Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2007-05-04 19:13:50

krollo
Member
Registered: 2007-05-04
Posts: 24

0.99...=1

Hi, krollo here.
This doesnt make sense.

0.9999....*10=9.99999....
9.9999-0.99999=9
9/9=1
1=0.99999....
yikes

Offline

#2 2007-05-04 19:39:06

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: 0.99...=1

why does it not make sense?

you have to remember that the 9... indicates there are an 'infinite' amount of 9's, in otherwords, multiplying by 10, does not make it have 'one less' 9 at the end that you would normally assume when multiplying a number by 10, the number of 9's after it, is still infinate, so when you subtract, they completely cancel out, leaving you with an integer value.

normally you would write the above proof as:

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

0.999... = 1

if you find it hard to think about that, think about it this way, if they were not the same number, then there would be a number that goes inbetween them, like 1 != 2, because there is for example, the number 1.5 that fits between them, but that isn't the case with 0.999... and 1, there cannot exist a number that goes between them, so they must be the same


The Beginning Of All Things To End.
The End Of All Things To Come.

Offline

#3 2007-05-05 01:00:16

Identity
Member
Registered: 2007-04-18
Posts: 934

Re: 0.99...=1

If you have x < 1, then x =/= 0.999... ?

Offline

#4 2007-05-21 06:11:48

Laterally Speaking
Real Member
Registered: 2007-05-21
Posts: 356

Re: 0.99...=1

In this same way, I suppose that 0.0000000000...001=0 if there are an infinite number of zeros before the 1, although anything with an infinite number is mostly theoretical (with the possible exception of pi).


"Knowledge is directly proportional to the amount of equipment ruined."
"This woman painted a picture of me; she was clearly a psychopath"

Offline

#5 2007-05-21 06:19:41

Devantè
Real Member
Registered: 2006-07-14
Posts: 6,400

Re: 0.99...=1

There isn't a need to create another thread.

Please discuss in this thread, in which there is a fair amount of controversy.

For extra information on the subject, click here, here, here, here, here, here and/or here.

Last edited by Devantè (2007-05-21 06:22:31)

Offline

#6 2007-05-21 07:18:40

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: 0.99...=1

This thread's been around for a while now, so you can't really complain about its creation.
I may be contradicting things I've said earlier, but I don't mind this thread. The posts are thoughtful, and we haven't descended into flaming, like in the one you linked to.

I think Identity's inequality is correct. It's obvious that no real number x exists such that 0.999... < x < 1, so if 0.999... < 1 then that means that the reals aren't dense. But they are.
Therefore, 0.999... = 1.


Why did the vector cross the road?
It wanted to be normal.

Offline

#7 2007-05-21 07:37:22

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: 0.99...=1

Laterally Speaking wrote:

In this same way, I suppose that 0.0000000000...001=0 if there are an infinite number of zeros before the 1, although anything with an infinite number is mostly theoretical (with the possible exception of pi).

Couldn't be further from the truth.  First off, many rational numbers have an infinite decimal expansion... it's just that this decimal expansion repeats.  But for irrationals, there are tons of constants known to be irrational, some which are believed to be, which are used every day for practical purposes.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Offline

#8 2007-05-21 09:36:11

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,555

Re: 0.99...=1

0.0...01 would not have a 1 at the end (even though I just wrote it that way!), because we are saying "there is no end to the 0s".

Sometimes during our discussions we have been confusing "infinite repetition" with infinity. 0.333... has an infinite repetition, but it is simply 1/3 which is not "infinite"


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Offline

#9 2007-05-21 12:55:46

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: 0.99...=1

whats the second decimal expansion of 0.3333.. ???


A logarithm is just a misspelled algorithm.

Offline

#10 2007-05-21 14:20:56

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: 0.99...=1

Every rational number has two decimal expansions, induced by the metric on the real numbers.

Every terminating decimal has two decimal expansions.  But how is this induced by the metric on the real numbers (absolute value)?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Offline

#11 2007-05-21 20:19:12

Sekky
Member
Registered: 2007-01-12
Posts: 181

Re: 0.99...=1

Ricky wrote:

Every rational number has two decimal expansions, induced by the metric on the real numbers.

Every terminating decimal has two decimal expansions.  But how is this induced by the metric on the real numbers (absolute value)?

Absolute value of a number would be the metric of that number with zero, the metric  d(1,0.9...) = 0 which follows pretty soon after the axioms.

Offline

#12 2007-05-22 02:33:56

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: 0.99...=1

That's just assuming that 1 - 0.999... = 0.  We want to show this.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Offline

Board footer

Powered by FluxBB