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is 2,99999....... = 3 ?

- Thread starter Little ant
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- #1

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is 2,99999....... = 3 ?

- #2

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Otherwise stated: Does 1/10^m = 0 ?

- #3

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- #4

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You seem to be satisfied if there is no real number between 2.999... and 3.

What is a real number?

- #5

Mentallic

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Any reason why you chose to use 2.99..=3 rather than the usual 0.999..=1?is 2,99999....... = 3 ?

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Someone asking the question in the original post is probably not familiar with extending the real numbers or with alternative formulations of analysis. They have also either not understood or have never seen a proper definition of the set of real numbers, using which, they would immediately see the truth of the statement. The purpose is to lead them towards a motivation of the definition.You seem to be satisfied if there is no real number between 2.999... and 3.

What is a real number?

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- #7

Borek

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Perhaps an attempt to avoid inevitable locking of the thread?Any reason why you chose to use 2.99..=3 rather than the usual 0.999..=1?

https://www.physicsforums.com/showthread.php?t=385883

https://www.physicsforums.com/showthread.php?t=215017

https://www.physicsforums.com/showthread.php?t=285351

https://www.physicsforums.com/showthread.php?t=130123

- #8

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Let x=2.9999....

Then 10x=29.9999....

Thus 10x-x=27

Which yields x=3.

So, we have 3=2.999...

- #9

HallsofIvy

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[tex]2.999...= 2+ 0.9+ 0.09+ 0.009+ \cdot= 2+ \frac{9}{10}+ \frac{9}{100}+ \frac{9}{1000}+ \cdot\cdot\cdot[/tex]

[tex]= 2+ 0.9(1+ 0.1+ 0.01+ \cdot\cdot\cdot)[/tex]

[itex]1+ 0.1+ 0.01+ \cdot\cdot cdot[/itex] is a geometric series, [itex]\sum_{n=0}^\infty ar^n[/itex] with a= 1, r= 0.1. The sum of such a serie is

[tex]\frac{a}{1- r}= \frac{1}{1- .1}= \frac{1}{0.9}[/tex]

So the series is

[tex]2+ .9\left(\frac{1}{.9}\right)= 2+ 1= 3[/tex]

- #10

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In terms of equalities:

2 + 1/3 = 2 + .333... - The original problem

1/3 = .333... - Subtract 2

1 = .999... - Multiply by 3

There are many much more rigorous proofs than this, but IMO, this is the most accessible.

- #11

Mentallic

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Perhaps if I created a thread asking if x+0.99...=x+1 for all x then we would have a general case and then there's no excuse in making more of these threads.Perhaps an attempt to avoid inevitable locking of the thread?

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Perhaps debating x + .999... = x + 1 is fun, and an excuse in itself.

- #13

gb7nash

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What's there to prove here though? Assuming x is a real number, this is equivalent to subtracting x on both sides and proving .999... = 1.Perhaps debating x + .999... = x + 1 is fun, and an excuse in itself.

- #14

HallsofIvy

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- #15

gb7nash

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Whoops, I'm looking far too into this.

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a friend told me than if a=b, then 0<|a-b|<E*. So, if we gues than a is not iqual b, then |a-b| is not equal 0. ok... the geat an E=|a-b|/2...the definition say 0<|a-b|<E=|a-b|/2. and that's not true. so a=b.

2,9999...->3

=>3-2,99....->0

the diference is so small.

- #17

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It is very difficult to tell what you are trying to say.

It is true that if a=b then a-b=0.

It is NOT true a=b implies 0<|a-b|<|a-b|/2

It is true if a=b then 0 ≤|a-b|≤|a-b|/2, because if a=b then |a-b| =0 so |a-b|/2 = 0 which gives us 0 ≤|a-b|≤0. Which implies |a-b| = 0, i.e. a-b=0.

I understand the notion you’re grasping at. You feel like 0 < 3 - 2.9999….. This isn’t true. Your problem is you are thinking of 2.999… as some progression or something that is growing. 2.9999… is just a symbol representation of a number, it is a fixed thing that doesn’t entail any type of change.

It is true that if a=b then a-b=0.

It is NOT true a=b implies 0<|a-b|<|a-b|/2

It is true if a=b then 0 ≤|a-b|≤|a-b|/2, because if a=b then |a-b| =0 so |a-b|/2 = 0 which gives us 0 ≤|a-b|≤0. Which implies |a-b| = 0, i.e. a-b=0.

I understand the notion you’re grasping at. You feel like 0 < 3 - 2.9999….. This isn’t true. Your problem is you are thinking of 2.999… as some progression or something that is growing. 2.9999… is just a symbol representation of a number, it is a fixed thing that doesn’t entail any type of change.

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- #18

gb7nash

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Is your friend a math major? Has he taken real analysis?how we can say if two numbers are equals?...

a friend told me than if a=b, then 0<|a-b|<E*

If a=b, then |a-b|= 0, it isn't greater than 0.

- #19

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[tex]a=b~\text{if and only if}~\forall \epsilon>0:~|a-b|<\epsilon[/tex]

This is indeed true, and show why 3=2.999....

- #20

Borek

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- #21

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I don't think they are that wrong in doing so. The Cauchy construction of the reals says that 2.999... is the equivalence class of the fundamental sequence (2.9, 2.99, 2.999, ...). But that very construction also says that 3 is the equivalence class of (2.9, 2.99, 2.999, ...). Thus 3=2.999...

So I have no problem with people that see real numbers as sequences. I have problems with people who work with sequences incorrectly. Specifically, saying that [tex]x_n<N[/tex] implies that [tex]\lim_{n\rightarrow +\infty}{x_n}< N[/tex]. It is this mistake that leads many people astray...

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