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:

$$\sum_{n=0}^\infty \left(\frac{1}{10}\right)^n=\frac{1}{1-\frac{1}{10}}=\frac{10}{9}=1.11..>1$$

 

[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:

$$\sum_{n=0}^\infty a r^n=\frac{a}{1-r}$$

If we start at n=1 we get:

$$\sum_{n=1}^\infty a r^n=\sum_{n=0}^\infty a r^n-\underbrace{a}_{\text{0th term}}=\frac{a}{1-r}-a$$

 

[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??