Is 0.9~ equal to 1 in mathematical calculations?

[Gara]

Okay so most of us know 0.9~ = 1, correct?

We find this by doing 1 devided by 3. This makes 0.3~

0.3~ * 3 = 0.9~ So 0.9~ = 1. As most of us know.

But would not 0.3~ = 0.4?

Then it would be 0.4 * 3 = 1.2? So 1 / 3 * 3 = 1.2?

¬_`

 

[Alkatran]

Okay so most of us know 0.9~ = 1, correct?

We find this by doing 1 devided by 3. This makes 0.3~

0.3~ * 3 = 0.9~ So 0.9~ = 1. As most of us know.

But would not 0.3~ = 0.4?

Then it would be 0.4 * 3 = 1.2? So 1 / 3 * 3 = 1.2?

¬_`

Does 1/3 = 4? No.

You need to learn the concept of a limit.

Consider this: You move half way across the street, then half of the remining distance, and keep going half way...

Wether or not you reach the end of the street isn't certain, BUT I can tell you you will NEVER get PAST the end point of the street.

Mathematically:
1/2+1/4+1/8+1/16 ... = 1

The same applies to the infinite 3s:
3/10 + 3/100 + 3/1000 ... = .3 + .03 + .003 ... = 1/3 = .333~

Consider this proof for 1 = .999~

.9 = 1 - 10^-1
.09 = 1 - 10^-2
.999~ = 1 - 10^-infinity
x^-y = 1/(x^y)
.999~ = 1 - 10^-infinity = 1 - 1/(10^infinity) = 1 - 1/infinity = 1 - 0 = 1

There. The difference is 0.

 

[cogito²]

Think of it this way [tex].\overline{999}[/tex] is the real number that the series [tex]\sum_{i=1}^\infty \frac{9}{10^n}[/tex] converges to. That series converges to 1 (due to a few simple proofs of properties of series of real numbers).

Edit: I just remembered I wrote up a pdf on this a while back. http://s89707303.onlinehome.us/mathwork/infinity.pdf it is.

 

[nolachrymose]

You could also use the formula for repeating decimals, I suppose. Since [itex].\overline{2}[/itex] is [itex]\frac{2}{9}[/itex], you could say [itex].\overline{9}[/itex] = [itex]\frac{9}{9}[/itex], or 1.

Cogito, shouldn't that infinite sum be this?

[tex]\sum_{i=1}^{\infty} \frac{9}{10^i}[/tex]

 

[matt grime]

0.4-0.33... > 0.4-0.34 = 0.06

so two allegedly identical real numbers are more than 6 hundredths apart on the real number line?

 

[cogito²]

You could also use the formula for repeating decimals, I suppose. Since [itex].\overline{2}[/itex] is [itex]\frac{2}{9}[/itex], you could say [itex].\overline{9}[/itex] = [itex]\frac{9}{9}[/itex], or 1.

Cogito, shouldn't that infinite sum be this?

[tex]\sum_{i=1}^{\infty} \frac{9}{10^i}[/tex]

yes it should. thank you for being perceptive.

 

[CJames]

1/9=.1~
2/9=.2~
3/9=.3~
3/9=1/3
...
9/9=.9~
9/9=1

 

[Gokul43201]

Now, where's that Math FAQ when you need it ?

 

[Leaping antalope]

Does 1/3 = 4? No.

You need to learn the concept of a limit.

Consider this: You move half way across the street, then half of the remining distance, and keep going half way...

Wether or not you reach the end of the street isn't certain, BUT I can tell you you will NEVER get PAST the end point of the street.

But in real life, don't we always get past the end point of the street easily??

Mathematically:
1/2+1/4+1/8+1/16 ... = 1

I have another question:
(1-1)+(1-1)+(1-1)+(1-1)...=0
We can change the above equation to~
1+(-1+1)+(-1+1)+(-1+1)...=1
Since both equatoins have infinite 1 and -1
Can we say that 1=0?

 

[liaoge]

I have another question:
(1-1)+(1-1)+(1-1)+(1-1)...=0
We can change the above equation to~
1+(-1+1)+(-1+1)+(-1+1)...=1
Since both equatoins have infinite 1 and -1
Can we say that 1=0?

OK, you went the way as Euler did.
Just look 1-1+1-1+... as a function, and it will always be changing, namely it is actually a variable. Then can you say a variable equals a constant?
....

 

[arildno]

I have another question:
(1-1)+(1-1)+(1-1)+(1-1)...=0
We can change the above equation to~
1+(-1+1)+(-1+1)+(-1+1)...=1
Since both equatoins have infinite 1 and -1
Can we say that 1=0?

The even partial sums of this series equal 0, whereas the odd partial sums equal 1.
A necessary condition for the sum of the infinite series to EXIST, is that ALL sequences composed of partial sums must converge to the SAME NUMBER.
Since one sequence of partial sums is the sequence of even partial sums, and another sequence of partial sums is the sequence of odd partial sums, we see that this condition is not fulfilled.
Hence, we say that the infinite series is divergent (i.e, it doesn't have a sum).

 

[JonF]

I have another question:
(1-1)+(1-1)+(1-1)+(1-1)...=0
We can change the above equation to~
1+(-1+1)+(-1+1)+(-1+1)...=1
Since both equatoins have infinite 1 and -1
Can we say that 1=0?

this is a very discrete quesiton...

 

[ReyChiquito]

a bit late, but i like this one

[tex]x=0.\overline{9}[/tex]

[tex]10x=9.\overline{9}[/tex]

[tex]10x-x=9.\overline{9}-0.\overline{9}[/tex]

[tex]9x=9[/tex]

[tex]x=1[/tex]

:)