does .9 repeating equal one?

[alofatti]

From Asher

Sounds like you're more into the theory aspect more? The math behind everything?

Funny thing to say that I am not even sure! Perhaps yes, but in the other hand I also like to see the application of it.

[KrazyHorse]

And I also was remembering the arrow going the wrong way on compactness and completeness.

Compactness->completeness, but completeness does not require compactness...

[alofatti]

Quoted from Frogger:

Yup...

Mixed up convergence and uniform convergence...



I'm not on the ball tonight.

I am the one who is to blame about. I started all this completeness stuff when I (incorrectly) stated that the decimal representable numbers were complete...

You are not alone!

[Sava]

No, as close as it gets, it never is 1.

Take any number and divide it in half an infinite amount of times, do you ever get to 0?

[Dauphin]

Ever heard of limits?

Zero is not just nothing, but also an infinitely small number.

[loinburger]

Originally posted by Sava
No, as close as it gets, it never is 1.

Did you just not bother reading the thread or something?

Anyway, here's the way I learned it back in Calc:

.999... == x, so 10 * .999... = 10x = 9.999...
9.999... - .999... = 9 = 9x
so x == 1, but since .999... == x, then .999... == 1.

Quick and dirty way to find the fractional representation of any rational number.

[Dauphin]

Which was pretty much what was posted in the first post.

[Sava]

loinburger, 1/3 does not equal .333 repeator, it is just a numerical approximation...

for all intents and purposes, .999 repeator is close to the value of 1 to say that substituting one for the other would be acceptable. But they are by no means equal.

[loinburger]

Originally posted by Sava
loinburger, 1/3 does not equal .333 repeator, it is just a numerical approximation...

It's only a numerical approximation if you truncate .333... somewhere along the lines, like you'd have to do when using a computer or something. But then you don't have .333... anymore.

.333... == 1/3, exactly.

[loinburger]

Originally posted by Sagacious Dolphin
Which was pretty much what was posted in the first post.

Methinks that all that talk about bijection and smegma confused the issue to holy hell.

[Sava]

If you told a computer to find the value for 1/3 without limiting a decimal point, the computer would never complete the operation.

1/3 does not equal .333 repeator exactly

[loinburger]

Originally posted by Sava
If you told a computer to find the value for 1/3 without limiting a decimal point, the computer would never complete the operation.

So what?

1/3 does not equal .333 repeator exactly

Yes it does. I'm still waiting for you to point out the error in my first post on this thread. Gainsaying is not a valid counterargument. You're just being contrary.

[loinburger]

Okay, here's another way to look at the problem. .999... is really just the sum of an infinite geometric series, with a0 = .9 and r = .1. The sum of an infinite geometric series (where -1 < r < 1) is a0 / (1-r), which is 1, thus .999... == 1. Substitute a0 = .3, and you get that .333... == 1/3.

[Sava]

it doesn't matter, it still isn't EXACTLY equal to 1/3, nothing you can say will change that.

[Dauphin]

The way I was informed that .999999 recurring is equal to 1 was by proving that 0.xxxx recuring in any base is identically equal to 0.yyyyy recuring in any other base. (x and y being the base number less 1.)

The only occasion on which such an identity would be true is if they are all equal to 1.

Not sure how accurate that is, as I've never seen it proven, but I'll take their word for it.