Is 1 = 0.9999......

[KydLynx]

Originally posted by: Fausto

Originally posted by: dullard
It depends, are you a mathematician or an engineer?

...or a philosopher?

:beer:

 

[RossGr]

Originally posted by: MadRat

Originally posted by: RossGr

Originally posted by: MadRat
Not really, it is basically the same relative argument.

To you, who thinks infinity is simply a large finite number, perhaps so.

You might say your limited concept of infinity limits you.

I've already shown you that you can take the argument to any finite place and so even at infinity it is true, too. You reverse polarity of the truth at infinity. I basically deny that the limit of infinity exists, which is the basis in your argument, but if it did then it would not be true anyhow.

You have only stated your beliefs, you have proved nothing. Your believes have no bearing on the outcome of a mathematical proof.

 

[MadRat]

Originally posted by: RossGr

Originally posted by: MadRat

Originally posted by: RossGr

Originally posted by: MadRat
Not really, it is basically the same relative argument.

To you, who thinks infinity is simply a large finite number, perhaps so.
You might say your limited concept of infinity limits you.

I've already shown you that you can take the argument to any finite place and so even at infinity it is true, too. You reverse polarity of the truth at infinity. I basically deny that the limit of infinity exists, which is the basis in your argument, but if it did then it would not be true anyhow.

You have only stated your beliefs, you have proved nothing. Your believes have no bearing on the outcome of a mathematical proof.

Sorry, but they are statements of truth I've repeatedly related to your argument, not opinions. So far the only .999...=1 arguments have been malformed arguments that have neither proven the statement .999...=1 nor that .999... can exist relative to any other number. The burden of proof is on you to make your argument true or not. Nothing as far as I've said about .999<>1, or that .999... is not relative to any definite number, is contradictory or wrong.

 

[Hector13]

Originally posted by: MadRat

Sorry, but they are statements of truth I've repeatedly related to your argument, not opinions. So far the only .999...=1 arguments have been malformed arguments that have neither proven the statement .999...=1 nor that .999... can exist relative to any other number. The burden of proof is on you to make your argument true or not. Nothing as far as I've said about .999<>1, or that .999... is not relative to any definite number, is contradictory or wrong.

when you say ".999... is not relative to any definite number", do you mean you can't tell me whether or not 2 > .9999..?
Or if 0 < .999....?

 

[JupiterJones]

Sorry, but they are statements of truth I've repeatedly related to your argument, not opinions. So far the only .999...=1 arguments have been malformed arguments that have neither proven the statement .999...=1 nor that .999... can exist relative to any other number. The burden of proof is on you to make your argument true or not. Nothing as far as I've said about .999<>1, or that .999... is not relative to any definite number, is contradictory or wrong.

This was proven hundreds of post ago. One simple proof (informally) is a proof by contradiction. You start with the assumption that .9999999999... is not equal to 1. Then for this to be true, there has to be some number Y such that .99999...<y<1 or 1<y<.9999... It is readily shown that for any Y this cannot be true. Therefore the original assumption is false, i.e. .99999.... is not equal to 1.

OR

1/3 = 0.333...

multiply by three to get
1 = 0.999...

OR

let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x = 1

Could it be any plainer?

 

[Krk3561]

its proved by calculus, 1 = 0.9999999...

&#8734;
&#8721; (9)(1/10)^n
n=1

by simply doing the series term by term you get:
1/9 + 1/90 + 1/900 + 1/9000 + ...... = 0.99999999999999.....

this is geometric series meaning its solution can be found by the formula:
(a)
(1-r)

(a = the first term in the series and r = what the series is being multiplied by)

which when applied to this series equals:
(9/10)
(1 - 1/10)

which when simplified equals:
1


This is a proof. There is no doubt! 0.9999999999..... = 1

 

[DrPizza]

Originally posted by: Krk3561
its proved by calculus, 1 = 0.9999999...

??
?? (9)(1/10)^n
n=1

by simply doing the series term by term you get:
1/9 + 1/90 + 1/900 + 1/9000 + ...... = 0.99999999999999.....

this is geometric series meaning its solution can be found by the formula:
(a)
(1-r)

(a = the first term in the series and r = what the series is being multiplied by)

which when applied to this series equals:
(9/10)
(1 - 1/10)

which when simplified equals:
1


This is a proof. There is no doubt! 0.9999999999..... = 1

And a correct proof, but before someone misinterprets your proof and say "it never reaches 1, that's the limit", I'll add (again

) Yes, the limit is 1. And, realize that .999.... ***IS*** equal to the limit, not approaches the limit.

 

[MadRat]

The halfway point between 0 and 1 is .5, but the same is not true for the halfway point between .999... and 0.

 

[abracadabra1]

Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

What do you think?

I've bolded the line I don't like. You just can't subtract 0.999... from 9.999...

The mathematicians will strike you down for doing so...or at least I will.

 

[silverpig]

Originally posted by: MadRat
The halfway point between 0 and 1 is .5, but the same is not true for the halfway point between .999... and 0.

Sure it is. Do the division.

 

[pennylane]

Simple Proof To End All Proofs

1/3 = .3333.......

1/3 + 1/3 + 1/3 = 1

.3333... + .3333... + .3333... = .9999......

.9999.... = 1

This has been posted before, but it's the simplest one I know, so for most people it should be believable.
But if for some reason you don't believe that 1/3 is a decimal followed by an infinite number of 3's... well then.... there is no hope for you.

 

[dighn]

Originally posted by: fanerman91
Simple Proof To End All Proofs

1/3 = .3333.......

1/3 + 1/3 + 1/3 = 1

.3333... + .3333... + .3333... = .9999......

.9999.... = 1

this has been raised many times

 

[pennylane]

Let's make a list of people that believe the two are equal versus people that don't. We learned the limit proof in calculus. If you've had calculus (and remember) you should know they're equal. I don't understand why so many think otherwise.

 

[AzNmAnJLH]

according to the math.... i totally agree

how about 1.0000.......... = 0.9999.............?

1.0000...... >= 1 >= 0.9999......

If 0.9999.... has been proved to = 1 then prove that 1.0000..... = 1

x = 1.0000......
10x = 10.0000........
10x - x = 10.0000...... - 1.0000.......
9x = 9
x = 1 and since 1 = 0.9999.....,


1.0000....... = 0.9999.........



remember 1.0000.... represents a number that is slightly larger than 1 but not really according to the math and that 0.9999.... represents a number that is slightly smaller than 1 but of course again not really since the math proved they all equals

do i make sense? tell me i'm on crack

 

[Chu]

Maybe this thread could die if someone could please convince Madrat that infinity is not a number

"Really Frigging Huge" != Infinite. You don't get to infinity in an algorithmic way.

-Chu

 

[JupiterJones]

The halfway point between 0 and 1 is .5, but the same is not true for the halfway point between .999... and 0.

Of course it is, because .499999999999... = .5

 

[HJB417]

but when u plot this on a graph
wont the line always go closer to 1 but never touch 1
I forgot the calc term for lines/functions like that.

 

[Krk3561]

Originally posted by: HJB417
but when u plot this on a graph
wont the line always go closer to 1 but never touch 1
I forgot the calc term for lines/functions like that.

asympote (sp?), but anyway it does reach 1 @ infinity

 

[Legendary]

Originally posted by: HJB417
but when u plot this on a graph
wont the line always go closer to 1 but never touch 1
I forgot the calc term for lines/functions like that.

The term you're thinking of is asymtotic, but the proof through series above proves .999... to equal 1 anyway.

 

[Hector13]

Originally posted by: Krk3561

Originally posted by: HJB417
but when u plot this on a graph
wont the line always go closer to 1 but never touch 1
I forgot the calc term for lines/functions like that.

asympote (sp?), but anyway it does reach 1 @ infinity

jesus.. did you guys even try to read the thread?
.9999r is a NUMBER (it is the same as the number one, but that is a different point all toghether). It does not approach anything.. it IS A NUMBER ALREADY!!! That is like saying the number .5 approaches something at infinite! The statement makes no sense.

If you "plot" .99r, it will just be a point (or a line if you are plotting something like y = .999r against X).

 

[HJB417]

Originally posted by: Krk3561

Originally posted by: HJB417
but when u plot this on a graph
wont the line always go closer to 1 but never touch 1
I forgot the calc term for lines/functions like that.

asympote (sp?), but anyway it does reach 1 @ infinity

Originally posted by: Legendary

Originally posted by: HJB417
but when u plot this on a graph
wont the line always go closer to 1 but never touch 1
I forgot the calc term for lines/functions like that.

The term you're thinking of is asymtotic, but the proof through series above proves .999... to equal 1 anyway.

Thx, It was worth a shot anyway.

 

[MadRat]

The same logic to say .999...=1 can be used to say .000...1 exists, but then that proves .999...<>1. How funny.

 

[Haircut]

OK, I said back in February that I would no longer take part in this thread. I've stayed away for nearly 11 months, but I can refrain from posting no longer.

Madrat (and any others that believe they are two different values), can I ask you to tell me what is the last digit of 0.999...

 

[Chu]

Originally posted by: MadRat
The same logic to say .999...=1 can be used to say .000...1 exists, but then that proves .999...<>1. How funny.

Madrat, again, you really need to think about what "..." means. If you are defining it in such a way that a number like .000...1 is possible, then yes, .999... != 1 because .000...1 means that at some point, the 0's stop, and hense under you definition at some point the 9's stop. Hense, you are saying "..." means "an unimaginable number," not infinite. This is *NOT* the definition that everyone else uses. You really need to review basic calculus, infinity is a concept that took ~300 years to rigrously define, and you are not using the same definition the rest of us are.

-Chu

P.S., in some way you are right, that if you abuse the definition of "..." then .000...1 does exist, but it is a limit point, which cannot be expressed under decimal notation unless you SEVERLY abuse notation, like the definition of "...".

 

[Krk3561]

Originally posted by: Hector13

Originally posted by: Krk3561

Originally posted by: HJB417
but when u plot this on a graph
wont the line always go closer to 1 but never touch 1
I forgot the calc term for lines/functions like that.

asympote (sp?), but anyway it does reach 1 @ infinity

jesus.. did you guys even try to read the thread?
.9999r is a NUMBER (it is the same as the number one, but that is a different point all toghether). It does not approach anything.. it IS A NUMBER ALREADY!!! That is like saying the number .5 approaches something at infinite! The statement makes no sense.

If you "plot" .99r, it will just be a point (or a line if you are plotting something like y = .999r against X).

Did you even read my post! I proved it equals one by calculus, you obviously didnt read this.

And ur too much of a dumbass to see were talking about a graph aproaching 1, not a number