What is the Intuition Behind .999... = 1?

[srfriggen]

Homework Statement

Using the geometric series formula, a/1-r , where l r l < 1, it is easy to see that:

1/10 + 1/10(1/10)^2 + 1/10(1/10)^3... = 1.


applying the same formula to say, 2/3 = .666... does not give the answer 7, but rather 20/27.

My question is, what is the intuition behind .999... = 1?

It is easy to prove, and seemingly obvious (my girlfriend said, "well sure, even if it is infinite you would always have a 9 at the end so all them would round up like a domino effect from the end". I said, "sure, but then why doesn't that work for .666?" and I'm stumped (obviously her intuition about rounding up isn't what matters. Can someone tell me what does?

Homework Equations

The Attempt at a Solution

 

[Cyosis]

applying the same formula to say, 2/3 = .666... does not give the answer 7

Luckily it doesn't. I suggest you grab a ruler and measure 0.67 cm then measure 7 cm to see the difference between the two. It is also not equal to 20/27. In fact the method you've tried to use (albeit incorrectly) to prove that 0.9..=1 works for 0.6.. as well. The answer you should get is 0.666.=2/3.

1/10 + 1/10(1/10)^2 + 1/10(1/10)^3... = 1.

This is also wrong and would in fact be equal to 1/9.

It is easy to prove, and seemingly obvious (my girlfriend said, "well sure, even if it is infinite you would always have a 9 at the end so all them would round up like a domino effect from the end".

There is no end, the 9s keep going on forever. You can represent such a decimal as 0.9+0.09+0.009+... and link it to the geometric series to see it converges to one.

 

[srfriggen]

Luckily it doesn't. I suggest you grab a ruler and measure 0.67 cm then measure 7 cm to see the difference between the two. It is also not equal to 20/27. In fact the method you've used to prove that 0.9..=1 works for 0.6.. as well. The answer you should get is 0.666.=2/3.

I don't have a ruler near me lol, but I must have used the formula incorrectly then?

my a = 2/3 and my r = 1/10...

a/1-r = (2/3)/(9/10) = (2/3)*(10/9) = 20/27

 

[D H]

I don't have a ruler near me lol, but I must have used the formula incorrectly then?

my a = 2/3 and my r = 1/10...

a/1-r = (2/3)/(9/10) = (2/3)*(10/9) = 20/27

That is the case (i.e., you have used the formula incorrectly). Why did you pick those values?

 

[Cyosis]

I don't have a ruler near me lol, but I must have used the formula incorrectly then?

Nevertheless I am sure you know that something that is slightly more than half a centimeter is smaller than 7 cm. So I am at a loss why you think that 0.6666..=7.

You can't set a=2/3, 0.6..=2/3 and therefore your suggested sum would be 2/3+2/30+2/300+...=2/3, which is obviously false.

Try to write 0.66... like I wrote 0.99... in post #4 (last line).

 

[srfriggen]

That is the case. Why did you pick those values?

I picked the values from the sequence of partial sums...

.6+.06+.006... = 2/3(1/10)^0 + 2/3(1/10)^1 + 2/3(1/10)^2...

When I picked those values for .999... I used .9 = 1/10 = .09 = (1/10)^2 = .009 = (1/10)^3...


I'm seeing that my values must be incorrect. Can you explain how then to prove that 2/3 = .666... by using the geometric convergence series formula?

 

[Cyosis]

You claim that 0.6=2/3, which is false (note that the 6 is not repeating here). That's why you get the wrong answer. You have made the same error in trying to prove that 0.9..=1.

 

[srfriggen]

You claim that 0.6=2/3, which is false (note that the 6 is not repeating here). That's why you get the wrong answer.

to quote myself, "I'm seeing that my values must be incorrect. Can you explain how then to prove that 2/3 = .666... by using the geometric convergence series formula?"

 

[srfriggen]

I picked the values from the sequence of partial sums...

.6+.06+.006... = 2/3(1/10)^0 + 2/3(1/10)^1 + 2/3(1/10)^2...

When I picked those values for .999... I used .9 = 1/10 = .09 = (1/10)^2 = .009 = (1/10)^3...


I'm seeing that my values must be incorrect. Can you explain how then to prove that 2/3 = .666... by using the geometric convergence series formula?

I'm sorry, I mean how 2/3 = 7

 

[srfriggen]

to quote myself, "I'm seeing that my values must be incorrect. Can you explain how then to prove that 2/3 = .666... by using the geometric convergence series formula?"

Sorry, I meant to say, can you show me how to prove 2/3 = .666... = 7 using the geo conv series formula, a/1-r

 

[Cyosis]

No I am not going to do your work for you. I have told you where you went wrong multiple times. It is up to you to do something with that information. Your idea that 0.66...=0.6+0.06+0.006 is correct, but you then claim that 0.6=2/3 *1/10, which is wrong. If you fix your claim of 0.6=2/3 with the correct fraction you're done.

 

[Cyosis]

Sorry, I meant to say, can you show me how to prove 2/3 = .666... = 7 using the geo conv series formula, a/1-r

Honestly are you pulling our leg here? How can 0.6, which is something smaller than 1, be equal to 7, which is something 7 times as large as 1?

To answer your question. I cannot, no one can, because it is not true.

 

[srfriggen]

No I am not going to do your work for you. I have told you where you went wrong multiple times. It is up to you to do something with that information. Your idea that 0.66...=0.6+0.06+0.006 is correct, but you then claim that 0.6=2/3 *1/10, which is wrong. If you fix your claim of 0.6=2/3 with the correct fraction you're done.

I don't like asking for direct help and know this forum is certainly not for that reason, so I appreciate your answer and thank you for pointing me in the right direction (that my fraction was incorrect). This isn't a homework question, it is just something I am stuck on.

I'll try some different methods...

 

[srfriggen]

Honestly are you pulling our leg here? How can 0.6, which is something smaller than 1, be equal to 7, which is something 7 times as large as 1?

To answer your question. I cannot, no one can, because it is not true.

Listen, I realize my mistake now. My question was flawed from the start. I was trying to actually show that 6.666 = 7.

Can you at least let me know if THAT is true so I can start working on it?

(I know questions can be frustrating sometimes, but please keep in mind that some of us are very novice when it comes to math. I only learned what a series was a week ago).

 

[D H]

Of course 6.666... is not 7.

 

[Cyosis]

srfriggen. It is very important that you read the replies to your thread thoroughly. As I have said on numerous occasions, your proof of 0.999=1, in post #1 is also wrong. Yet you use that proof as the basis to prove the other 'equalities' in this thread. As a result all the other things you want to proof will be wrong too.

As for.

Listen, I realize my mistake now. My question was flawed from the start. I was trying to actually show that 6.666 = 7.

I really suggest you grab that ruler. You are now claiming, using a different analogy, that six whole pizzas plus two thirds of a pizza equals 7 whole pizzas. That's not true either.

(I know questions can be frustrating sometimes, but please keep in mind that some of us are very novice when it comes to math. I only learned what a series was a week ago).

You don't have to worry about this. However it is not the series that go wrong, but very simple arithmetic. If you fix your arithmetic you will obtain the correct answers. I refer once again to the last line in post #11.

 

[Mentallic]

If 0.999...=1 then 6+0.999...=6.999...=6+1=7

Obviously 6.666...= 7 is false.

 

[srfriggen]

srfriggen. It is very important that you read the replies to your thread thoroughly. As I have said on numerous occasions, your proof of 0.999=1, in post #1 is also wrong. Yet you use that proof as the basis to prove the other 'equalities' in this thread. As a result all the other things you want to proof will be wrong too.

Why is that proof wrong?

.999 = \sum (1/10)^n , and using a/1-r, we see that converges to 1. you are saying that is not a correct way to show .999 = 1?

 

[Cyosis]

You can definitely use geometric series to obtain the proof you desire. However what you're saying is wrong.

Using the geometric series:

<br /> \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n=\frac{1}{1-\frac{1}{10}}=\frac{10}{9}=1.11..&gt;1<br />

 

[D H]

Why is that proof wrong?

.999 = \sum (1/10)^n , and using a/1-r, we see that converges to 1. you are saying that is not a correct way to show .999 = 1?

That sum is 0.111..., not 0.999...

 

[Cyosis]

To avoid further confusion. The answers in post #19 and #20 differ from each other because the sum in post #19 runs from 0 to infinity while DH let it run from n=1 to infinity.

 

[srfriggen]

srfriggen. It is very important that you read the replies to your thread thoroughly. As I have said on numerous occasions, your proof of 0.999=1, in post #1 is also wrong. Yet you use that proof as the basis to prove the other 'equalities' in this thread. As a result all the other things you want to proof will be wrong too.





You don't have to worry about this. However it is not the series that go wrong, but very simple arithmetic. If you fix your arithmetic you will obtain the correct answers. I refer once again to the last line in post #11.

Ok, I've read through everything (sorry but there were a lot of replies in a short amount of time... just having an off day).


Yes I was wrong, it should have been .999 can be shown by a = 9/10, and r = 1/10, with the series obviously \sum 9 (1/10)^n

So that proves .999 = 1 by using the formula a/1-r (am I correct to this point?)


Further now, .666... can be represented by \sum 6(1/10)^n (correct?).


So when you use the formula, a=6/10 and r=1/10, you wind up with it converging to 2/3.

I think I have a grasp now. Can you confirm my work?

 

[Cyosis]

Yes the posts came fast after each other, which is why I already assumed you either skipped most of it or didn't have any time to digest their information.

Ok, I've read through everything (sorry but there were a lot of replies in a short amount of time... just having an off day).Yes I was wrong, it should have been .999 can be shown by a = 9/10, and r = 1/10, with the series obviously LaTeX Code: \\sum 9 (1/10)^n

So that proves .999 = 1 by using the formula a/1-r (am I correct to this point?)Further now, .666... can be represented by LaTeX Code: \\sum 6(1/10)^n (correct?).So when you use the formula, a=6/10 and r=1/10, you wind up with it converging to 2/3.

I think I have a grasp now. Can you confirm my work?

For both problems the a and r you have selected are now correct. However in your sums you, for some reason, use a different a, which makes them incorrect (assuming your sum runs from n=0 to infinity).

 

[srfriggen]

Yes the posts came fast after each other, which is why I already assumed you either skipped most of it or didn't have any time to digest their information.



For both problems the a and r you have selected are now correct. However in your sums you, for some reason, use a different a, which makes them incorrect (assuming your sum runs from n=0 to infinity).

no, my sums run from 1 to infinity (sorry I did not specify). Does that correct everything now?

 

[Cyosis]

Then it would be correct yes. Although looking at your first post in particular

Using the geometric series formula, a/1-r , where l r l < 1, it is easy to see that:

I have some doubts as to how you obtained that expression. I say this because the geometric series formula a/(1-r) belongs to a geometric series that starts at n=0.

 

[srfriggen]

Then it would be correct yes. Although looking at your first post in particular



I have some doubts as to how you obtained that expression. I say this because the geometric series formula a/(1-r) belongs to a geometric series that starts at n=0.

I'll look it up in my textbook a bit later. I have to get some work done here at the office!

But not sure why it should matter. If I choose n=1 everything works out. If n=0, then I am completely lost as to how to set up my geo series and would kindly ask that you show me.

 

[Cyosis]

It matters quite a bit. If you start at n=0 you add the 0th+1th+2nd+3rd+.. terms together. If you start at n=1 you add the 1th+2nd+3rd+... terms together. In other words if you start at n=1 you skip the first term.

Geometric series |r|<1:

<br /> \sum_{n=0}^\infty a r^n=\frac{a}{1-r}<br />

If we start at n=1 we get:

<br /> \sum_{n=1}^\infty a r^n=\sum_{n=0}^\infty a r^n-\underbrace{a}_{\text{0th term}}=\frac{a}{1-r}-a<br />

 

[Martin Rattigan]

1/10 + 1/10(1/10)^2 + 1/10(1/10)^3... = 1

This is also wrong and would in fact be equal to 1/9.

Strictly speaking 1/9-1/10²=91/900.

 

[Cyosis]

Your equality is correct yet has no bearing on this topic. No one here ever tried to calculate 1/9-1/10^2, which isn't even an infinite sum.

Edit: I see what you're getting at. He skipped the (1/10)*(1/10)^1 term.

 

[srfriggen]

So what would the closed form of .666 look like if you start n at zero??

 

[Mindscrape]

Hah, man srfriggen, I don't want to be mean, but you've got to use your brain! Analyze what people have told you, use some logic.

.6666... = \sum_{n=1}^{\infinity} 6 (\frac{1}{10})^n - 6=6\sum_{n=1}^{\infinity} (\frac{1}{10})^n -1 = a(\frac{1}{1-r}-1)

 

[srfriggen]

You know that is a little mean. I'm already down on myself because my 100 average in my calc II class got messed up by an 80 on the last test, which was all series. I appreciate you and everyone trying to push me in the right direction (teach a man to fish ) but I'm just stuck. You have the answer. I can't get it. That's frustrating and really making me doubt myself. I'm trying to use my brain and I hasn't been working lately. Maybe I'll see where I'm going wrong but I can't unless I see the answer.

Sorry I'm just terribly stressed. I can't figure It out okay?!?

 

[Mindscrape]

It's okay that you're not getting it, but slow down a bit. You're a bright person if you've been doing well in CalcII until now, something just happened between you and power series. You'll get it eventually, but in the mean time, you've got all the answers and explanations in front of you. Start over from the beginning, understand the .9999... example, move to the .66666... example and you'll get it. Once you get it, you'll look back and wonder why you didn't get it in the first place. Cheer up mate :)

 

[Cyosis]

So what would the closed form of .666 look like if you start n at zero??

Obviously the same. We're talking about a number here, a number doesn't just randomly change. Let this be clear once and for all 2/3=0.666.. period. Whatever method you're using to prove that equality, they must all give the same answer.

I suggest you start with the definition of a geometric series and to what expression it equals to (general case).

 

[srfriggen]

It's okay that you're not getting it, but slow down a bit. You're a bright person if you've been doing well in CalcII until now, something just happened between you and power series. You'll get it eventually, but in the mean time, you've got all the answers and explanations in front of you. Start over from the beginning, understand the .9999... example, move to the .66666... example and you'll get it. Once you get it, you'll look back and wonder why you didn't get it in the first place. Cheer up mate :)

Thanks mindscrape. I get down on myself but thank god the effect of that just makes me more determined rather than mope and give up. I need to figure out how to not doubt my abilities so quickly when I don't grasp something. My teachers all think I'm the best in my classes (I was told I had the highest numerical grades and that my approaches are always "more sophisticated"), but I'm still so hard on myself when I don't understand something. My younger sister, a soon to be doctor, is the same way. 3.9 avg at georgetown, high honors in her classes at downstate medical center, yet the moment she gets a bad grade or thinks she gets a bad grade she throws everything out the window.

Hopefully this doesn't get moved into the philosophical section. I'm 28 with a degree in Economics from Villanova University, worked for Merrill Lynch, hated it, now work for my family, and am taking undergrad courses as prereq's for a graduate program in math. I wonder, if I get so stressed now what is it going to be like when I am doing this full time?!

Do you guys have any advice on how to be successful in this field? (It's the only thing I am passionate about, so there is a lot of stress to make it work, you know? If not this, then what?, you know?)

 

[jbunniii]

Do you guys have any advice on how to be successful in this field? (It's the only thing I am passionate about, so there is a lot of stress to make it work, you know? If not this, then what?, you know?)

Everyone who studies math has hit roadblocks many many times. Sometimes it just takes some time to digest a new idea before you get the "ah ha!" moment. It's completely normal and it doesn't mean that you're no good at it or anything like that.

Someone else had a great quote the other day, unfortunately I don't remember who so I can't attribute it, but it was something like "even if you're exceptionally good at math it just means you have an SUV so you can get stuck in more exotic places than the rest of us."

If possible, it's often useful to just take a break from whatever problem you're banging your head against, and do something completely unrelated instead. Then come back to the problem with a clear mind. Sometimes what seemed inscrutable before suddenly seems obvious. Even if not, at least you will be rested and less stressed, and therefore better positioned to make some progress.

 

[jbunniii]

To address the problem at hand, (why is 6.666... not equal to 7), try thinking about it this way. We will start with 6, then start adding 6's to the right of the decimal point, one at a time. At each step, we will check how far away we are from 7. If 6.666... really equaled 7, we would expect to see that distance shrink to zero as we add more digits.

Step 1: 6 -- distance from 7 is 1

Step 2: 6.6 -- distance from 7 is 0.4

Step 3: 6.66 -- distance from 7 is 0.34

Step 4: 6.666 -- distance from 7 is 0.334

Step 5: 6.6666 -- distance from 7 is 0.3334

See the trend? We are clearly NOT going to get arbitrarily close to 7. In fact it's clear that even if we added infinitely many digits, we would still be at a distance of 0.333... from 7.

Now, 0.333... is clearly half of 0.666...

So what fraction of the way are we between 6 and 7? Well, whatever fraction it is, we would have to increase it by half of the same fraction to get all the way to 7. Call the fraction x. Then x has to satisfy

x + (1/2) x = 1

(3/2) x = 1

x = 2/3

Thus 0.666... must equal 2/3, so 6.666... equals 6 and 2/3rds.

 

[srfriggen]

Obviously the same. We're talking about a number here, a number doesn't just randomly change. Let this be clear once and for all 2/3=0.666.. period. Whatever method you're using to prove that equality, they must all give the same answer.

I suggest you start with the definition of a geometric series and to what expression it equals to (general case).

It matters quite a bit. If you start at n=0 you add the 0th+1th+2nd+3rd+.. terms together. If you start at n=1 you add the 1th+2nd+3rd+... terms together. In other words if you start at n=1 you skip the first term.

Geometric series |r|<1:

<br /> \sum_{n=0}^\infty a r^n=\frac{a}{1-r}<br />

If we start at n=1 we get:

<br /> \sum_{n=1}^\infty a r^n=\sum_{n=0}^\infty a r^n-\underbrace{a}_{\text{0th term}}=\frac{a}{1-r}-a<br />

ok i think think think i have it...

starting at n=0, and using my values for a and r, the closed form would be:

\sum 6/10^n+1

now I can still have my a = 6/10 , and r = 1/10, I can use the formula a/1-r , and the series converges to 2/3 = .666

What makes sense now (correct me if I am wrong), is that to keep the series equal, so that you are not missing any terms, if you decrease or increase the starting point n, you much add or subtract that value to all the n's in the closed form before continuing... ex, \sum 6/10^n+1 (where n = 0) is equal to \sum 6/10^n-11 (where n = 12)

 

[srfriggen]

It's okay that you're not getting it, but slow down a bit. You're a bright person if you've been doing well in CalcII until now, something just happened between you and power series. You'll get it eventually, but in the mean time, you've got all the answers and explanations in front of you. Start over from the beginning, understand the .9999... example, move to the .66666... example and you'll get it. Once you get it, you'll look back and wonder why you didn't get it in the first place. Cheer up mate :)

Check out my last post, have I figured it out? (sorry for the life story etc.)

 

[Mentallic]

Yes I was wrong, it should have been .999 can be shown by a = 9/10, and r = 1/10, with the series obviously \sum 9 (1/10)^n

Ok from the looks of it, it seems as though they're racking your brains about the sum being from 0 to infinite rather than from 1 to infinity.
There's a quick and easy fix for that. Change your power as so:

\sum_{n=1}^\infty 9\left(\frac{1}{10}\right)^n=\sum_{n=0}^\infty9\left(\frac{1}{10}\right)^{n+1}

Can you see why this works? For the right side of the equality if we put n=0 in, the power becomes 1, put in n=1 the power becomes 2. This is the same as what you had, but now we have a sum from 0 to infinity

Ok but anyone would look at that power and think, why isn't it simplified to a power of n rather than having n+1 there?

Can you fix this? Think about the rule a^{x+y}=a^xa^y

 

[srfriggen]

Ok from the looks of it, it seems as though they're racking your brains about the sum being from 0 to infinite rather than from 1 to infinity.
There's a quick and easy fix for that. Change your power as so:

\sum_{n=1}^\infty 9\left(\frac{1}{10}\right)^n=\sum_{n=0}^\infty9\left(\frac{1}{10}\right)^{n+1}

Can you see why this works? For the right side of the equality if we put n=0 in, the power becomes 1, put in n=1 the power becomes 2. This is the same as what you had, but now we have a sum from 0 to infinity

Ok but anyone would look at that power and think, why isn't it simplified to a power of n rather than having n+1 there?

Can you fix this? Think about the rule a^{x+y}=a^xa^y

Yes I see now. Check out post #69. I seem to have it figured out :)

and I see what you mean... something like 5^n+1 can be expressed as 5^n X 5. From there you can start simplifying things our pulling out constants from sigma.

 

[Mentallic]

Well there is no post #69 (unless there's a joke to be had here, I don't get it).

Yep you got it.