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

  • Thread starter srfriggen
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    Convergence
In summary: Just like 0.9.. is not 1. Also I don't believe you that you only learned what a series was a week ago. You seem to have a good grasp of it and you have been using that formula all along.
  • #36
srfriggen said:
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.
 
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  • #37
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.
 
  • #38
Cyosis said:
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 [itex]2/3=0.666..[/itex] 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).

Cyosis said:
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:

[tex]
\sum_{n=0}^\infty a r^n=\frac{a}{1-r}
[/tex]

If we start at n=1 we get:

[tex]
\sum_{n=1}^\infty a r^n=\sum_{n=0}^\infty a r^n-\underbrace{a}_{\text{0th term}}=\frac{a}{1-r}-a
[/tex]


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:

[tex]\sum[/tex] 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, [tex]\sum[/tex] 6/10^n+1 (where n = 0) is equal to [tex]\sum[/tex] 6/10^n-11 (where n = 12)
 
  • #39
Mindscrape said:
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.)
 
  • #40
srfriggen said:
Yes I was wrong, it should have been .999 can be shown by a = 9/10, and r = 1/10, with the series obviously [tex]\sum[/tex] 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:

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

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 :smile:

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 [tex]a^{x+y}=a^xa^y[/tex]
 
  • #41
Mentallic said:
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:

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

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 :smile:

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 [tex]a^{x+y}=a^xa^y[/tex]

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.
 
  • #42
Well there is no post #69 (unless there's a joke to be had here, I don't get it).

Yep you got it.
 
<h2>What is the intuition behind .999... = 1?</h2><p>The intuition behind .999... = 1 is that the decimal representation of .999... is a mathematical concept known as a "limit". This means that as the number of 9s after the decimal point increases, the value gets closer and closer to 1. In fact, it can be proven that .999... is exactly equal to 1.</p><h2>How can .999... equal 1?</h2><p>While it may seem counterintuitive, the concept of limits in mathematics allows for .999... to equal 1. This is because as the number of 9s after the decimal point increases, the difference between .999... and 1 becomes smaller and smaller. In the limit, this difference becomes infinitesimally small, making .999... and 1 essentially the same value.</p><h2>Can you provide a real-life example of .999... = 1?</h2><p>One real-life example of .999... = 1 is the concept of time. We measure time in seconds, but there are also smaller units such as milliseconds and microseconds. In the limit, as the time interval becomes smaller and smaller, the difference between .999... seconds and 1 second becomes infinitesimally small, making them essentially the same value.</p><h2>Why is it important to understand that .999... = 1?</h2><p>Understanding that .999... is equal to 1 is important in mathematics and science because it allows for more precise calculations and proofs. It also helps to understand the concept of limits and the idea that numbers can approach a value without ever reaching it.</p><h2>Is there a visual representation of .999... = 1?</h2><p>Yes, there are several visual representations of .999... = 1. One example is a number line, where the distance between .999... and 1 becomes smaller and smaller as the number of 9s after the decimal point increases. Another example is a geometric representation, where a line segment with a length of .999... can be divided into smaller and smaller segments, approaching a length of 1.</p>

What is the intuition behind .999... = 1?

The intuition behind .999... = 1 is that the decimal representation of .999... is a mathematical concept known as a "limit". This means that as the number of 9s after the decimal point increases, the value gets closer and closer to 1. In fact, it can be proven that .999... is exactly equal to 1.

How can .999... equal 1?

While it may seem counterintuitive, the concept of limits in mathematics allows for .999... to equal 1. This is because as the number of 9s after the decimal point increases, the difference between .999... and 1 becomes smaller and smaller. In the limit, this difference becomes infinitesimally small, making .999... and 1 essentially the same value.

Can you provide a real-life example of .999... = 1?

One real-life example of .999... = 1 is the concept of time. We measure time in seconds, but there are also smaller units such as milliseconds and microseconds. In the limit, as the time interval becomes smaller and smaller, the difference between .999... seconds and 1 second becomes infinitesimally small, making them essentially the same value.

Why is it important to understand that .999... = 1?

Understanding that .999... is equal to 1 is important in mathematics and science because it allows for more precise calculations and proofs. It also helps to understand the concept of limits and the idea that numbers can approach a value without ever reaching it.

Is there a visual representation of .999... = 1?

Yes, there are several visual representations of .999... = 1. One example is a number line, where the distance between .999... and 1 becomes smaller and smaller as the number of 9s after the decimal point increases. Another example is a geometric representation, where a line segment with a length of .999... can be divided into smaller and smaller segments, approaching a length of 1.

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