0.999... = 1 (?)

In statistics, for all around purposes, .99999 does equal 1. The difference between .99999 and 1 are so small that it really wont make a difference.

Going off this one a bit:

If we take out the infinities, here's what happens:

lim (lim (E (1/x) from 0 to y) as y approaches b) as x approaches c

Then by this theorum:
E (1/x) from 0 to y = y/x

You get:
lim (lim (y/x) as y approaches b) as x approaches c
= lim (b/x) as x approaches c
= b/c

So then we end up with what we had before:
lim (b/c) as b and c approach infinity

As you said, differentiate.

lim db/dc as b and c approach infinity
= lim 1/1 as b and c approach infinity
= 1

*Still confused.*
Maybe a visual example like a graph may help.

If we take out the infinities (I'll use multiplication, as it is the same thing)


Nevermind what I said about differentiation though. It's not going to apply in this scenario anyway, as we have two limits. You have to evaluate one of the limits first, then the other. Period. Therefore, as we have infinity-limits here, we just have to evaluate those.

(I'm kind-of doing something illegal though: Infinity isn't really a number, so differentiating it does not really make sense. For this equation, it does not really matter, as we already found out that this equals to 0 from the other limit.)

Noah

Taking what Noah was doing:

when you get to (lim 1/x, x->infinity)*(lim y, y-> infinity) = 0
you have solved your problem already, no need to develop both sides since what you did on the right side was simply arranging the expression in order to solve it.

Also, you can't really evaluate in a different order (going with the y limit first) since that leads to indetermination (infinity*zero).

I haven't read the whole thread but some things caught my attention.

 Milelke

Nope, a number is anumber.
In the technical world, .99999... is very different from 1.

Ok.
Technically no.
Because clearly all are eyes can see 0.999... not 1.

In the world of making things simple 0.999... =1
Because if it goes on forever, there is no number in between, essentially making 0.999... the equivalent to 1.

But in terms of are they different numbers, of course. 0.999 is not 1, it gets terribly close, but, no, it is not 1, if it it were, it would just be written as 1 (unrounded).

Because the improper fraction 8/2 is obviously very different from 4.

 ShiKage

1 is an integer. 0.9999999 repeated isn't. Therefore 1 != 0.9999999999999. 0.9999999999999 is a number which is missing a fragment of its ability of becoming a whole number. Whether that be impossible to see the difference in, it's still not the same as the integer "1." It's still not a whole number.

No proof. If 8/2 can be an integer, so can 0.9 repeated. Your missing fragment is infinitely small and therefore no fragment at all.

 ClawofBeta

They have some faulty premises.

"For any two different real numbers, you can pick a third number which is between them."

Who said that?

If a is a real number and b is a real number, and a < b, let c = (a+b)/2. Then a < c < b, and c is a real number because it's a quotient of real numbers.

Applied to our situation we have this (assuming that a and b are actually different)
a = 1
b = 0.99999999...
c = 0.99999999...5 (???)
But c can't exist. The 9's in b go on forever. If c has any digit less than 9, that implies that it must be less than b, which is a contradiction.

If the difference between two numbers is zero, then they are equal. For example, 5 - 5 = 0 because 5 = 5.

The difference between 1.0000... and 0.9999... is:

1.0000... - 0.9999... = 0.0000...
= 0

Therefore, they are equal.


Their arguement is bs. Sure, "..." might mean going on forever, but 1.0000... - 0.9999... = 0.0000(keep on going) until 1. This applies to the 3rd problem, too.

Keep on going how long? If I lend you a dollar and you say you'll pay me back at the end of time, is that really any different than giving you the dollar outright?

 2147483647

There IS a way to dispute it. Whether you want to accept it or not. Just like there's proof of 0.999... = 1, there's proof that 0.999... =/= 1. Point is: don't go around bashing other people like this is your religion.

Proof below:

Take these two definitions:

1 = lim (1 + 1/x) as x approaches infinity
1 = lim (1 - 1/x) as x approaches infinity

Then:

lim (1 + 1/x) as x approaches infinity - lim (1 - 1/x) as x approaches infinity
= lim (2/x as x) approaches infinity
= 0.

This is generally accepted by mathematicians.

But what happens when you repeat this process infinite times? (This is elaborated by my post below.)
What you end up with is the following:

1 = lim (1 - x/x) as x approaches infinity
1 = lim ((x-x)/x) as x approaches infinity
1 = lim (0/x) as x approaches infinity
1 = 0 ?

Since that is impossible, 1/x accounts for something, and 1 =/= 0.999...

Elaboration:

lim 1/x as x approaches infinity = 0
lim 2/x as x approaches infinity = 0
lim 3/x as x approaches infinity = 0
.
.
lim 10/x as x approaches infinity = 0
lim 10000000000/x as x approaches infinity = 0

Then if you do this infinite times, you end up with:

lim x/x as x approaches infinity = 0, which is false, because:

lim x/x as x approaches infinity
= lim dx/dx as x approaches infinity (L'Hopital's rule)
= lim 1/1 as x approaches infinity
= 1

Can't do this. lim (1-x/x) as x approaches infinity is not 1.

By the same argument:
If x is finite, then x + 1 is finite.
0 is finite.
Repeating this infinite times, infinity is finite.

Your equation holds when the numerator is finite, but not when it's infinite. The reason that all those previous limits are 0 is that the denominator "outpaces" the numerator. If they both scale equally, that no longer holds.

Another approach, going off some of what Noah said. You have lim y/x as x and y approach infinity. However, they may not approach at the same rate. What if y is defined as x^2? Then you have lim x^2 / x as x approaches infinity. Then your whole limit's just lim x, which is infinity, not 1 or 0 or anything else.

By replacing y with x you're essentially asserting that y = x, which isn't valid. lim y = lim x does not imply that y = x.

I like Noah's example better, but I get the point. It'd still help if I had a visual example though, if someone's willing to provide such. This reminds me of Gabriel's Horn, which I would never be able to conceptualize without seeing the graph itself.

Not sure if this is what you're looking for...

 1/x

 2/x

 100/x

No matter how high you make the numerator, as long as it's a constant (horizontal line), y = x will eventually overtake it infinitely. However...

 x/x

 x^2/x

No longer a constant.

Something like f(x) = 1/x and x = 0 drawn on the same graph. The line x = 0 is the asymptote of f(x) = 1/x, so the only point they can meet is at (infinity) or (- infinity). Since infinity occurs at x = 0, then 1/infinity = 0.

However, the problem I have with this graph is that f(x) = 1/x joins the line x = 0 at two different locations on opposite ends. Also, x = 0 is just a line and is undefined when you try writing it as g(x), because then you'd have to write it as g(x) = lim kx as k approaches infinity, which would cease to make sense.

Parametric anyone?

Something like f(x) = 1/x and x = 0 drawn on the same graph. The line x = 0 is the asymptote of f(x) = 1/x, so the only point they can meet is at (infinity) or (- infinity). Since infinity occurs at x = 0, then 1/infinity = 0.

However, the problem I have with this graph is that f(x) = 1/x joins the line x = 0 at two different locations on opposite ends. Also, x = 0 is just a line and is undefined when you try writing it as g(x), because then you'd have to write it as g(x) = lim kx as k approaches infinity, which would cease to make sense.

Parametric anyone?

Parametric:


Is that what you mean?

No. That's the exact same thing as what I mentioned before and therefore it's not making any difference in the explanation.

I've seen this question posted numerous times - this is the only answer that makes sense:

Does 0.999... = 1?
Geometry - No
Algebra - Yes
Calculus - Yes
Achilles and the Tortoise - No

Achilles and the Tortoise - No

Doesn't this introduce the concept of limit, though? That means if I treat the limit of a number and its actual value differently, doesn't that make the 2 numbers completely separate?

No. That's the exact same thing as what I mentioned before and therefore it's not making any difference in the explanation.

Why not use y = 0 instead? Slightly nicer to deal with and you get the same thing, you just have to look at an infinite x-value.

It makes perfect sense that the line approaches 0 on both ends. What is 1/-infinity? By algebra, -(1/infinity). You stated that 1/infinity is 0, so 1/-infinity is -0, which is also 0.

Of course this is a rather casual use of "infinity" which really shouldn't be treated as a number.

Just realized in sets, it's different:
[1,0] is a closed set. In this set all real numbers between 1 and 0 are included.
(1,0] is an open set. In this set 1 is not included, but .999.... is.

find it through google.

and em kinda off topic.... http://www.basilmarket.com/forum/1430035...99_1.html#

More people on Basil are correct...

I'm going to be optimistic and say that that means SPers are more critical thinkers.

Just realized in sets, it's different:
[1,0] is a closed set. In this set all real numbers between 1 and 0 are included.
(1,0] is an open set. In this set 1 is not included, but .999.... is.

This is a circular argument though. The set notation [0, 1) simply means that any number x is in the set if 0 <= x < 1.

Then you're back to having to prove that 0.999... < 1.

More people on Basil are correct...

I'm going to be optimistic and say that that means SPers are more critical thinkers.

I'm assuming that wasn't you being cynical. Anyone can be correct, I guess. If one person who sounds smart or has a reputation of knowing what they're doing agrees to something, everyone else notices and decides to agree because they assume it's a correct answer. So it's basically a pack-leader leading a group, but the rest of them don't know why they stumbled upon that answer except that "Oh, he's always right, so he must be right again".

It's better to debate logic than to assume the correct answer. I actually encourage threads like these and as such as long there aren't any negative after-effects such as arguments, and mindless member-bashing.

I'm assuming that wasn't you being cynical. Anyone can be correct, I guess. If one person who sounds smart or has a reputation of knowing what they're doing agrees to something, everyone else notices and decides to agree because they assume it's a correct answer. So it's basically a pack-leader leading a group, but the rest of them don't know why they stumbled upon that answer except that "Oh, he's always right, so he must be right again".

It's better to debate logic than to assume the correct answer. I actually encourage threads like these and as such as long there aren't any negative after-effects such as arguments, and mindless member-bashing.

Aside from 2147483647 who picked "yes" but have valid arguments why it might be no. Everyone else was like, no because it is no by changing the definition of mathematical terms.

So how about having a little bit of fun...

x = 1 + 2 + 4 + 8 + 16 + 32 + ...
2x = 2 + 4 + 8 + 16 + 32 + 64 + ...
2x - x = x = (2 + 4 + 8 + 16 + 32 + ...) - (1 + 2 + 4 + 8 + 16 + ...) = -1 + 0 + 0 + 0 + 0 + ... = -1
1 + 4 + 8 + 16 + 32 + ... = -1?

If the series are infinite, how can you subtract term n in x from term n+1 from 2x? (2-2, 4-4, etc.) Both series diverge in the end, so... o-o
If you follow the parenthesis, it comes out as 1 + 2 + 4 + 8 + 16 + 32 +..., which diverges to infinity.
Infinity = -1? I think you made an oopsie. :|

More people on Basil are correct... I'm going to be optimistic and say that that means SPers are more critical thinkers.

People, myself included, like to post these kinds of things on Basil to see them fail. Sooner or later, they build up resistance to them. :| I think there's three kinds of people who voted yes at Basil: Some tiny minority of people who know what they're talking about, some number of people who've seen this thread before, and people who think "I don't understand what this says, but it looks legit/probably a troll" after seeing the OP and comments.
Boils down to your conclusion, really. xP

Let me rewrite it.
x = 1 + 2 + 4 + 8 + 16 + 32 + ...
2x = 2 + 4 + 8 + 16 + 32 + 64 + ...
x = 1 + (2 + 4 + 8 + 16 + 32 + 64 + ...) = 1 + 2x
x - 1 = 2x
-1 = x
1 + 4 + 8 + 16 + 32 + ... = -1?

But yes, you're right; the proof is fallacious because divergent series can't be algebraically manipulated like that.

I've seen this question posted numerous times - this is the only answer that makes sense:

Does 0.999... = 1?
Geometry - No
Algebra - Yes
Calculus - Yes
Achilles and the Tortoise - No

I haven't really seen how you could apply this to geometry.

As for Achilles and the Tortoise as well, I cannot see how you do it here either. Maybe I'm just stupid today.

And yeah, you cannot add, multiply or remove infinity, as that does not make sense.