0.999... = 1 (?)

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.