.99999 does not equal 1

The thing is that it doesn't work that way if you have an infinite amount of nines. Infinity plus one is just infinity again. You can't put something at the end of infinity. Your sig does ring true., cbarnett.

If the naysayers are ever actually able to back their position up let me know.  Otherwise, it's clearly been sufficiently proven that .999... = 1. 

Quote from: Wendigo on August 11, 2008, 05:13:30 PM

The thing is that it doesn't work that way if you have an infinite amount of nines. Infinity plus one is just infinity again. You can't put something at the end of infinity. Your sig does ring true., cbarnett.

Same Idea different numbers:  F(x)=1/x when does x=0

Or f(x)=1/(1-x) when does x=1

Quote from: cbarnett97 on August 11, 2008, 05:34:17 PM

Quote from: Wendigo on August 11, 2008, 05:13:30 PM

The thing is that it doesn't work that way if you have an infinite amount of nines. Infinity plus one is just infinity again. You can't put something at the end of infinity. Your sig does ring true., cbarnett.

Same Idea different numbers:  F(x)=1/x when does x=0

Or f(x)=1/(1-x) when does x=1

x equals one at an undefined point in the equation. That has nothing to do with .99 repeating.

Quote from: Raist on August 11, 2008, 05:40:17 PM

Quote from: cbarnett97 on August 11, 2008, 05:34:17 PM

Quote from: Wendigo on August 11, 2008, 05:13:30 PM

The thing is that it doesn't work that way if you have an infinite amount of nines. Infinity plus one is just infinity again. You can't put something at the end of infinity. Your sig does ring true., cbarnett.

Same Idea different numbers:  F(x)=1/x when does x=0

Or f(x)=1/(1-x) when does x=1

x equals one at an undefined point in the equation. That has nothing to do with .99 repeating.

what set of numbers makes this equation valid?

Welcome to make a point or stfu.

Quote from: cbarnett97 on August 11, 2008, 05:41:18 PM

Quote from: Raist on August 11, 2008, 05:40:17 PM

Quote from: cbarnett97 on August 11, 2008, 05:34:17 PM

Quote from: Wendigo on August 11, 2008, 05:13:30 PM

The thing is that it doesn't work that way if you have an infinite amount of nines. Infinity plus one is just infinity again. You can't put something at the end of infinity. Your sig does ring true., cbarnett.

Same Idea different numbers:  F(x)=1/x when does x=0

Or f(x)=1/(1-x) when does x=1

x equals one at an undefined point in the equation. That has nothing to do with .99 repeating.

what set of numbers makes this equation valid?

You gave a domain of 1 number. Therefore no numbers in this set give f(x) a defined value. Welcome to make a point or stfu.

Quote from: Raist on August 11, 2008, 05:44:36 PM

Quote from: cbarnett97 on August 11, 2008, 05:41:18 PM

Quote from: Raist on August 11, 2008, 05:40:17 PM

Quote from: cbarnett97 on August 11, 2008, 05:34:17 PM

Quote from: Wendigo on August 11, 2008, 05:13:30 PM

The thing is that it doesn't work that way if you have an infinite amount of nines. Infinity plus one is just infinity again. You can't put something at the end of infinity. Your sig does ring true., cbarnett.

Same Idea different numbers:  F(x)=1/x when does x=0

Or f(x)=1/(1-x) when does x=1

x equals one at an undefined point in the equation. That has nothing to do with .99 repeating.

what set of numbers makes this equation valid?

You gave a domain of 1 number. Therefore no numbers in this set give f(x) a defined value. Welcome to make a point or stfu.

Let me slow down for you.........

Could x be .99999repeating forever and still give a defined result?

Quote from: cbarnett97 on August 11, 2008, 05:47:00 PM

Quote from: Raist on August 11, 2008, 05:44:36 PM

Quote from: cbarnett97 on August 11, 2008, 05:41:18 PM

Quote from: Raist on August 11, 2008, 05:40:17 PM

Quote from: cbarnett97 on August 11, 2008, 05:34:17 PM

Quote from: Wendigo on August 11, 2008, 05:13:30 PM

The thing is that it doesn't work that way if you have an infinite amount of nines. Infinity plus one is just infinity again. You can't put something at the end of infinity. Your sig does ring true., cbarnett.

Same Idea different numbers:  F(x)=1/x when does x=0

Or f(x)=1/(1-x) when does x=1

x equals one at an undefined point in the equation. That has nothing to do with .99 repeating.

what set of numbers makes this equation valid?

You gave a domain of 1 number. Therefore no numbers in this set give f(x) a defined value. Welcome to make a point or stfu.

Let me slow down for you.........

Could x be .99999repeating forever and still give a defined result?

Yes, when you use the proof I supplied.

so how can .99999....=1

3/3 = 1

1/3= .33333...

Therefore, (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1

Quote from: cbarnett97 on August 11, 2008, 05:51:08 PM

so how can .99999....=1

Quote from: Hara Taiki on August 09, 2008, 07:18:17 PM

3/3 = 1

1/3= .33333...

Therefore, (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) roughly equal 1

Fixed

If the naysayers are ever actually able to back their position up let me know.

Dear God! Make it stop!

Still trying to convince me that my calculator is wrong.

In fact, my calculator just proved that my proof is correct.

Divide 1 by 3, so you get (.3333333...). Don't clear the screen. Multiply that result by 3, and tell me what you get.

Then, remember that basic maths tell us that multiplying (.333333...) by three is the same as performing (.33333333....) + (.33333333....) + (.33333333....). So if the result of multiplying (.333333...) by 3 = 1, so does (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1.

Go ahead, call all calculators liars.

Still trying to convince me that my calculator is wrong.

In fact, my calculator just proved that my proof is correct.

Divide 1 by 3, so you get (.3333333...). Don't clear the screen. Multiply that result by 3, and tell me what you get.

Then, remember that basic maths tell us that multiplying (.333333...) by three is the same as performing (.33333333....) + (.33333333....) + (.33333333....). So if the result of multiplying (.333333...) by 3 = 1, so does (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1.

Go ahead, call all calculators liars.

In mathematics, the recurring decimal 0.999…, which is also written as 0.\bar{9} , 0.\dot{9} or \ 0.(9), denotes a real number equal to 1. In other words, the notations "0.999…" and "1" represent the same real number. The equality has long been accepted by professional mathematicians and taught in textbooks. Various proofs of this identity have been formulated with varying rigour, preferred development of the real numbers, background assumptions, historical context, and target audience.

.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.

You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1, and here's the key:

THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO.

Thus, if you are going to assign a value to .9999... (going on
forever), the only sensible value is 1.

There is nothing special about .999...  The idea that 1/3 = .3333...
is the same. None of .3, .33, .333333, etc. is exactly equal to 1/3,
but with each 3 added, the fraction is closer than the previous
approximation. In addition, 1/3 is the ONLY number that the series
gets arbitrarily close to.

As before, 0.999... = 0.9 + 0.09 + 0.009 + ... = 9/10 + 9/100 + 9/1000 + ... The expression 0.999... implies an infinite sum, i.e., a sum of an infinite number of terms. In mathematics, such infinite sums are called series. Series are introduced and studied rigorously in Calculus, where a distinction is made: some series are convergent, some are divergent. Every convergent series has a unique number associated with it, its sum. Divergent series are not so lucky. All infinite decimal fractions, like 0.999..., are shown to correspond to convergent series (which converge to their respective sums.) 0.999... converges to 1 which is expressed simply as 0.999... = 1. Similarly, 0.333... is a convergent series whose sum is 1/3: 0.333... = 1/3.

Quote from: Hara Taiki on August 11, 2008, 05:57:50 PM

Still trying to convince me that my calculator is wrong.

In fact, my calculator just proved that my proof is correct.

Divide 1 by 3, so you get (.3333333...). Don't clear the screen. Multiply that result by 3, and tell me what you get.

Then, remember that basic maths tell us that multiplying (.333333...) by three is the same as performing (.33333333....) + (.33333333....) + (.33333333....). So if the result of multiplying (.333333...) by 3 = 1, so does (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1.

Go ahead, call all calculators liars.

This isn't technically proof since the calculator rounds the number off after a certain number of digits....

Quote from: Raist on August 11, 2008, 06:00:15 PM

Quote from: Hara Taiki on August 11, 2008, 05:57:50 PM

Still trying to convince me that my calculator is wrong.

In fact, my calculator just proved that my proof is correct.

Divide 1 by 3, so you get (.3333333...). Don't clear the screen. Multiply that result by 3, and tell me what you get.

Then, remember that basic maths tell us that multiplying (.333333...) by three is the same as performing (.33333333....) + (.33333333....) + (.33333333....). So if the result of multiplying (.333333...) by 3 = 1, so does (.33333333....) + (.33333333....) + (.33333333....) = (.99999999...) = 1.

Go ahead, call all calculators liars.

This isn't technically proof since the calculator rounds the number off after a certain number of digits....

Only in the display. But in memory, the number is fully crunched. Else, when I did the multiplication, it would not have told me simply "1". It would have been less than 1. But the fact that even though it only showed me a limited number of 3's and showed me the result as "1" is proof enough that my proof is 100% correct.

I have a video displaying my proof. Once it's done processing, I will link to it. Then you can try and disprove it and the three sources Roundy posted.

You're not seeing the point. If the calculator did, in fact, round it off, why did I get 1 and not some rounded version of .99999999999999?

My calculator displayed 0.33333333333333333333333333333333. So add that to itself 3 times, and you would get 0.999999999999999999999999999999.

But I got 1 when multiplying it in the calculator. You can see the steps I took in the video when I link to it.

EDIT: " class="bbc_link" target="_blank" rel="noopener noreferrer">Video. Watch fullscreen.

You're not seeing the point. If the calculator did, in fact, round it off, why did I get 1 and not some rounded version of .99999999999999?

My calculator displayed 0.33333333333333333333333333333333. So add that to itself 3 times, and you would get 0.999999999999999999999999999999.

But I got 1 when multiplying it in the calculator. You can see the steps I took in the video when I link to it.

EDIT: " class="bbc_link" target="_blank" rel="noopener noreferrer">Video. Watch fullscreen.

"Thus, if you are going to assign a value to .9999... (going on
forever), the only sensible value is 1."

Now where does it say that they are equal?

It says basically that while it does not equal 1 it is just a whole lot easier to write one instead of an infinte number of 9's