0.99 infinite = 1?

[Xog]

I read somewhere that the infinite of 0.999999999 etc. etc. etc. is equal to 1?

is this true? what if I use 0.99\Bar{9} instead?

[genocidal]

Take calculus.

[DoTheFandango]

There's some formula... s = t1 / (1 - r) where t1 is the first term in an infinite series. Take .99 repeating to be .9 + .09 + .009, so each is 1/10 times (r) the last one.

So you take 9/10, r=1/10

.9 / (1-.1) = .9/.9 = 1

[Xog]

what about this, which has been brought up many times:

??

x = 0.999...
10x = 9.999...
10x - x = 9.999... - x
9x = 9.999... - 0.999...
9x = 9
x = 1

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Does it seem logical to subtract an infinite from an infinite?

[DoTheFandango]

I just gave you the answer. Shut up and look.

[Xog]

But I just don't get how infinity can be multiplied, added, divided, or subtracted by another inifinity.

edit:

to explain myself more thoroughly, dont view them as numbers, think of 1 line.

1 line that goes on for infinity on a plane (also parallel to below)

now think of 3 parallel lines in a seperate group gathered together that go on for infinity, still growing, and can't stop, ever. same rule applies for the above line.

you can't move one line from the 3 lines to the single other line and make it 2 sets of 2 parallel lines. The line will keep going from where it left off and will sprout another line, creating another line. Now you still have 2 sets of lines, but one of them is 2 lines and 3 lines instead of 2 lines and 2 lines or if u didnt move any at all you'd have 1 line and 3 lines.

[DoTheFandango]

It's not, its taking a geometric sequence, finding the pattern (.1x) and then finding the sum of all the numbers when added together.

[Xog]

so is the following formula incorrect?

x = 0.999...
10x = 9.999...
10x - x = 9.999... - x
9x = 9.999... - 0.999... *** You're subtracting a smaller infinite from a larger infinite. ***
9x = 9
x = 1


in essense, they're still both "infinites"..

[PjOtTeR]

what about this, which has been brought up many times:

??

x = 0.999...
10x = 9.999...
10x - x = 9.999... - x
9x = 9.999... - 0.999...
9x = 9
x = 1

----------

Does it seem logical to subtract an infinite from an infinite?

This isn't correct and I'll show you why. if you have 0.999... etc with lets say thousand 9's behind the dot... 10 times 0.999... would make 9.999... with 999 9's behind the dot...
so this 9.999... minus the x would make 8.999...
so that calculation is flawed. and it doesn't work this way.

You mean that if you have T(n) = T(n-1) + T(0)*0.1^n
With T(0)= 0.9
and you have to find a solution for n = infinite...?
lil hard to do in this text or to write out on comp for me.. but that basicly is your problem... could explain and look it up for you tommorow if I have somewhat more time..

[Spider]

Neeeeeeeeeeeerrrrddddddddds.

[Kim]

Neeeeeeeeeeeerrrrddddddddds.

Spider of course knows the real answer and doesn't want to release the inner nerd from within.

[DoTheFandango]

so is the following formula incorrect?

x = 0.999...
10x = 9.999...
10x - x = 9.999... - x
9x = 9.999... - 0.999... *** You're subtracting a smaller infinite from a larger infinite. ***
9x = 9
x = 1


in essense, they're still both "infinites"..

No, you're saying that .99 repeating is = .9 + .09 + .009 ... + .9*(.1n), not .9999999... + .0999999... etc. Sum of finites.

[Noah]

I'm one of the stupid nerds.

[Ewan]

0.999... is infinitely close to 1, so is the same as 1.

[Bioture]

There's some formula... s = t1 / (1 - r) where t1 is the first term in an infinite series. Take .99 repeating to be .9 + .09 + .009, so each is 1/10 times (r) the last one.

So you take 9/10, r=1/10

.9 / (1-.1) = .9/.9 = 1

and you got all of this from a prof you couldn't understand. my man.