.999~ = 1
 

ok, bear in mind you need to comprehend at least 8th grade math to understnd this(stolen from LUE on gamefaqs.com quite some time ago, but good for conversation)


1/3 =.333~
2/3 =.666~
.333~+.666~=.999~
1/3 + 2/3 = 1
.999~=1

or

.999~=1
x = .999
10x= 9.999~
9x = 9
x=1
1=.999~

discuss.

Re: .999~ = 1
 

1/3 as .333 is an estimate and since it goes on forever u cant say it equals 1 exactly


but i no what u mean and its clever

Re: .999~ = 1
 

That's a pretty flawed proof since .333~ and .666~ are both decimal approximations of a fraction. 1/3 is really a (theoretically) neverending sequence of .33333~ (you get the idea) and the same is true for 2/3.

So, in reality ~1/3 + ~2/3 = ~1. (~ is the sign for approximation [I think]). ;)

Re: .999~ = 1
 

He's using the ~ to represent a repeating a repeating decimal, I think.

1/9 = .111~
2/9 = .222~
7/9 = .777~
9/9 = .999~ = 1

This can also be done with

1/11 = .0909~
2/11 = .1818~
5/11 = .4545~
10/11 = .9090~
11/11 = .9999~

So, yes. .999~ = 1

I <3 math.

Re: .999~ = 1
 

while we're bending the rules of mathematics, how about this one?

1 = 1

-1 = -1

-1/1 = -1/1

-1/1 = -1/1(-1/-1)

-1/1 = 1/-1

(-1/1)^.5 = (1/-1)^.5

((-1)^.5) / (1^.5) = (1^.5) / ((-1)^.5)
cross multiply, and you get

((-1)^.5) * ((-1)^.5) = (1^.5) * (1^.5)

-1 = 1

discuss that one why don't ya

Re: .999~ = 1
 

Not quite ;)

1 = 1

-1 = -1

-1/1 = -1/1

-1/1 = -1/1(-1/-1)

-1/1 = 1/-1

(-1/1)^.5 = (1/-1)^.5

((-1)^.5) / (1^.5) = (1^.5) / ((-1)^.5)
cross multiply, and you get

Did you forget order of operations? :p
Exponents first :)

(1/-1) / (1/1) = (1/1) / (1/-1)

(-1) / (1) = (1) / (-1)

-1 != 1
-1 = -1

Atleast I'm pretty sure that's right... :confused:

Re: .999~ = 1
 

What do you mean flawed math? A lot of people don't like this..

.999~ = 1 (~ means "going on forever")

...but its true. He already gave this proof, so I don't know why I'm showing it again.

x = 0.9~
- 10x = 9.9~
___________ (subtract and the .9~'s cancel out
-9x = -9
x = 1

Therefore: 1 = -.9~

Like it or not, its true.

Re: .999~ = 1
 

There are acctually slight errors that can be found in all of those. Try this one, in calculus, you'll deal with "conditionally convergent series". which, if summed to infinity, can equal any number.


(Infinity)
(Sigma) (-1)^n * (1/n)
n=1

or in other words, plug 1 in for N, plug 2 in for N and add it, plus 3 in..... to infinity. And the number you get? Whatever you want, it can be 5, it can be 0, it can be 450005.4343. Crazy, huh?

Re: .999~ = 1
 

Fair enough the second proof works, but I was referring to the first one.

If you check the math on the second one, though, you end up with 9 * .999~ = 8.999999999999999999999999999991 (give or take .000000000000000000000000000001) but it never equals 9. I could graph it and show where the line never intersects, but I'm wayyyy lazy to do that. :p

In calculus they always tell you there's more than one calculus for every problem, and think these are definitely proofs of that atleast ;)

Re: .999~ = 1
 

How about this proof:

given: e^(i * pi) = -1

e^(3 * i * pi) = e^(i * pi) * e^(i * pi) * e^(i * pi) = -1 * -1 * -1 = -1

therefore:
e^(i * pi) = e^(3 * i * pi)

ln e^(i * pi) = ln e^(3 * i * pi)

(i * pi) ln e = (3 * i * pi) ln e

i * pi = 3 * i * pi

1 = 3

Re: .999~ = 1
 

Back when I was 229 Team Leader, I got a random email from some mathematician out to prove that dividing something by zero actually equaled zero.

http://members.lycos.co.uk/zerobyzero/

It's pretty weird.
It was definitely strange to get an email from him, apparently he just searched the net randomly for people to hear his theory, and our team name ("Division by Zero") led him to me.

Re: .999~ = 1
 

Haha that site makes me laugh...

Some of the theories on there are cool...but most of them are just kind of strange. He really likes the number zero... :(

His samples really don't add up to me, though.

Ex:

(1/2) / (0/2) = (1/2) * (2/0) = 0

You can't divide something into portions of zero! I know he's using the basis of cross multiplication but yeah...it's still wrong.

Back to the topic at hand, the last proposed proof (the e^x one) has officially blown my mind. I think it's time for a cookie.

Re: .999~ = 1
 

Mathematicians would agree that .999... and 1 represent the same real number. In fact, from my real analysis book, "Elementary Analysis" by Kenneth Ross, it says:
[short proof, similar to above]
"Thus 0.9999... and 1.0000... are different decimal expansions that represent the same real number!"
Later on the book proves that the only case where this can occur is when the two expansions for the same real number end in an infinite series of nines and infinite series of zeros.

Re: .999~ = 1
 

major probelm: you cannot take the ln of a negative number. therefore, the term ln e^(i * pi) cannot exist. i have another one of those phony proofs, its kinda fun:

let a=1, b=1
a = a
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
1 + 1 = 1

have fun =D

Re: .999~ = 1
 

square rooting then squaring again cause a loss of information. consider:
x = 3, y = -3
x^2 = y^2
sqrt(x^2) = sqrt(y^2) sqrt = square root of
however, x != y even though sqrt(x^2) = x and sqrt(y^2) = y and sqrt(x^2) = sqrt(y^2).

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.]

Re: .999~ = 1
 

I'm glad to see so many people discussing mathematics and proofs. When doing so, please don't refer to your "trusty TI-89" calculator, or even the latest PC or Apple supercomputer. All computers run in binary and have problems handling the easy math problems that a 4th grader can understand (can a computer really express exactly that 1/3 + 1/3 + 1/3 equals 1?). Besides the inherent flaws with computers using binary, you also have to remember that someone had to program your calculator or computer and they are very capable of making mistakes just like you and I. Programming algorithms and software for math relies heavily on a ridgid set of rules (can you divide by zero, or take the ln of a negative?).

Mathematical proofs are carefully constructed arguments based on either known (and accepted truths) like axioms and definitions, and also theorems which have been previously proven. All of the above "proofs" or mysterious equations have a flaw in them somewhere which is not readily apparent to the reader. Unless you know the rules, you do not know what can or cannot be done. This is like the FIRST competition, is it not? If you don't know an obscure rule that other teams have read about in a team update or the manual, you are at a distinct disadvantage.

Over the years, math has evolved and our eyes have been opened to things our ancestors never thought possible. Let me cite a few examples.

Negative numbers. When caveman first started counting their rocks (1,2,3...), they didn't have a need for negative numbers. Did they exist? Sure, but it wasn't until people started borrowing sheep from their neighbor and they "owed 2" did the concept of being in the hole with a negative 2 sheep did this make any sense.

Irrational numbers. Pythagoras nearly made himself go crazy trying to find a fraction that would satisfy the hypotenuse of a right triangle with the two legs equal to 1. The ancient greeks thought all numbers could be represented by fractions. Of course, we know that the hypotenuse is √2, which is not rational. The introduction of rational numbers changed our way of viewing numbers and the associated rules. Transcendental numbers rocked our world in a similar way.

Non-integer exponents. A fifth grader can understand that x²=x*x and that x³=x*x*x. If you asked them, what is x^6, they would say "multiply 6 x's together". If you asked them, what is x^(-2) or what is x^(2.154), they would say "you can't do that!". A Calculus student wouldn't even flinch.

How about 0/0? Is this 1 or 0? Isn't there a rule that says anything divided by itself is equal to 1? But isn't there also a rule that says 0 times anything is 0, and since 0 is in the numerator, this must be 0. Hmmm. Go find L'Hopital and ask him what to do.

Imaginary numbers? No need to say anything. People still don't understand them.

Quaternions? These are extensions of Imaginary numbers. Think i²=-1 is hard to understand. Wait until you find out that j²=-1 and k²=-1 but that i≠j≠k.

Again, once you know the rules, your eyes become opened. If you don't understand the rules of the game, someone that does is sure to either pull the wool over your eyes or beat you at the game.

Sorry for the long post, but its been awhile since I've had the opporunity to talk about math with such a willing audience.

Let me end with my favorite quote of all time:
"God made the integers; all the rest is the work of man." - Leopold Kronecker

Dave...

Re: .999~ = 1
 

These things are pretty accurate when it comes to knowing about certain things. The TI-83's actually know when a fraction is non termaniting like 1/3. That is one of the very first things that I thought was odd about it. Though some of the algorithims are inaccurate like most if not all of the calculus functions. They are always off by a little bit.

Re: .999~ = 1
 

One would believe that .999 is not equal to one.


In order to make .999 equal to 1, you first have to declare that your tolerance for the project is +/- .0005. This would allow you to round any number greater than .9995 up to 1.

You would have to look at what type of calculation you are doing. If it was calculating the length of a peice of angle iron, I think your tolerances could be within +/- .005 or even +/- .05. If you were calculating the molar mass of a particle of angle iron though, you might want to have a tolerance of +/- 5.0 x 10^-29 (just to use as an example.)

Finally, remember that if you design something in CAD that the CNC machine or Machinist that makes the part will have some degree of error (hence the +/-). So if you send a part that is measured to be .999 with a tolerance of +/- any thing, you will most likely get a part that measures 1.

Just my 25 cent worth:)

Gabe Goldman
Prez and Founder of VCU Robotics Club

Re: .999~ = 1
 

This is exactly why the proof that I posted with e^(i*pi) doesn't work.

Re: .999~ = 1
 

e^(pi*i) IS actually -1. The problem is somewhere else...

Re: .999~ = 1
 

yes it is actually -1 but the problem is when i compare e^(3*i*pi) to e^(i*pi) cause even though they are both -1 they are not equal

Re: .999~ = 1
 

cut it down for space reasons, but no reason to apologize, in fact it is wholeheartedly welcomed by me, and the fact that you mentioned mathamatical terms i can't understand at this time intruiges me. now if only i could do better in pre calculus....

Re: .999~ = 1
 

thats easy, you cant divide by zero..we did that one in my algebra 4 class

Re: .999~ = 1
 

a problem here - S(n) is not true for all n. The proof of S(n+1) involves at least 3 distinct persons, and thus is not applicable for S(2). Thus, the entire sequence breaks down, as the S(n+1) proof therefore does not work for S(3), et cetera, ad infinitum.

Re: .999~ = 1
 

Right, but the TI-89 isn't solving a problem like ln[e^(3*i*pi}] by first multiplying 3, i, and pi together, then raising e to that result, then taking the ln of that. That kind of calculation would introduce error due to storing numbers in binary. Instead it applies known mathematical rules to produce the correct answer with little to no actual computation. For example, if you type "sin(x + 90)" (in degrees) where x is some unknown variable and press enter, the TI-89 will display "cos(x)". Obviously it knows the rules. It either instantly recognized the +90 as a shift to cos, or it expanded sin(a + b) to sin(a)cos(b) + sin(b)cos(a) and simplified.

That's basically what a human does when they prove something. They go step by step, and each step must be backed by a known and proven mathematical rule. If a calculator was programmed to know all the rules, why couldn't it produce a similar proof? How would checking an answer on the calculator be any different than applying those rules yourself? From my experience, I'm pretty sure the TI-89 is programmed to know all the rules. If you've never used one, it is quite an impressive calculator.

Obviously there is an error in the proof, since we all know that 3 != 1. I was simply using my TI-89 to figure out the location of the error, not the specific rule that was broken.

Re: .999~ = 1
 

also, a note on the 3 = 1 proof -
given a value X,
e^(X*i) = i*sin(X) + cos(X) (eulers law or something like that... forget the exact name)

thus, we are working on a unit circle, and in radians. As far as reference angles go, 3*pi does actually equal pi. However, this is only in the reference angle form. Essentially, when computing e^(X*i), the operation is one-way. You cannot definitively find X given the value of e^(X*i) - you can only find one or more possible X's.

Re: .999~ = 1
 

Exactly what I was saying earlier. This is related to sines and cosines.

sin(0) = sin(2*pi) does not imply 0=2*pi. It implies 0 rad = 2*pi rad. Just as pi rad = 3*pi rad.

Re: .999~ = 1
 

not quite - 0 rad does not equal 2pi rad - they are different angles. However, the key lies in the fact that sine and cosine are not definitively reversible operations.

Re: .999~ = 1
 

Hmmm... What are those acos and asin buttons on my calculator for? Sine and cosine are "reversible" operations since there are inverse functions.

I think that what you really mean is that the sine and cosine functions are not one-to-one mappings.

Re: .999~ = 1
 

hence definitively reversible. also known as not being one-to-one, as you said. Essentially, the inverse function is not actually a function. it's all good :)

Re: .999~ = 1
 

Hi guys,
never posted before because i'm afraid i'm not as technically proficient with FIRST robotics as a lot of you, but, if there's one thing i know, it's calculus.

.999~ (.999 recurring or repeating as its commonly known) does in fact equal one. The "Multiply by Ten" Proof is true, but difficult to grasp for the common mind, so suffice to say that this proves it once and for all: (granted someone already did mention this)

A infinite geometric series is a series of numbers where one begins with a set point (the First element) and finds the next point by multiplying the First element by some number X. Then, to find the next number in the sequence, you keep on taking the number you have and multiplying by X (X is called the common ratio, in fact). Odd as it may seem, such infinite series of numbers do in fact add up to definite finite real values. The sum of an infinite geometric series is A / (1 - R) where A is the First element of the series and R is the common ratio. In this case, The series is .9 + .09 + .009 + . 0009 ad infinitum. Thus, the First element "A" is clearly .9 and the Common Ratio "R" is clearly .1 , or One-Tenth. Now, consider: in this case, A / (1 - R) = .9 / (1 - .1) which equals .9/.9 which of course equals 1. Ta da! For those of you who don't believe this geometric series sum rule i am using, there are myriad other ways to show the fact, such as the calculus' Ratio test, etc.

Meanwhile, this is a fascinating discussion, and I enjoyed reading everyone's thoughts. Someone mentioned the idea of 0/0. There's a name for it, and I seem to forget exactly what the name is (the name does make sense once you hear it). Anyway it's not 0; 0/0 is equal to INF/INF or 1^INF or 0^0 or any other myriad devilish expressions : it simply does not exist. I think it might be called an infinite discontinuity but that's just a guess. anyhow, there's math for ya!

captain dan
team 1168
www.friarbot.com rocks my world too hard

Re: .999~ = 1
 

note - only for R < 1.

One of the best known applications of this is L'Hopital's rule, which is used for finding the integrals of certain functions that can be transformed into something that resembles one of those patterns - 0/0, inf/inf, inf-inf, 1^inf, 0^0, et cetera.

Re: .999~ = 1
 

L'Hopital's Rule is actually usually used for computing limits. For example, to compute

lim x/sin(x) as x-->0, we can apply L'Hopitals rule to find that

lim x/sin(x) = lim 1/cos x = 1.

Re: .999~ = 1
 

I believe you are looking for "indeterminate form." Check out mathworld

Re: .999~ = 1
 

also thus used in calculus for finding the limits of infinite integrals (happened to be doing that a couple of days ago in calculus, so I mentioned it in that context). :)

Re: .999~ = 1
 

a couple of things
Firstly:
i can prove that 1=0


ok, assume its an identity (cant do that on a keyboard but assume the = is an identity sign)
if you say 1=0
if you take 1 from each side
0=-1
multiply by 2
0=-2
add one to both sides
1=-1
Square
1=1
voila :D
or do it backwards to get it if you like.

also, i can prove mathamatically the first part
1/9 = 0.1111... (recurring)
10/9 = 1.1111... (recurring)
10/9 - 1/9 = 9/9 = 0.9999999999, or 1 as 9/9 is 1 but 9 x 0.11111111111 = 0.9 revcurring. therefore 0.9999.... doesn't exist


its marvalous that e^(pi i )=-1

Re: .999~ = 1
 

squaring is not a 1 to 1 operation, thus you cannot do the proof backwards, and thus it is not valid.

x^2 = y^2 --/--> x = y
(x squared = y squared) does not imply (x = y)

Re: .999~ = 1
 

You can't get the square root of 1 equal to -1 so it fails.

Re: .999~ = 1
 

At least when it is not defined. arccos x is defined from 0 to 1 and arcsin x from -pi/2 to pi/2. This makes it one to one and gives it an inverse function.

Re: .999~ = 1
 

can't divide by zero....3rd to last step

Re: .999~ = 1
 

at least someone found the correct reason for it being incorrect :)


What is the sqrt of 1?

Consider that if you square -1 you get 1 and so the sqrt of 1 is -1 or 1

Re: .999~ = 1
 

Anyone want to help me with this one?

Find the fallacy in the following argument that 0 = 1.

dv = dx --> v = Integral (dx) = x
u = 1/x --> du = -1/(x squared) dx

0 + Integral (dx/x) = (1/x)(x) - Integral ((-1/(x squared)(x)) dx = 1 + Integral (dx/x)

So, 0 = 1

It is integration by parts ( integral (u dv) = uv - integral (v du)). I should be figuring it out myself but when I saw the problem I thought of all of you.

Re: .999~ = 1
 

Given that you're working with reals, it's true that you can square a negative number, but the square root function only outputs positive numbers.

sqrt(1) /= 1 or -1
+/-1^2 = 1

Re: .999~ = 1
 

constant of integration.

Re: .999~ = 1
 

Everyone's proving stuff...
Well... 2 = 1
Why ??

Check this out... nothing illegal and you can go backwards :)

a = b
ab = b^2
ab - a^2 = b^2 - a^2
a(b-a) = (b-a)*(b+a)
a = b+a

So if a=1, b=1 too (since the first thing I assumed was a = b )

1 = 1+1 ???

Have fun...
Don't flame... I am putting this to have some fun :p

Re: .999~ = 1
 

What do you mean with that ?? Explain please...

Re: .999~ = 1
 

Nothing's illegal, eh? To get from
to
you have to divide by (b-a) which is zero! Which is the DEFINITION of illegal! ;)

Re: .999~ = 1
 

Re: .999~ = 1
 

yep. oldest trick in the false-proof book. second oldest is trying to argue that x^2 = y^2 implies that x = y.

Re: .999~ = 1
 

essentially, an indefinite integral will always be equal to the same indefinite integral plus a constant. it would have to be, as the following example shows:

f(x) = 2x + 1
g(x) = 2x

f'(x) = 2
g'(x) = 2

if we didn't have a constant of integration,
integral(f'(x)) = 2x = integral(g'(x)) = 2x
and thus f(x) would be implied to be = to g(x)

however, the correct form is this:
integral(f'(x)) = 2x + C1
integral(g'(x)) = 2x + C2

thus, f(x) = g(x) + C2 - C1, where C2 and C1 are integers.

essentially what this all means is that you can't simply "take out" an indefinite integral from an equation, and thus the fact that 0 + integral(f(x)) = 1 + integral(f(x)) does not imply that 0 = 1.

Re: .999~ = 1
 

well said...
whenever you integrate an indefinate you must always have a constant. in definate integrals, this doesn't matter as when you solve it you get
(.........+k)-(.........+k) and they cancel out. (where k is a constant)

Re: .999~ = 1
 

No. This is quite different. Quaternions are an extension of the real numbers and you don't need to even mention them to talk about e^(i*pi). With quaternions, ab != ba in some circumstances. There are even octonions with a(bc) != (ab)c.

The reason why your e^(i*pi) thing doesn't work is that you are taking the log of a negative/complex value incorrectly. Why does log(exp(3pi*i) = i*pi? Because exp(3pi*i) = exp(pi*i) = -1 and in the natural extension of the log to the complex plane log(-1) = i*pi. So one must be very careful when using natural logs on complex arguments. There are several good books on the subject, in fact MIT has a complex analysis class available entirely free. This is necessary before one "proves" anything involving complex arguments using functions defined on the reals.

Re: .999~ = 1
 

*wipes a single tear from his eye*

just when i think i was some hot ****, i have never been so proud to be proven that i am not by far as much of a math geek

<3 the topic

Re: .999~ = 1
 

alright i thought of a different way to prove that .999...=1

first we have to prove that if (X+Y)/2=X or (X+Y)/2=Y then X=Y
this is simple

(X+Y)/2=X
X+Y=2X
0=X-Y
Y=X

alright now we move on

1 (.9999...+1)/2 ?= .999...
2 (1.999...)/2=
3 .999...=.999...
now you might be thinking: "whoa...how do you get from 2 to three...that looks crazy"

well 1.9/2=.95
1.99/2=.995
1.999/2=.9995

so 1.999.../2=.999...5

however i have given all you non believers false hope with that last 5

1.999.../2=.999...

because the nines will go on FOREVER, they wont stop, so that five does not exist the five never happens because the nines never stopped...thus .999...=1