity and carrying out routine operations, when actually you can't really define infinity in the first place.

Just one of those mysteries of math I guess.

So I would stick to the humble number line and say that 0.99..... < 1.

Abhay Kesheorey

This is reply to Bhavan's post - Agreed that 1 - 0.99999..... = 0.1111... and so this itself proves that 1 > 0.999.... BUT then the following is absurd:

Let us have a hypothesis that

1 > 0.999....(infinity)

Multiply by 10 on both sides

10 > 9.9999.... (infinity)

10 - 1 > 9.9999... - 0.9999....

9 > 9 ---> This is absurd as we know 9 = 9 and contradicting to our ooriginal hypothesis.

This is pretty much Amit's solution which goes against your reasoning

Abhay,

Good reasoning but then look at the law of infinities - you can definitely have an infinity supply of integers. Theoroetically we can imagine such a number that never ends, but practically our thinking is only finite. If you start with a line of any length, observe that you can divide it into half, then divide one of the pieces into half and keep repeating this process. Observe that no matter how many times you repeat the process, you will always have another piece that you can divide in half again.

Think about Hotel Infinity - where you can accomodate any # of guests because infinity + 1 = infinity and infinity - 1 = infinity. Isn't that what the logarithms and summationn series formulae based on? The law of limits and calculus? So lets think more hard and get more ideas.

Good job brainstorming, keep the creative juices flowing!

I am inclined to think along Abhay's lines.

consider N=0.99999...(upto infinity)

10N= 9.9999....(upto infinity-1)

so when we subtract,

9N=(9.9999...0)-(0.99999....)

(a zero at the end of a decimal doesnt change it's value..ryt and N has one decimal place more than 10N)

so 9N = 8.9999.....1

so we get back N=0.9999....

so 0.9999...<1

Kavya,

There is a fundamental flaw in your calculation.

0.9999.... * 10 is not equal to 0.9999...0

Infinity times something is still infinity. There is no bounds for it.

So 0.999.... * 10 = 0.999....(infinity # of nines again)

okay, how about this?

1/3=0.3333.....(upto infinity)

3*(1/3)=1

i.e 3*0.333...=0.9999....=1

Kavya,

Thats an interesting thing you have found there. Good job! This looks pretty logical to me. Lets hear more voices on what people have to say about this. Good one though, keep thinking like that for the other puzzles too.

i don't know if this solution have already benn post, but i usualy do this:

0.99999... = 0.9 + 0.09 + 0.009 + 0.0009 ....

so we have a geometric progression,

the first number is 0.9 and the common ratio is 0.1

using the geometric series sum formula we have:

A1*(q^n - 1) / q-1

0.9*( 0.1^infinite - 1 )/0.1-1

[ 0.1^infinite converges to zero ]

0.9*(-1)/-0.9

-0.9/-0.9 = 1

[sorry if me english was not perfect XD]

Victor Matheus

Your english is just fine, so dont worry about that and by the way welcome aboard by posting your first comment to the puzzle.

Your reasoning looks good. Different flavors of the same have been posted by different folks here. good. Using Geometric progression and summation series to express the answer is one good way to look at it.

Any more takers on this? So is it final that 0.9999.... = 1?

Definitely should close this one out. 0.9999... = 1. I think there are enough proofs that have been posted to convince anyone!!

Ok Guys! We have agreed that 0.99999.... = 1 until someone comes along and argues otherwise!. Wow its going to take some time to figure out who is the taker of this puzzle. There are multiple folks out there. Let me see:

Amit Sarda

Ram

Kavya

Victor Matheus

get the credits for this puzzle for solving it in different ways and methods. If I have missed anyone's name please let me know!

This may be late and in the worng place but since "What's the deal with 0.999999999999...." has been closed I am posting here.

0.999999999999.... will always be less than 1.

Each of the proofs that I saw for 0.999... = 1 had flaws.

Panda,

I put the comment in the right place. So why dont you pour in your views and we start discussing the flaws in the previous ones. cool! good!

This is sometimes referred to as "Xeno's Paradox" though it is not the same thing, it is the same principle. Xeno's paradox states:

In order to leave the room I am in, I need to first walk halfway to the door. From this point, I need to walk half the remaining distance. At any given time, I still need to walk half the distance ... how do I ever get there?

This is similar, in that you start at 0 and cover 90% of the way to 1 (0.9), then another 90% (0.99), then another 90% (0.999), and after infinite times, you actually reach 1. This isn't a rigorous proof, just a helpful analogy.

I agree with Panda. 0.9999999... will always be less than 1 no matter how many 9's there are because the number on the left of the decimal point is 0, which is less than 1. Oops, sorry for commenting on a closed/solved puzzle. Please check out my blog. Thanks!

Hi,

I am posting a comment on the "The bad king's wine cellar" problem which I happened to chance on recently. I am posting here since that thread was closed. Please bear with me.

The answer as posted by Karthik (http://puzzleguru.blogspot.com/2006/04/bad-kings-wine-cellar.html?showComment=1146614100000#c114661413696979682) is incorrect. Why?

Let us assume that the poisoned wine bottle was #121.

On the 31st day, prisoner #1 would die, next day prisoner #2 would die and the third day, none would die (because prisoner #1 is already dead!) and the king would not know if it was 121 or 122.

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