[gunnar]

Is it just my machine and my Calculator app, or does adding 9.11 to any number result in the addition of not 9.11 but 9.109999 etc.? Any idea why it doesn't just use the whole number?

[Peabo]

yet more WTC censorship rearing its ugly head

[MacGorilla]

Works fine on my computer. if I add 9.11 to 0, I get 9.11

[Developer]

Calculator also thinks that 4^0,5 is 1.
(please note the decimal delimiter)

+9.11 is working for me though.

[mflavin]

Originally posted by gunnar:
Is it just my machine and my Calculator app, or does adding 9.11 to any number result in the addition of not 9.11 but 9.109999 etc.? Any idea why it doesn't just use the whole number?

It isn't just 9.11. This seems to occur with many numbers.

To show this in effect, take 0.1, and add 0.1 to it repeatedly. You end up with 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.799999999999999999, 0.89999999999999, etc.

Not sure what the reasons for this are.

[Developer]

What processor?

[iJed]

This is simply a floting point rounding error. It happens on every processor and is not a problem with your Mac. I would try and explain why this happens but my FP binary is a little rusty. Anyway, Apple should have really taken this into account when they developed the calculator app.

[mchladek]

Hmm ... not getting that problem with my calc.app at all. Like iJed said (hehe, a rhyme ) it's probably just a fp error with certain processors. Send Apple feedback. Their fab() function or its Obj-C or C++ counterpart probably wasn't compiled properly for some processors. Very interesting.

[wataru]

Originally posted by Developer:
Calculator also thinks that 4^0,5 is 1.
(please note the decimal delimiter)

Not on mine.

But adding 0.1 repeatedly did get me 0.799999^

[Developer]

Originally posted by wataru:

Not on mine.

That's why I said "note the decimal delimiter". Set the number formatting to use the comma, and you'll see that bug.

I don't have a problem with repeatedly adding 0.1 - independent of the number formatting. That's why I asked for the processor (for the lack of a better hypotheses).

[gunnar]

Here's a screen shot adding 9.11 to 4 and 7.11 to 4.

Calculator Image

[Developer]

Originally posted by gunnar:
Here's a screen shot adding 9.11 to 4 and 7.11 to 4.

Calculator Image

Ah! Thanks for the clarification. It's wrong in the paper tape, but the result is still correct. Yes, I have that too.

[chabig]

You are right about the 4^0,5 bug. Did you report it?

Chris

[Developer]

Originally posted by chabig:
You are right about the 4^0,5 bug. Did you report it?

No, I didn't.

[3R1C]

This is embarrasing. Are you all saying that our calculator doesnt calculate?
Im mortified to be a mac user at this moment.

[Seb G]

Bad. Very bad. And I thought it was a problem that the calculator was dog slow. Ah well, I don't use it anyway (Pari/GP does all I need).

[villalobos]

Originally posted by Seb G:
Bad. Very bad. And I thought it was a problem that the calculator was dog slow. Ah well, I don't use it anyway (Pari/GP does all I need).

Does that affect only the new calculator in 10.2? It works fine under 10.1.5.

villa

[techweenie1]

The results differ between Advanced and Basic view for me, when in basic view I get 9.11 + 0 = 9.11 but in Advanced I get 9.1099999999999...

[Drizzt]

there is no problem, since mathematically 0.999999999999999999999999999999999999999999999999 9999999999999999999999999999999 (9 to infinity and beyond..) equals 1

Why it shows like that, I don't know, but the reality is that it's correct.

A bug? Yes
A bug that will make you do bad calculations? No

[tooki]

I hope Apple invested more QA effort in core components of the OS than they did in the calculator!

tooki

[Boochie]

I don't know what type of "mathematics" they teach up there in Quebec, but in the rest of Canada (and the world) 0.9999999... does not equal 1.

Let's not confuse mathematics with roundoff error. The reality is that the number being shown in the Calculator display is incorrect, and it's inexcusable. Anybody remember the Pentium math bug from about five years ago?

Originally posted by Drizzt:
there is no problem, since mathematically 0.999999999999999999999999999999999999999999999999 9999999999999999999999999999999 (9 to infinity and beyond..) equals 1

Why it shows like that, I don't know, but the reality is that it's correct.

A bug? Yes
A bug that will make you do bad calculations? No

[brown monk]

Originally posted by Boochie:
I don't know what type of "mathematics" they teach up there in Quebec, but in the rest of Canada (and the world) 0.9999999... does not equal 1.

Mathematics is the same everywhere. It's the language of nature remember .

Anyway, he is correct. 0.9999 (infinitely many 9's) does equal 1. I'm not sure how much math you know so I won't go into a proof of it. Any math major in college who has taken analysis can tell you this.

[Drizzt]

Originally posted by Boochie:
I don't know what type of "mathematics" they teach up there in Quebec, but in the rest of Canada (and the world) 0.9999999... does not equal 1.

Let's not confuse mathematics with roundoff error. The reality is that the number being shown in the Calculator display is incorrect, and it's inexcusable. Anybody remember the Pentium math bug from about five years ago?

I don't remmember how.. but I have a math teacher that prooved us that something like 0.9 where the 9 is preiodical is equal to 1.

It was proven mathematically.. I know it sounds silly but it's true

[Earth Mk. II]

Originally posted by Drizzt:
there is no problem, since mathematically 0.999999999999999999999999999999999999999999999999 9999999999999999999999999999999 (9 to infinity and beyond..) equals 1

Why it shows like that, I don't know, but the reality is that it's correct.

A bug? Yes
A bug that will make you do bad calculations? No

Wouldn't 0.999999... be the limit of x/9 as x approaches 9? Though .9 repeating is very close to 1, .99999... != 1.

I doubt it would throw any common calculations off much, and if you need that level of precision you'd probably have a better calculator already.

report it though... looks annoying.

[JKT]

Originally posted by Boochie:
I don't know what type of "mathematics" they teach up there in Quebec, but in the rest of Canada (and the world) 0.9999999... does not equal 1.

Conceptually you can think of it this way:

1/9 = 0.1 recurring
2/9 = 0.2 recurring
3/9 = 0.3 recurring
...
8/9 = 0.8 recurring
9/9 = 0.9 recurring and as we all know, any x/x = 1

That isn't the real proof, but as everyone else is saying, it is true, 9/9 does indeed = 1

[JLL]

Originally posted by JKT:

Conceptually you can think of it this way:

1/9 = 0.1 recurring
2/9 = 0.2 recurring
3/9 = 0.3 recurring
...
8/9 = 0.8 recurring
9/9 = 0.9 recurring and as we all know, any x/x = 1

That isn't the real proof, but as everyone else is saying, it is true, 9/9 does indeed = 1

x = 0.9 recurring

10x = 9.9 recurring

10x - x = 9

9x = 9

x = 1

[Morpheus]

Claim:
0.999... = 1

Proof:
By geometric series formula: sum{1/10^n , n=1..infinity} = 1/(1-(1/10)) - 1 = 1/9.

Since 0.999... = 9 * sum{1/10^n,n=1..infinity} the claim follows.

(and most of the math before in this thread is bs)

[Drizzt]

Originally posted by Morpheus:
Claim:
0.999... = 1

Proof:
By geometric series formula: sum{1/10^n , n=1..infinity} = 1/(1-(1/10)) - 1 = 1/9.

Since 0.999... = 9 * sum{1/10^n,n=1..infinity} the claim follows.

(and most of the math before in this thread is bs)

Thanks guys.. I knew I wasn't crazy..

Or am I? Are we?

[Boochie]

I stand corrected. Perhaps the Quebecers are on to something.

In any case, what the calculator does shouldn't happen.

Originally posted by Drizzt:


Thanks guys.. I knew I wasn't crazy..

Or am I? Are we?

[ErsatzTom]

It sounds like the reason for this rounding error is that the calculator is doing its calculations in floating point binary (read base 2) math not decimal (base 10... what we use). In binary, there is no direct respresentation of 1/10th, the same way there is no direct representation of 1/3rd in decimal floating point. Instead, for 1/3 we get .333333...

In base 10, after the decimal point you get 1/10th, 1/100th, 1/1000th, etc. In binary, you get 1/2, 1/4, 1/8, 1/16, 1/32, etc. To get 1/10 you need 1/16 + 1/32 + 1/256 + 1/512 + etc. (.00011001100110011...). There is no exact representation.

Having said all that, though, apple really should have used decimal math to produce the results people expect.

[wingdo]

Originally posted by JLL:


x = 0.9 recurring

10x = 9.9 recurring

10x - x = 9

9x = 9

x = 1

If x = 0.99 recurring
then 1x = 0.99 recurring
then 2x = 1.99 recurring
.
.
then 9x = 8.99 recurring

therefore x != 1

It is however, commonly accepted that 0.99 infinite recurring = 1

[wingdo]

Originally posted by JLL:


x = 0.9 recurring

10x = 9.9 recurring

10x - x = 9

9x = 9

x = 1

If x = 0.99 recurring
then 1x = 0.99 recurring
then 2x = 1.99 recurring
.
.
then 9x = 8.99 recurring

therefore x != 1

[Drizzt]

Originally posted by wingdo:


If x = 0.99 recurring
then 1x = 0.99 recurring
then 2x = 1.99 recurring
.
.
then 9x = 8.99 recurring

therefore x != 1

It is however, commonly accepted that 0.99 infinite recurring = 1

Since we know that both statements are true, this is why 8.99 recurring = 9

[ego]

Originally posted by wingdo:


If x = 0.99 recurring
then 1x = 0.99 recurring
then 2x = 1.99 recurring
.
.
then 9x = 8.99 recurring

therefore x != 1

That's BS math you're doing here.

If I remember enough of my maths / topology (13 years ago) the real numbers define a metric space with a canonical distance d(x,y) = |x-y|.

A distance is a function such that (among other properties) d(x,y) = 0 iff x=y

So x != y iff d(x,y) > 0
So x != y iff there exists an eps, eps > 0, d(x,y) >= eps

Since there isn't an eps so that,
|1 - 0.9999... (inf rec)| > eps,
then 1 = 0.9999....
QED

1 and 0.9999... are just alternative representations of the _same_ real number. Real numbers are defined as an extension of the rational numbers to the limits of rational series, 1/0.99... are just two ways to represent the same limit. You could also use 1.00... as a representation (eg as the limit of a serie 1.1, 1.01, 1.001, ...).

Logic, "pure" algebra and topology are the basis for all maths :-)

(Now I admit I don't remember all the details on the definitions of a metric space and a distance, but hopefully I didn't botch it too much).


To come back on topic, yes this "defect" is the result of using a floating-point representation and not rounding aggressively enough when displaying.

[dukope]

i realize this horse is dead, but here's a very simple version of the proof that's easy enough to write on a napkin when it's time to collect on your bet.

1/3 = .33333~
2/3 = .66666~
----------------- +
3/3 = .99999~

[ego]

Originally posted by dukope:
i realize this horse is dead, but here's a very simple version of the proof that's easy enough to write on a napkin when it's time to collect on your bet.

1/3 = .33333~
2/3 = .66666~
----------------- +
3/3 = .99999~

Sorry but this isn't a valid proof. You are using the relations "1/3=.333~" "2/3=.666~" as axioms, plus the fact that you can add two numbers expressed as infinite series.

This is an "ok" explanation for high-school level math. For anybody who had to study more advanced math, you'd have to prove your axioms. And really proving that "1/3=.333~" is equivalent to proving "1=.999~".

Here is the proof:
"1/3" is the limit of two rational series, one finite (1/3), the other infinite (0.3, 0.33, 0.333, ...). For all eps, there is an n, so that d(limit, xn) < eps, so both limits are the same real number.

[TheMosco]

So has anyone reported this stuff to Apple?

[Millennium]

Technically, 0.999999999999999 repeating is equal to 1. This can be proven using a field known as "Discrete Mathematics".

I can't prove it myself, unfortunately. I loathed my Discrete Mathematics courses in college. On one of the final exams, we had to mathematically prove that in any number divisible by 9, the sum of the digits would be divisible by 9. We also had to prove that the positive rational numbers were countable. Yes, this field of mathematics will screw up everything you've ever thought about counting, even.

In any case, however, this is an issue of rounding error. It comes from the nature of expressing floating-point numbers in a binary system, given a finite number of bits in which to do it. It really doesn't work too well. Your computer will compute the number only to a certain number of significant digits, and will then literally make the rest of the digits up. This is true for all processor architectures. Specific mathematics programs, such as Mathematica and bc (a command-line calculator with arbitrary precision), get around this in a wide variety of ways, but evidently the Calculator does not.

Does that make this bug acceptable? Heck no. Apple's sitting on a base of very good calculation code from bc, and they really ought to be using it for Calculator. But at least this should give you an idea of where the bug comes from.

[yukon]

i'll pay you 99.9999~� for every dollar I recieve from you.

thats either called a bad scheme, or really bad taxes...

[MacOSR]

Originally posted by brown monk:


Mathematics is the same everywhere. It's the language of nature remember .

Anyway, he is correct. 0.9999 (infinitely many 9's) does equal 1. I'm not sure how much math you know so I won't go into a proof of it. Any math major in college who has taken analysis can tell you this.

I AM a math major and no, 0.999.. != 1.000..

In calc 2,3 or 4 (don't remember which any more) we proved that 0 = 1 using mathmatics rules and priciples. Does this make it true? Nope. The purpose was to illustrate the fact that our hypothese's and principles aren't always perfect.

Regardless, this bug needs reported to Apple. Unless of course you believe that "technically" 0.999.. = 1.000..; Then there is no bug to report

[Seb G]

Specific mathematics programs, such as Mathematica and bc [...] get around this in a wide variety of ways, but evidently the Calculator does not.

Like rounding rather than truncating what you've got. Takes maybe 5 minutes to implement.

My go at 0.9 recurring = 1:

Code:
                                   n 
0.99999 recurring =     lim       Sum 9*10^(-i)
                     n -> infty   i=1
 
and
 
  lim	f(n) = a  <=>  for all e > 0 there exist an N 
n -> infty             such that, for all n >= N, |f(n)-a| < e.

Let f(n)=the sum above, a=1, and let e>0 be given.
The set N to be the smallest integer bigger than
-log(e) (where the log is to the base 10).

Anything missing here? Now, the new calculator displays 9.109999999999999 where it should display 9.11. By the usual rounding conventions, a number rounded to 9.109999999999999 is strictly less than 9.1099999999999995, so it is still a bug. There's no mathematical proof that shows 9.11 = 9.109999999999999, or that 9.11 < 9.1099999999999995.

I AM a math major and no, 0.999.. != 1.000..

Care to give a proof? Care to give your definition of 0.999..?

[gorgonzola]

FWIW, the real numbers are defined as Cauchy series converging to specific values; you can't use the real number system without saying that 0.999... == 1 and the equivalent for the rest of them.

Also, the proof that JLL gave is perfectly valid:

x = .999...
10x = 9.999...

10x - x = 9x = 9
x = 1

And proving that 0 = 1 is a very famous fallacy, not a proof. Depending on which "proof" you were shown, the solution depends on the fact that you divide by zero at one point, which makes it invalid.

My 2�.

[whee]

This is all standard behavior in any application that follows the IEEE standards for floating-point arithmetic. Getting the 'correct' result from a program following these standards might be a bit of a stretch.

If you'd like higher precision, I suggest using a program that utilizes an arbitrary precision system, such as calc.

[ego]

Originally posted by gorgonzola:

Also, the proof that JLL gave is perfectly valid:

x = .999...
10x = 9.999...

10x - x = 9x = 9
x = 1
My 2�.

It is only valid if you have previously proven that you can add two converging series, scale them by a factor and that applying the same operations on their limit will give you the limit of the new serie. Once you have done this, then sure, go ahead and manipulate them that way.

This is clearly relying on results a bit more advanced than simply relying on the defition of real numbers.

BTW .9999... is really a shortcut notation for lim(n->inf) 1-1/10^n .

So the proof above is:
x = lim 1-1/10^n
10x = 10 . lim 1-1/10^n
-> ability to scale a serie
10x = lim 10(1 - 1/10^n)
= lim (10 - 10/10^n) (9.999...)

abitlity to add/substract two converging series:
10x -x = lim (10 - 10/10^n - (1 - 1/10^n)
= lim (9 -9/10^n)
Again sustracting a constant (ie a converging serie) from the serie gives:
10x-x = 9 - lim(9/10^n)
or 9x = 9 - 0
or x = 1
so lim(1-1/10^n) = 1

Just this to show that the "proof" relies on properties that may seem obvious but aren't. Mathematical proofs that rely on "obviousness" are only acceptable when the "obvious" facts are well-known, proved theorems. Intuition doesn't make a proof.

[ego]

Originally posted by MacOSR:

I AM a math major and no, 0.999.. != 1.000..

Mmm, sorry MacOSR but you won't be making a career in Maths unless you can follow the proofs above.

0.999... = lim(1-1/10^n)
= lim(1)
= 1
= lim(1+1/10^n)
= 1.000...

I may be uncertain about my use of distances/topology in the first "proof" I offered. However reals as extension of rationals to the limits of convering series is basic knowledge for any college-level math student. The only "insight" is in recognizing that 0.999... is just a notation to express the limit of some infinite rational serie.

Does it show so much that my math training was heavily influenced by the Bourbaki movement :-) ?

[ego]

Originally posted by Millennium:
On one of the final exams, we had to mathematically prove that in any number divisible by 9, the sum of the digits would be divisible by 9.

Mmm, I knew how to prove this at some point :-(

We also had to prove that the positive rational numbers were countable. Yes, this field of mathematics will screw up everything you've ever thought about counting, even.

This one is easy. There is a trival injection N->Q+ (f1 : n -> n)
There is also an easy injection Q+->N
(f2: q -> 2^p*3^n (where p/n is the canonical representation of q) (any two numbers prime between themselves would do)

Since there is an inject Q->N and one N->Q, then there exists a bijection between Q+ and N.
So Q+ is countable. QED.

Note that you can use
f2-> 2^(s+1).3^p.5^n
to construct an injection Q->N (q=s.p/n, s=1/-1). So Q is countable also.

[Morpheus]

Originally posted by Millennium:
On one of the final exams, we had to mathematically prove that in any number divisible by 9, the sum of the digits would be divisible by 9.

Proof by induction:

9 has the property, so everything is ok.

if n is divisible by 9 and has the property, then n+9 has it too (3 cases):
(1) If the last digit of n is 0, when adding 9, it becomes 9 so the property is there.

Be n_1 the last digit of n, n_2 the second but last etc.

(2) If n_1 is not 0 and n_2 is not 9, when adding 9, n_1 goes to n_1 -1 and n_2 goes to n_2 +1. So we're set.

(3) If n_1 is not 0 and n_2=...=n_k= 9, these digits become 0 and n_(k+1) becomes n_(k+1)+1 and n_1 goes to n_1 -1 and we're done.

Example: 61899993 + 9 = 61900002

(BTW this and the "Q ist countable" problem have nothing to do with 0.9999..=1)

[Brazuca]

You guys remind me of why I like math.

Have you all discovered TeX yet? and Equation Editor? You could all be writing your proofs and posting them as a small pdf (the size of your equation) which makes them a lot easier to see.

http://www.esm.psu.edu/mac-tex/

[Seb G]

Brazuka, here you go:

\[0.99999\dots = \lim_{n\rightarrow \infty} \left(\sum_{i=1}^{n} 9 \times 10^{-i}\right)\] and \begin{eqnarray*} \lim_{n\rightarrow \infty} f(n) = a & \iff & \mbox{for all }\epsilon > 0 \mbox{ there exist an } N \in \mathbb{Z}\\ & & \mbox{such that, for all } n \geq N, |f(n)-a| < \epsilon. \end{eqnarray*} Let $f(n) = \sum_{i=1}^{n} 9 \times 10^{-i}$, $a=1$, and let $\epsilon>0$ be given. Set $N = \lceil -\log_{10}(\epsilon) \rceil$.

OR



Happy?

[Ghoser777]

I wrote an app called MathRPN and it has the same problem... stupid IEEE. Essentially, to fix the problem you have write all your own addition, subtraction, multiplication, division and create your one number class blah blah blah. Aka, it's a pain. It's much easier to just do built in math (aka num1 + num2), and I'm betting apple developers aren't going to be missing much sleep over something as simple as a round off error. Should they fix it? Probably. Is it a big issue? Nah.

4^(0.5) == 2 on my machine (G3), don't know how you got 1.

And yes, as a math major, I can verify that 0.9999... = 1. I believe in one of my class a while back we proved that every number has one decimal form except for those ending in repeating 9's, in which case there are two forms. We used the simple argument of 10x = 9.9999... as stated above, but of course we proved that these types of operations on infinite sums was valid first.

Matt Fahrenbacher