Topic: Typing math

[aznkidzx]

[tex]0+0=1[/tex]

Thats how I roll!

[Mchl]

[tex]0+0=1[/tex]

Thats how I roll!

My mums Excel spreadsheets do this kind of thing all the time.

[ShadeSlayer]

[tex]0.\overline{999} = 1[/tex]

[tex]\frac3{9} = 0.\overline{333} [/tex]

[tex]\frac6{9} = 0.\overline{666} [/tex]

[tex]so[/tex]

[tex]\frac9{9} = 0.\overline{999}[/tex]

[tex]but[/tex]

[tex]\frac9{9} = \frac1{1} = 1[/tex]

Lol.

[corbin]

And that's why patterns shouldn't be assumed based off of 2 terms ;p.

[Daniel0]

Actually corbin, it's widely accepted among mathematicians that .999... = 1. ShadeSlayer's proof may be said to be flawed in the sense that it doesn't explain how the pattern he uses works.

A better version might be:
[tex]\frac{1}{3} = 0.\overline{333}[/tex]
[tex]3 \cdot 0.\overline{333} = 3 \cdot \frac{1}{3} = \frac{3}{3} = 1[/tex]

Therefore:
[tex]0.\overline{999} = 1[/tex]

[corbin]

Hrmmm....  Yeah....  Strange to me though since .999 (repeating) would never actually equal 1.  Then again, if I look at it that way, repeating .333 would never actually equal 1/3....

[Daniel0]

Dunno if it helps your understanding, but 0.99 is closer to 1 than 0.9 is, so for each 9 you add, you get closer to 1. If you have an infinite amount of 9's then you are getting infinitely close to 1. Or if you want it in math terms:

[tex].9 = \frac{9}{10^1}[/tex]
[tex].09 = \frac{9}{10^2}[/tex]
[tex].009 = \frac{9}{10^3}[/tex]
etc.

[tex].9 + .09 + .009 = \frac{9}{10^1} + \frac{9}{10^2} + \frac{9}{10^3} = .999[/tex]

So:
[tex]\sum_{n=1}^{\infty} \frac{9}{10^n} = 0.\overline{999} = 1[/tex]

You see that as [tex]n \to \infty[/tex] (analogous to "you are adding more 9's on the end") you are getting closer to 1.

[corbin]

Yeah, we actually discussed this in class one day when talking about geometric sequences a while back.


And yes, as n -> inf., the number gets closer to 1, but it would never actually reach one.


Wouldn't .999 (repeating) = 1 - 1/inf, not 1?



(I do realize that obviously I'm arguing pointlessly since people accept .999 to be 1, just like .333 is assumed to be 1/3 [both repeating].  I guess  I just think that things should be left in fractional form ;p.)

[Daniel0]

Wouldn't .999 (repeating) = 1 - 1/inf, not 1?

Yeah, but that's the same thing because [tex]\frac{x}{\infty} = 0[/tex] and 1 - 0 = 1

[corbin]

Wait....  x/inf is always assumed to be 0?


Ok then...  I shall now shut up ;p.

[Daniel0]

Yeah, if you think about it then if you split something up an infinite amount of times, then there is nothing left (a.k.a. zero).

[corbin]

Hrmmm....  Idk....


By that theory... 

10 * (1/inf) * inf = 0?

(Parenthesis just for looks.)

10 * (0) * inf = 0.

Or:

10 * 1 / inf * inf

10 / inf * inf = 0

But....

1/x * x/1 always = 1, yes?  Or does infinity just break every rule?



(I should just read the wiki on infinity and stop arguing against things known to be true ;p.)

[Daniel0]

10 * (1/inf) * inf = 0?

(Parenthesis just for looks.)

10 * (0) * inf = 0.

Or:

10 * 1 / inf * inf

10 / inf * inf = 0

The results of all that is undefined because you cannot multiply 0 with infinity.

[GingerRobot]

I'm pretty sure the way we were shown that 0.9999.. = 1 was something along the lines of:

let [tex]x = 0.\overline{999}[/tex]

then [tex]\frac{x}{10} = 0.0\overline{999}[/tex]

and [tex]x - \frac{x}{10} = 0.\overline{999} - 0.0\overline{999} = 0.9[/tex]

so [tex]\frac{9x}{10} = 0.9[/tex]

Therefore:
[tex] x = 1[/tex]

[corbin]

10 * (1/inf) * inf = 0?

(Parenthesis just for looks.)

10 * (0) * inf = 0.

Or:

10 * 1 / inf * inf

10 / inf * inf = 0

The results of all that is undefined because you cannot multiply 0 with infinity.

I now hate infinity.