User talk:Justin545

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Welcome!

Hello, Justin545, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful:

I hope you enjoy editing here and being a Wikipedian! Please sign your messages on discussion pages using four tildes (~~~~); this will automatically insert your username and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or ask your question and then place {{helpme}} before the question on your talk page. Again, welcome! Have fun editing Wikipedia !! ReluctantPhilosopher (talk) 17:13, 16 February 2008 (UTC)

[edit] Question about template "helpme"

{{helpme}} What type of questions can I ask when I use {{helpme}}? I have some questions about quantum mechanics to discuss but I am not sure if it is suitable to use {{helpme}}...

Justin545 (talk) 23:54, 22 February 2008 (UTC)

To answer your question, {{helpme}} should only be used for questions such as using wikipedia. For content questions like physics you should go to Wikipedia:Reference desk. Hope this helps, if you have any questions feel free to ask me on my talkpage.--Sunny910910 ^{(talk|Contributions|Guest)} 23:57, 22 February 2008 (UTC)

Well, thanks! Wikipedia:Reference desk looks like a good place for me to discuss the questions! - Justin545 (talk) 00:06, 23 February 2008 (UTC)

[edit] Response in quantum computing thread

I just wanted to mention that I finally posted a reply in the quantum computing thread. It might not show up in your watchlist because the thread is now in the archives at Wikipedia:Reference_desk/Archives/Science/2008_February_23. -- BenRG (talk) 16:25, 27 February 2008 (UTC)

Thank you for the mention and reply. I will take some time to understand it and give some response with repsect to your reply :) - Justin545 (talk) 01:07, 29 February 2008 (UTC)

Thanks for the barnstar. :-) Although I don't think I understand quantum mechanics all that well... -- BenRG (talk) 14:36, 14 March 2008 (UTC)

[edit] 0.999999.... and 1

Having seen you post several times in the mathematics section recently, I decided to take a look at your user page and was surprised to learn that you believe 0.99999.... doesn't equal 1. Considering that your a skeptic, I'm not terribly surprised by that, and it is good to question things, even generally excepted ideas. So to justify why 0.999...=1 one needs to thoroughly understand the ... notation's definition. The ellipsis notation (...) is used to indicate that a pattern in the digits of a decimal notation is repeated without end (infinite). so 0.9999... mean that the pattern of 9's in decimal places of successively decreasing value is repeated without end. Given the definition of ... it is easy to show either through limits or a converging summation, that 0.999...=1. There are also ways of showing it without using "higher" mathematics but those ways aren't always so convincing, it is certainly not the most intuitive result in mathematics. (One thing I must point out about the ... notation is that when it is used on a finite decimal that does not contain two or more repetitions of a pattern, it indicates that the decimal is infinite in length and does not imply that there is or is not a pattern) A math-wiki (talk) 23:29, 20 March 2008 (UTC)

Let

$g(c)\equiv\frac{f(x+c)-f(x)}{c}$

Then

$\lim_{c\rightarrow 0}g(c)=\lim_{c\rightarrow 0}\frac{f(x+c)-f(x)}{c}=f'(x)$

But, according to your opinion

$\lim_{c\rightarrow 0}g(c)=\lim_{a\rightarrow 1^-}g(1-a)=g(1-0.999999...)=g(1-1)=g(0)=\frac{f(x+0)-f(x)}{0}=\infty$

which is seems ~~pretty~~ odd to me. - Justin545 (talk) 12:37, 21 March 2008 (UTC)

Two errors here. First, you can't (in general) evaluate limits by plugging in the limiting value, otherwise there'd be no need for the concept of a limit. And that is what you've done here, since one of your stated assumptions was that 0.999999... = 1. Second, 0/0 is not infinite but rather indeterminate; it's the solution of 0x = 0, which is satisfied by every x. In contrast, 0x = 1 is not satisfied for any (finite) x, so 1/0 has no (finite) value instead of having all values. Getting 1/0 at the end of a calculation often means that there's no solution, but getting 0/0 just means that you can't find the solutions, if any, by that approach. -- BenRG (talk) 14:47, 21 March 2008 (UTC)

>> "Second, 0/0 is not infinite but rather indeterminate"

Apologies. I was making the response in a hurry. It should be corrected as

$\lim_{c\rightarrow 0}g(c)=\lim_{a\rightarrow 1^-}g(1-a)=g(1-0.999999...)=g(1-1)=g(0)=\frac{f(x+0)-f(x)}{0}=\frac{0}{0}$

>> "since one of your stated assumptions was that 0.999999... = 1"

My opinion is $0.999999...\ne 1$ and A math-wiki's opinion is

0.999999... = 1

. I was trying to show that $0.999999...\ne 1$ .

>> "you can't (in general) evaluate limits by plugging in the limiting value, otherwise there'd be no need for the concept of a limit."

I'm not sure I understand that. I think it should be able to plug in the limiting value to evaluate limits since it's the basic concept of limits. Suppose we want to evaluate

f'(x 0)

where

x 0

is some fixed constant, we should be able to plug in

Δ x = 0.001

$\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$		(1)

The more does

Δ x

approaches to zero, the more accurate limit value we will get. So plug

Δ x = 0.000001

into (1) will get a more accurate value than

Δ x = 0.001

. Plug

Δ x = 0.00000000001

into (1) will get a more accurate value than

Δ x = 0.000001

...and so on. And we will end up with an infinitely small value

Δ x = 0.000...1

, plug that value into (1) will get the true value (with infinite accuracy) of

f'(x 0)

. According to A math-wiki's opinion, however, he may also say

Δ x = 0.000...1 = 0

and we will get $\frac{0}{0}$ by plug

Δ x = 0

into (1). - Justin545 (talk) 00:01, 22 March 2008 (UTC)

First off, I want to mention that x=0.000...1 would technically imply an unknown finite number of zeros proceeding the 1, however I understand your meaning in this context. So let me write 0.000...1 as u mean it in a different form $x=\lim_{k \rightarrow \infty}0+\frac{1}{k}$ , (I assume u would agree that, 0=0.0=0.00 etc. From this limit is it quite clear that the limit is 0 since 1/k becomes arbtrarily small as k becomes arbitrarily large.

As for >> "you can't (in general) evaluate limits by plugging in the limiting value, otherwise there'd be no need for the concept of a limit." (above) to clarify on what makes a limit different than a straight forward computation is the limit poses the question what is the function's output approaching as it's input is approaching ___. Whereas removing the limit statement, your absolutely right that 0/0 would be the 'value' (though technically, one would say the function is undefined at __in the above equation but that's not what the limit is asking, it's asking what is it approach not what is the actual value. A really good way of showing this is on a rational function, say you have $f(x)=\frac{x^2+6x+5}{x+1}$ and you are asked what the limit as x approaches 1 is. The limit in this case is defined whereas the value of f(x) is not at x=1 . In this case, f(x) is said to have a removable discontinuity at x=1 since you could cancel the factor x=1 on both the numerator and the denominator to get $\frac{x+6}{1}=x+6$ . Finally, directly applying the limit in the definition of the derivative will also yield 0/0 not $\infty$ as you position would imply. $f'(x)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$ , substitute, $f'(x)=\frac{f(x+0)-f(x)}{0}=\frac{f(x)-f(x)}{0}=\frac{0}{0}$ that's exactly what you got when you attempted to show that my assertion that 0.999...=1 was incorrect. One last thing, there does come a point for some f(x)'s where one can substitute the limit directly and $x=\lim_{k \rightarrow \infty}0+\frac{1}{k}$ is an example, this will yield $x=0+\frac{1}{\infty}$ , and the reason that is a valid substitution is because 1/ $\infty$ is a creative form of zero since 1 is arbirarily small compaired to infinity. thus $x=0+\frac{1}{\infty}=0$ A math-wiki (talk) 09:34, 22 March 2008 (UTC)

>> " The ellipsis notation (...) is used to indicate that a pattern in the digits of a decimal notation is repeated without end (infinite)...I want to mention that x=0.000...1 would technically imply an unknown finite number of zeros proceeding the 1"

Ur prior statement seems to contradict ur latter statement.

>> "say you have $f(x)=\frac{x^2+6x+5}{x+1}$ and you are asked what the limit as x approaches 1 is. The limit in this case is defined whereas the value of f(x) is not at x=1 ."

I think

f (x)

has also defined value at

x = 1

since $f(1)=\frac{1^2+6+5}{1+1}=\frac{12}{2}=6$

>> "In this case, f(x) is said to have a removable discontinuity at x=1 "

Excuse me...removable discontinuity?

>>you could cancel the factor x=1 on both the numerator and the denominator

Yep, because $f(x)=\frac{x^2+6x+5}{x+1}=\frac{(x+1)(x+5)}{x+1}=x+5$

>> "1/ $\infty$ is a creative form of zero since 1 is arbirarily small compaired to infinity"

Correct,

1

is so small compaired to infinity so we can ignore the difference between 1/ $\infty$ and

0

without changing the results for "some" cases. But we can not "always" ignore the difference between 1/ $\infty$ and

0

for "all" cases, since the difference between 1/ $\infty$ and

0

is significant for some cases. For example, we should NOT ignore the difference between 1/ $\infty$ and

0

for the case $f'(x)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$ , substitute the limit directly by 1/ $\infty$ and

0

will yield two different results for the case, which means the difference between 1/ $\infty$ and

0

is significant for the case. For another example, $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is NOT "always" equivalent to $\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$ since substitute the limit directly by $x\rightarrow a$ and

x = a

will not always yield the same results for all of kind of

f (x)

's and

g (x)

's. $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is "said" to be equivalent to $\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$ only when substitute the limit directly by $x\rightarrow a$ and

x = a

will yield the same results for some

f (x)

's and

g (x)

's. When $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is "said" to be equivalent to $\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$ , it means the difference between $x\rightarrow a$ and

x = a

is trivial and therefore can be ignored for the case. - Justin545 (talk) 08:11, 23 March 2008 (UTC)

>> " The ellipsis notation (...) is used to indicate that a pattern in the digits of a decimal notation is repeated without end (infinite)...I want to mention that x=0.000...1 would technically imply an unknown finite number of zeros proceeding the 1" (...) with no symbols after it means that is infinite number of digits, if there is a number or several numbers afterwards then it is an unknown finite number of digits.

>> "say you have f(x)=\frac{x^2+6x+5}{x+1} and you are asked what the limit as x approaches 1 is. The limit in this case is defined whereas the value of f(x) is not at x=1 .", sorry that was a typo on my part, try x=-1 instead of x=1.

>> "In this case, f(x) is said to have a removable discontinuity at x=1 ", a removable discontinuity can be defined as follows, assume there exist two function f(x) and g(x) such that f(x)=g(x) if and only if $x\ne k$ , the functions in my previous argument were meant to demonstrate this concept.

$\frac{1}{\infty}=0$ regardless of circumstances assuming the standard definitions for those symbols. I suggest your read Limit of a function it will help you more clearly understand limits. A math-wiki (talk) 13:31, 23 March 2008 (UTC)

Take a look at this approach instead.

$0.999...=\lim_{n \rightarrow \infty}\sum_{i=1}^n {\frac{9}{10^i}}$

Take that as a definition of 0.999... and then evaluate that sum. A math-wiki (talk) 23:37, 21 March 2008 (UTC)

When

f

is a continuous fuction, we can write

$\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a^-}f(x)=\lim_{x\rightarrow a^+}f(x)=f(a)$

$f(x)\equiv x$

Then we can ensure

$\lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1^-}f(x)=f(1)=1$

where $x\rightarrow 1^-$ means that

x = 0.999999...

U can say

f (0.999999...)

is "symbolically" equivalent to

1

. But I doubt

f (0.999999...)

is also "semantically" equivalent to

1

since there are subtle differences. For example, if $\lim_{x\rightarrow a}f(x)$ is also "semantically" equivalent to

f (a)

, we should be able to say that $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is "always" equivalent to $\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$ . But actually $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ can be said to be equivalent to $\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$ only if the denominator

g (x)

is non-zero. That subtle difference makes me believe

0.999999...

is not "semantically" equivalent to

1

. - Justin545 (talk) 12:53, 22 March 2008 (UTC)

$x \to 1^-$ does not mean

x = 0.999...

or any particular value, it is a specification on the limit, telling us where and if applicable in which direction we are to be evaluating the limit. Consider $\lim_{x \to p}f(x)=L$ this limit exists if and only if for every real

ε > 0

there exists a real

δ > 0

such that

0 < | x - p | < δ

implies

| f (x) - L | < ε

. Note that the value of the limit does not depend on the value of

f (p)

, and a more general definition applies for functions defined on subsets of the real line. Let

(a, b)

be an open interval in $\mathbb{R}$ , and

p

a point of

(a, b)

. Let

f

be a real-valued function defined on all of

(a, b)

except possibly at

p

. We then say that the limit of

f

x

approaches

p

L

if and only if, for every real

ε > 0

there exists a real

δ > 0

such that

0 < | x - p | < δ

and $x \in (a,b)$ implies

| f (x) - L | < ε

. Note that the limit does not depend on

f (p)

being well-defined.

Firstly, there is no such thing as symbolically vs semantically equivalent, either something is equal to something else or not. 0.999....=1 can be verified very easily by the convergence of the sum I provided. As for you limit argument about g(x), that really has no bearing here, as the denominator is non-zero. If you have a sum which appears to be 0/0 your again dealing with an indeterminate form. And the expression should be rewritten to so that you answer will look different than 0/0 because 0/0 tells you f(x) has a discontinuity at x=k where f(k)=0/0. The best way to clarify what limits mean and how they work would be to look at the tangent line problem. Say we have a function f(x), and we want to know what the equation for a line that touches f(x) at x=3 is. We define the secant line, a line which cuts through

x = 3

and

x = 3 + ε

. Now the secant line intersects f(x) at

(3, f (3))

and

(3 + ε, f (3 + ε))

, but we want the tangent line not the secant line, the tangent line just touches f(x) at x=3 so it's intersection points are superimposed at (3,f(3)) and (3,f(3)), the slope of the secant line is $m=\frac{y_2-y_1}{x_2-x_1}$ . if we directly substituted the intersection point(s) (3,f(3)), and (3,f(3)), it's quite obvious that we would see 0/0 as the slope which is merely indeterminate, obviously m has some value if f is indeed a continuous function on an interval containing x=c (x=3 in our case) as is assumed in the Tangent line problem. Indeterminate tells us that direct substitution will not work. but we know that the answer must exist and it must be possible to find it. This is where the concept of limits was born, we use the secant line's slope to approximate the slope of the tangent line, then we apply a limit to increase the accuracy to arbitrarily high levels (e.g. perfect). The slope of the secant line is $m=\frac{f(x+\epsilon)-f(x)}{(x+\epsilon)-x}$ (this looks a lot like a derivative now doesn't it?) Now to get the exact slope of the tangent line one must invoke a limit. $m=\lim_{\epsilon \to 0}\frac{f(x+\epsilon)-f(x)}{(x+\epsilon)-x}$ . (looks even more the definition of the derivative). the final step is to simplify the denominator and what you have is the derivative of f(x) at c for whichever c you choose, if you don't choose a c as I didn't then it gives f'(x).

And another really key point to make is that $\lim_{x \to c}f(x)=f(c)$ if and only if f(x) is continuous on an interval [a,b] that contains c. Otherwise f(x) may not be defined at x=c as is the case of the rational function I gave an example from in one my posts above. In general it is incorrect to assume $\lim_{x \to c}f(x)=f(c)$ A math-wiki (talk) 22:19, 22 March 2008 (UTC)

I also get the inpression that you do not fully understand limits and their definition, so reading up on Limit of a function may help clarify your understanding. A math-wiki (talk) 22:27, 22 March 2008 (UTC)

There is an article here on wikipedia, that specifically addresses why 0.999...=1, it is here 0.99999... A math-wiki (talk) 22:59, 23 March 2008 (UTC)

>> "Note that the value of the limit does not depend on the value of $f (p)$ "

It just means

f (x)

has a removable discontinuity at

x = p

as you said.

>> "Firstly, there is no such thing as symbolically vs semantically equivalent, either something is equal to something else or not."

Agree. But it's a matter of interpretation of the math language. Basically, math itself has no meaning at all. Math is just a set of inference rules. You can make any interpretation of the math as long as you don't break the rules and you can assure of the consistency. I was using some inappropriate words, such as symbolically vs semantically equivalent, to explain the concept. But I think my latter interpretation:

1

is so small compaired to infinity so we can ignore the difference between 1/ $\infty$ and

0

without changing the results for "some" cases. But we can not "always" ignore the difference between 1/ $\infty$ and

0

for "all" cases, since the difference between 1/ $\infty$ and

0

is significant for some cases. For example, we should NOT ignore the difference between 1/ $\infty$ and

0

for the case $f'(x)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$ , substitute the limit directly by 1/ $\infty$ and

0

will yield two different results for the case, which means the difference between 1/ $\infty$ and

0

x = a

will not always yield the same results for all of kind of

f (x)

's and

g (x)

x = a

will yield the same results for some

f (x)

's and

g (x)

's. When $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is "said" to be equivalent to $\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$ , it means the difference between $x\rightarrow a$ and

x = a

is trivial and therefore can be ignored for the case.

should be able to satisfy the rules and definitions of limits. You can provide any counterexample to proof the interpretation is wrong if you disagree.

>> "0.999....=1 can be verified very easily by the convergence of the sum I provided."

Frankly, I'm not good at math as you are since I have no such perfect educational background as you have. I think the convergence of the sum is not so obvious to me. - Justin545 (talk) 09:04, 24 March 2008 (UTC)

>> "And the expression should be rewritten to so that you answer will look different than 0/0 because 0/0 tells you f(x) has a discontinuity at x=k where f(k)=0/0."

Consider a function $h(x)\equiv 1$ then we can conclude that

h (0) = 1

. As you said we can cancel the the value on both the numerator and the denominator, which implies $\frac{x}{x}=1$ which implies $h(x)=\frac{x}{x}$ . Think again, we conclude that $h(0)=\frac{0}{0}$ —– an indeterminate form! Does that mean $\frac{x}{x}\ne 1$ or $\frac{0}{0}=1$ ? If $\frac{x}{x}=1$ , how come we could get two different results

h (0) = 1

and $h(0)=\frac{0}{0}$ ? How to solve this problem? - Justin545 (talk) 09:30, 24 March 2008 (UTC)

In that case 0/0=1, but consider the function

h (x) = 2

then you could conclude through the same argument you used that $\frac{0}{0}=2$ , the point I'm making is, that 0/0 doesn't tell you what the value is, so it is indeterminate. Technically, making the substitution $1=\frac{x}{x}$ would imply $x \ne 0$ as a condition, for the very reason that you would obtain 0/0, which, because of it's indeterminate nature, would result in the answer not being well-defined for

x = 0

. 0/0 is considered to be undefined, don't confuse undefined with infinite, they are not the same thing! so technically, the latter form of h(x), which I shall call H(x) is undefined for H(0), but h(0)=1, so with the exception of x=0, H(x)=h(x). You have introduced a removable discontinuity at x=0, by the substitution you used. The limits as x approaches 0 of both h and H exist and are equal, because both h, and H meet the definition a Limit of a function when p=0. A math-wiki (talk) 23:29, 24 March 2008 (UTC)

And for a simply proof that 0.999...=1,

$0.333...=\frac{1}{3}$ , I assume you agree that this is a true mathematical statement, then

$3 \cdot 0.333...=3 \cdot \frac{1}{3}$ If 0.333... is interpreted as base 10 would imply, then it's equivalent to $3 \cdot 10^{-1}+3 \cdot 10^{-2}+3 \cdot 10^{-3}+...$

$0.999...=\frac{3}{3}$

$0.999... = 1$ A math-wiki (talk) 23:47, 24 March 2008 (UTC)

>> "making the substitution $1=\frac{x}{x}$ would imply $x \ne 0$ as a condition"

According to your response, you were saying $\frac{x}{x}\ne 1$ ? Isn't it? Therefore, $\frac{x}{x}$ is not equal to

1

except we write

$\frac{x}{x}=1$ , $\forall x\ne 0$

correct? Thus, we can not simply cancel the the values on both the numerator and the denominator but rather we should also add $\forall x\ne 0$ behind the cancelling, right?

>> $\frac{1}{9}=0.111...$ therefore $9*0.111...=0.999...=9*\frac{1}{9}=1$

Accept. That is the proof from the article 0.99999.... But recall what you said

"I decided to take a look at your user page and was surprised to learn that you believe 0.99999.... doesn't equal 1. Considering that your a skeptic, I'm not terribly surprised by that, and it is good to question things, even generally excepted ideas."

I think questioning and skepticize are basic attitude to handle scientific problems. A statement can't be said to be true until we can proof it. That is what I did when I put

!=	This user believes that 0.99999... does NOT equal 1.

on my user page since I didn't see the proof of 0.99999....=1 before so I doubt their equivalence. There is no need to surprise. If you carefully read the user box "This user believes that 0.99999... does NOT equal 1." you will realize the box is just telling my faith but not the fact that 0.99999... does NOT equal 1. And there are some students have the same faith as me since there is a section from the article 0.99999... :

Students of mathematics often reject the equality of 0.999… and 1, for reasons ranging from their disparate appearance to deep misgivings over the limit concept and disagreements over the nature of infinitesimals. There are many common contributing factors to the confusion:

Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal." Seeing two manifestly different decimals representing the same number appears to be a paradox, which is amplified by the appearance of the seemingly well-understood number 1.^[1]
Some students interpret "0.999…" (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity".^[2]
Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value, since a sequence need not reach its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999…" as meaning the sequence rather than its limit.^[3]
Some students regard 0.999… as having a fixed value which is less than 1 by an infinitely small amount.
Some students believe that the value of a convergent series is an approximation, not the actual value.

which means it's not so odd to believe 0.99999....=1 and the fact of 0.99999....=1 is not a common sense for all students! - Justin545 (talk) 03:56, 25 March 2008 (UTC)

$\frac{x}{x}=1$ , if and only if, $x \ne 0$ so if we have $f(x)=\frac{x}{x}$ , then $f (x)$ is defined on $(-\infty,0),(0,\infty)$ but if we consider $\lim_{x \to p}f(x)$ that is defined for $p \in (-\infty,\infty)$ that includes p=0!! The limit to exists at p=0 because f is defined for all x in an interval (a,b), except at p=0 which is admissible under the definition of the Limit of a function. So the limit exists, despite f not being well-defined for x=0. This limit is evaluated as follows, $\lim_{x \to 0}\frac{x}{x}$ , direct substitution yields 0/0, so instead we define another function which is equal to f(x) on (a,b) except at x=0, and is continuous on [a,b], g(x)=1, since then the limit may be evaluated by direct substitution since g is defined at x=0, the reason this works conceptually, is since g is continuous, then we know that it will answer the question what is f approaching as x approaches 0 (in both directions). And using graphical reasoning, the graph of f is a line with m=0, that that passes through (1,1)

It should be noted that for most practical purposes (especially applications, canceling without keeping the conditions that prevent the original denominator from becoming zero) is really not necessary. Where it can become an issue is in things like the formal proof of the chain rule, and other such formal proofs, where absolutely every detail must be included to ensure accuracy and correctness of theoretical results.

It should also be noted that the proof of 0.999...=1 has no variables, and all of the quantities involved are non-zero, so it is perfectly ok, even under fully formal conditions to cancel with any conditions, since there are no variables to worry about. So the proof above in without error in that regard.

I was surprised that you believed $0.999... \ne 1$ due to the fact that your posts had shown considerable confidence with at least Calculus level mathematics, so I presumed that you would have justified why 0.999...=1 for yourself already. You're absolutely right that you should approach science and especially mathematics in the mind set of a skeptic. I too questioned that result when it was first presented to me, but after completing a proof like one I provided with fractions, I was convinced of the validity of that claim. I was even more skeptical of the chain rule when it was first presented to me, but I eventually studied the proof in my text with sufficient thoroughness to assure myself of it's validity. I decided to start this post since from what I knew of your mindset, I suspected that you would be able to remain rational thorough the discussion, and you have much to my delite and appreciation. A math-wiki (talk) 10:11, 25 March 2008 (UTC)

User talk:Justin545

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[edit] Question about template "helpme"

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[edit] 0.999999.... and 1

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