Re: .999~ = 1
 

you just don't know how to deal with imaginary numbers. plus if you look at how i display it i pull the e and i out of it making it ln e which is a positive number. Although there are other flaws. you just have to find them.

Re: .999~ = 1
 

A little while ago I wrote a long calculus proof about why .999... = 1 but I can't find it.

But it's true. Think about it this way:

1 - .999... = 1/inf

1/inf = 0 (by definition).

For the "proof" that 3 = 1, my trusty TI-89 says that ln[e^(3*i*pi}] = i*pi not 3*i*pi. I don't know much about imaginary numbers, so don't ask me why.

For the "proof" that 2=1, that's easy. At one point you divide both sides by (a-b). a = b therefore a-b = 0 and you are dividing by 0.

Re: .999~ = 1
 

Although, I think it would actually go like this:
let a=1, b=1
a = a
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)

This line gives you a division by zero error when you divide both sides by (a-b), because a-b = 0.

Also, e^i*pi is a perfectly valid expression that doesn't take the ln of a negative number. e^i*pi = cos(pi) + i sin(pi) = -1

Re: .999~ = 1
 

Don't think .999999999... = 1?

Think about it! Between any two numbers, there must needs be another number. For example, between 1 and 3 there is 2; between .5 and .6 there is .55; you can satisfy yourself if you don't buy it. Anyway, try to find a number between .999999... and 1 ... you can't, as they are the same number (just as you probably couldn't find a number between 2 and, well, 2).

Anyway, check out the authority on the matter: Dr. Math

Re: .999~ = 1
 

This is related to sines and cosines. To say cos(pi) = cos(3*pi) proves that 1=3 would be incorrect.

Re: .999~ = 1
 

no. order of operations only matters when you are actually evaluating something. you are not yet evaluating at this point, only at the end.

The reason why this little tidbit of confusion can exist is because the laws governing the use of the subdivision of fractions involving roots do not extend into the realm of imaginary numbers.

Re: .999~ = 1
 

Actually, if (-1/1)^.5 = (1/-1)^.5, then (+-)((-1)^.5) / (1^.5) = (+-)(1^.5) / ((-1)^.5)

Therefore, the final step of the proof will state (+-)1=(+-)1, which is true.


Here is another interesting one: a proof by induction that everyone who reads Chief Delphi is the same age!

First, a little refresher of proof by induction. In an inductive proof, we prove that statement S(n) is true when n=1, and then prove that if S(n) is true, then S(n + 1) is true. If we prove these two things, we have then proven that S(n) is true for all values of n.

Statement S(n): In any group of n Chief Delphi readers, everyone in that group has the same age.

First I prove that S(1) is true:

1. In any group that consists of just one Chief Delphi reader, everybody in the group has the same age, because after all there is only one person!
2. Therefore, S(1) is true.

Next, I prove that if S(n) is true, then S(n + 1) must also be true.

1. Let G be an arbitrary group of n+1 Chief Delphi readers; we just need to show that every member of G has the same age.
2. To do this, we just need to show that, if P and Q are any members of G, then they have the same age.
3. Consider everybody in G except P. These people form a group of n Chief Delphi readers, so they must all have the same age (since we are assuming that, in any group of n Chief Delphi readers, everyone has the same age).
4. Consider everybody in G except Q. Again, they form a group of n Chief Delphi readers, so they must all have the same age.
5. Let R be someone else in G other than P or Q.
6. Since Q and R each belong to the group considered in step 3, they are the same age.
7. Since P and R each belong to the group considered in step 4, they are the same age.
8. Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.
9. We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.

The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n it is also true for n+1, so by induction it is true for all n.

Re: .999~ = 1
 

Also, the infinite decimal expansion 0.99999... is the geometric series .9+.09+.009+...

a = 0.9 (the first term)
r = 0.1 (the constant multiple).

Geometric series with |r|<1 converge to a/(1-r). In this case, that's 0.9 / (1-0.1) = 0.9/0.9 = 1

So 0.9999... = 1.

Re: .999~ = 1
 

Real number...yes.

That's because you can't have decimals and fractions in a real number...so I guess you could say yes, .999~ and 1 are the same real number.

[Edit: Wait...just kidding. Im dumb.]

Re: .999~ = 1
 

well, for those who still have doubts about the difference between the number. first of all, because i am too lazy to find the alt + some odd number to find a 9 with a repeating symbol above it, yes i did use .999~ to represent an infinite amount of 9's, so that the only numerical difference between .999~ and 1 is .000~1, which is an infinitesimly small amount, and therefore in math, ignored.

also when i said infinitesimly, if i made up a word there, sorry, but it sounds good....

Re: .999~ = 1
 

That works for all sciences. I forgot the method of rounding but it really would not matter what decimal place .999999999999 is rounded to since it rounds directly up to 1.

Re: .999~ = 1
 

It's it's not a new word, but it's spelled infinitesimally.

Re: .999~ = 1
 

Theres a flaw in your theory. If x=.999, then 10x is gonna equal 9.99, not 9.999~. that would mean that 9x=8.991, and x would equal .999, which makes sense, since this is what you started out with as x.

Re: .999~ = 1
 

Ah, but x != .999, rather x = .999 ... (repeating for ever and ever, or as long as you'd like to count). Refer to my previous post, and click on the link (it explains things to skeptics quite well, I think). And it's not a "theory" -- it's actually quite established that .9 repeating is the same number as 1.

Re: .999~ = 1
 

I guess Pyro could have documented his steps a little more clearly. ;)

His 9x=9 didn't come from multiplying 9*.999.... (Which would indeed give you 8.999...1.) Rather it came from subtracting the two equations above it.

[Actually, if you were to multiply out 9*.999..., since the .999... never ends, the 8.999...1 would be 8.999.... You'd never get to the 1. According to the postulate, this would be equal to 9, but we can't rely on that, since it's the postulate that we're trying to prove.]