.9~=1

[Jayferd]

You've got a point there, and the concept is basically the same. You said:

As my teacher said: You can continue to draw this graphic as long as possible, you'll never reach 0, you'll only get closer.

That's exactly correct, BUT you can only draw the graph out to a finite distance. If you could draw infinitely (which would be an awesome superpower! lolz), then yes, you'd hit 0.

Think of it this way: Look at your graph again.

Suppose you draw a horizontal line through a y-value that is strictly greater than 0. And say you try to make it as close to 0 as possible. The fact is that no matter how close it gets to 0, the top half of the graph is always going to intersect your horizontal line eventually. We define this to mean that the limit of your graph at infinity is 0. This is a simplified version of the epsilon limit.

This is the concept of a limit. When we say "0.9~", we like to say that means "Zero point infinitely many 9's", but it's actually the limit of a sequence. "0.9~" literally means the limit of the sequence:

0, 0.9, 0.99, 0.999, 0.9999, 0.99999, ...

So... Yes. This sequence will never include the value 1. But! It's limit is 1. Just like the limit of the sequence "0, 0.3, 0.33, 0.333, 0.3333" is 1/3, but the sequence never includes a value equal to 1/3.

But you ask, where did I get the "limit"? We're talking about a number, not a limit, right? But that's where you'd be wrong. The "~" (or overbar, as it should be) means there's a limit in there. You had it partly right when you said:

0.999~ is actually, not a solid number. That means it can't be treated in such ways, meaning it can't be in an equation.

It's not a "solid number" at all. It's a limit. Limits can be in equations, you just have to... be careful with them.

Anyways. I hope I've clarified that a little.

 

[Killer Tree]

I'm taught that 1/x will never reach 0, even if you'll go on infinitively(grammar?). As my teacher said: You can continue to draw this graphic as long as possible, you'll never reach 0, you'll only get clos

1/infinite = 0

 

[Jayferd]

...but only because it's defined explicitly to be so. That's a sticky one, even for mathematicians.

 

[Killer Tree]

...but only because it's defined explicitly to be so. That's a sticky one, even for mathematicians.

It logically makes sense too. Since the limit of 1/x as x approaches infinite = 0

 

[GoldShadow]

Yeah, it works according to basic calculus.

 

[AltF4]

Very nice, Jayferd. Well laid out.

Btw, something strikes me as unsatisfactory with the "sum of a geometric series equation" proof for .999~=1. Because that equation has to be derived using calculus. And the reason people have a hard time grasping .999~=1 is because they don't accept that an infinite series can sum to a finite value. So it's kind of using a principle that we're hoping to prove. The proofs of this I find more satisfying are the purely algebraic ones.

Oh also, I rememeber this example being explained to me a long time ago by my pre-calc teacher:

Imagine having a pie. You cut that pie in half, then cut the remaining half in half, then again, and so on for an infinite number of cuts. So what you've got is 1/2 of a pie + 1/4 of a pie + 1/8 of a pie ... all the way down.

So how much of a pie do you have? Well, duh! You've still got a whole pie. You didn't lose any of it just by cutting. 1/2 + 1/4 + 1/8 + 1/16 + ... has to sum to 1, you see.

 

[Killer Tree]

Another way to think of it is...


For a number to be different there needs to be numbers in between them.


What number is inbetween .9repeating and 1?

 

[inside]

Wow, I never really liked math. I don't get how people can have fun with equations and such.

 

[Jayferd]

Very nice, Jayferd. Well laid out.

Btw, something strikes me as unsatisfactory with the "sum of a geometric series equation" proof for .999~=1. Because that equation has to be derived using calculus. And the reason people have a hard time grasping .999~=1 is because they don't accept that an infinite series can sum to a finite value. So it's kind of using a principle that we're hoping to prove. The proofs of this I find more satisfying are the purely algebraic ones.

Thanks, AltF4. The geometric series formula is derived using a very basic limit equation, I wouldn't say it was quite "calculus". The finite formula for geometric series, which is

S_n = a*(1-r^(n+1))/(1-r)

can be derived using a very simple algebraic trick, as such (stolen almost directly from wikipedia http://en.wikipedia.org/wiki/Geometric_series):

S_n = \sum_{k=0}^n ar^k
(1 - r)*S_n = (1 - r)*\sum_{k=0}^n ar^k
= \sum_{k=0}^n ar^k - \sum_{k=1}^{n+1} ar^k <-- index change!
= a - r^(n+1)

Since (1 - r)*S_n = a - r^(n+1), you can just divide both sides by (1 - r) to get S_n = (a - r^(n+1))/(1-r).

Then, if -1 < r < 1, we know that r^(n+1) --> 0 as n --> \infty, so that

\lim S_n = \lim (a - r^(n+1))/(1-r) = a/(1-r)

so... In short, yes. A limit is involved, but the question at hand *is* a limit! As I said before, the "~" in "0.9~" means "limit" by definition.




Actually, that geometric series approach is a handy way to turn repeating decimals into fractions. Take 0.13131313..., for example. We can write that as a geometric sequence with common ratio .01:

0.1313... = 0.13 + 0.0013 + 0.000013 + ...

The first term, a, is 0.13, and the common ratio, r, is .01, so...

0.1313... = 0.13/(1 - 0.01) = 0.13/0.99 = 13/99!

aaand, calculator check... reveals that 13/99 is in fact 0.1313~!

 

[psicicle]

@ the bf.com guy saying math is wrong with infinite sums converging... heard of newton? It's an approximation of real life but by using calculus (which you are trying to disprove indirectly) and specifically integration (and derivative stuff), people were able to send people to the moon.

You need prove that math is wrong, which is frankly impossible because it is based on a few axioms which are assumed to be true.

 

[mark.]

.9999~+x=1 under the conditions of
x=.1111~, so
.9999~+.1111~=1.

they just compliment each other.
im so good :x
LOL

 

[GoldShadow]

.9999~+x=1 under the conditions of
x=.1111~, so
.9999~+.1111~=1.

they just compliment each other.
im so good :x
LOL

Haha, actually x=0.000~01

 

[HeadISBAgent]

.9999~+x=1 under the conditions of
x=.1111~, so
.9999~+.1111~=1.

they just compliment each other.
im so good :x
LOL

.9999~+.1111~ does not equal 1

Haha, actually x=0.000~01

x=0

 

[Killer Tree]

^
HeadISBAgent is correct.

.999~ + .111~ = 1.111~

 

[GoldShadow]

.9999~+.1111~ does not equal 1


x=0

If you take it to infinity, but I didn't!

 

[AltF4]

saying .000~1 is nonsense. The ~ notation means "on to infinity". You can't have a number placed after an infinite string of 0's, because there's no end.

 

[inside]

^That's true.

 

[GoldShadow]

saying .000~1 is nonsense. The ~ notation means "on to infinity". You can't have a number placed after an infinite string of 0's, because there's no end.

That's what she said.

 

[Killer Tree]

If you take it to infinity, but I didn't!

the ~ means you tried to take it to infinite.


edit: woops didn't see AltF4Warrior post on the next page.

 

[Jayferd]

Proof:

0.9~ + 0.1~ > 0.99 + 0.11 = 1.1

Thus 0.9~ + 0.1~ > 1.

Q.E.D.

Also, GoldShadow, I challenge you to define the concept of "0.00~01". Does that mean "Infinitely many zeroes and then a 1"? The decimal terminates at the 1 ... but it still has infinitely many digits. Is it a rational number? Or irrational?

Either way, if it's a real number (which is what I hope we've been talking about this whole time), you should be able to construct a sequence of rational numbers that converges to it. An idea would be the sequence

1, 0.1, 0.01, 0.001, 0.0001, 0.00001, ...

but unfortunately, this sequence converges to 0. ...which fits nicely with the fact that the sequence

0, 0.9, 0.99, 0.999, 0.9999, 0.99999, ...

converges to 1, which was the original subject of this topic.

 

[Wikipedia]

I didn't read the whole thread. I was just wondering if this thread has been Goldshadow'd yet.

 

[GoldShadow]

Hahaha.

Alas, AltF4 and HeadISB are the resident mathematicians, not me. Consequently, this thread has been AltF4'd and HeadISB'd.

 

[Wikipedia]

Oh, that's right, you are our resident cellular biologist. Got it, thanks.

 

[Killer Tree]

Biololollolgist

 

[AltF4]

I think you mean Bio-lol-ogist.

I prefer the term: Mathemagician