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

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

Topic closed

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

Hi, krollo here.

This doesnt make sense.

0.9999....*10=9.99999....

9.9999-0.99999=9

9/9=1

1=0.99999....

Offline

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

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

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

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

Offline

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

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

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

Offline

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

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

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

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

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

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

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

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

A logarithm is just a misspelled algorithm.

Offline

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

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

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

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

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

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

Pages: **1**

Topic closed