Archimedes
Member
True in base 10, but there's nothing special about base 10. In base 3, 1/3 is equal to 0.1 with no infinitely repeated long division.
-
Features
Paradox!
- Thread starter beero1000
- Start date
10 X 0.999... is not defined in algebra. You can not algebraically add, subtract, multiply, and divide infinites.
if a variable results in 0.9999... you must pick a finite truncation point. In numerical methods the truncation point can affect error propagation.
x = .999
10X = 9.99
10X - x = 9.991
x(10 - 1) = 9.991
x = .991
In a real world calculation the number of digits for .999.. is selected to maintain the required number of significant digits in the results. in this case if all that is needed is auacury to xx.x it works.
A baker is contracted to make 9 cakes for a party. His trainee accidentally drops 6 cakes loading them into the delivery van. What is the ratio of delivered cakes to ordered cakes?
.333...
A contractor brings 300 bricks to a worksite. 150 bricks are used to build a barbecue in his client's back yard. Expressed in base 5 what is the ratio of remaining bricks to the starting amount of bricks?
.222...
I can't disabuse you of any neopythagorean hangups you may have about math, but the limitations of your chosen applied math field do not carry to all math.
If I define x to equal .999... or .333... there's no ambiguity about what the value is, which is distinct from identifying a finite number of digits as a repeating decimal. If the value is defined or calculated then I have no qualms about using the operations I have. Just like the conditional convergence example using the harmonic series.
Like Archimedes said - at noon. That is to say, at step \(\omega\)
10 x 0.999... is not defined in algebra as you attempted to do. You can make up any definitions you like, but your example resulting in 0.999.. = 1 is not algebraically correct. Quantitatively 0.999... has no meaning.
algebraically correct-
i = .999, .9999, .99999..
j = 10 x.
k = 10x -x
l = k/10
Defines l = .991, .9991. .99991....
Bases? hex and binary are second nature.
You still have not have provided a quantitative answer to beero's problem with your infinites.. either that or say there is no finite answer.
1/3 = .333... ok, but so what? How does that apply to the op?
(1/3) * 5 = ?
.333... x 5 has no value. .3 x 5 = 1.5, .33 x 5 = 1.65, .333 x 5 = 1.665
So I initially thought we had a difference of opinion on how (or whether) infinite sets could be described and manipulated, but I'm aghast at this system of math you've invented and are advocating. There is absolutely no issue with assigning an algebraic variable with a repeating decimal value (that is to say specific rational numbers). I'm not even able to fathom what axioms your system of math is based upon, and I don't think a term exists to describe this school of thought with regards to permitted mathematical operations. The operations I've described are done every day in pre-algebra classes around the world. No mathematician alive or dead, I'd wager, would side with you on this point.
At least the Pythagoreans just threw Hippasus overboard. You've nuked the whole boat.
Archimedes
Member
But i isn't equal to 0.9 or 0.99 or 0.999 or any of the values you've given. It's explicitly defined as having an infinitely long string of 9s after the 0. part of the number.
Therefore the 1 at the end you seem to object to is entirely irrelevant as it doesn't exist. What does it mean for a 1 to exist at the "end" of an number with an infinitely long sequence of 9s after the decimal point? There is no 1. The 9s never end therefore the 1 doesn't exist so what's the problem?
The "1" at the "end" is surely non-existent.
Disclaimer: I am not a logician or a mathematician and have no formal training in this kind of thing so I may be talking through my hoop and actually acknowledge that I am, but I still like the intellectual exercise taking part in subjects like this.
This actually isn't even an issue with computing unless you're only employing naive programmers.
Symbolic computation packages like Mathematica and Mathcad have no issue working with repeating decimals.
EDIT: RPN Scientific does it on my Windows phone! 3[enter]1[enter]3[enter][divide][multiply] shows 1. Not bad for a free application.
One more EDIT: Found an infinitely long video on converting repeating decimals to fractions using operations not defined in algebra
https://www.youtube.com/watch?v=FPNhCVZlbJs
Symbolic computation packages like Mathematica and Mathcad have no issue working with repeating decimals.
EDIT: RPN Scientific does it on my Windows phone! 3[enter]1[enter]3[enter][divide][multiply] shows 1. Not bad for a free application.
One more EDIT: Found an infinitely long video on converting repeating decimals to fractions using operations not defined in algebra
https://www.youtube.com/watch?v=FPNhCVZlbJs
Last edited:
I would say that if you go at 1/2 intervals of time you would get to planck time barrier, then what?Start with an empty vase at 11:00am. At 11:30am, place 10 balls in the vase, and remove one ball. At 11:45am, place 10 more balls in the vase and remove one ball. Continue repeating the procedure of adding 10 balls and removing one, but at each step reduce the time between steps by half. Question: At noon, how many balls are in the vase?
My answer: At noon, there are exactly 42 balls in the vase.
Of course, there are many other possible answers; this is a paradox, after all. Thoughts?
What does RPN have to do with it? That affects nothing. You may be looking at rounding and truncatioinerrors in a hand held app due to small variable size or the floatingpoint algorithms. .
Yes math tools can do symbolicmanipulation but when it comes to actually performing a calculationan infinity an only be approximated.
Math packages today generally meet thefloating point standard. It doesn't matter how it is done in software, you are limited by the number of bytes or precision allocated to the variables. .999... computationally does not exist.
Symbiotically a tool may accept the infinity symbol, but in the background there will be code to reduceit to a computational form. Yes, you cam convert repeating decimals to other rprentaions but you can not algebraically calculate 10 X 8 .999...
When you enter a number like .999 it isstored similar to scientific notation in binary as a mantissa andexponent. In software .999 is a text string symbol. The process of going from floating point digital binary storage to a graphicdisplay or text form of a number is called unpacking ordenormalization. it is what the function printf() does in C.
Your decimal numbers are stored and processed algebraically as binary integers. Decimal multiplication, division, addition, and subtraction are done as binary integers. It is converted to base 10 representation for the user interface.
in a pocket calculator or an app or amath tool when you do 1/3 there is always a truncation limit and arounding depending on the rounding convention used when you make acalculation.
In digital computation 10* 0.999... isanother paradox or a singularity. It would take infinite time andinfinite memory to resolve.
http://en.wikipedia.org/wiki/Floating_point
http://en.wikipedia.org/wiki/IEEE_754-2008
'…..' Is a symbolic notation for aninfinity. ∞
Sameas limit x→ 0 1/x. We say the limit is infinity and give it theinfinity symbol, but it is not a number. For any finite x 1/x has afinite value.
Like wise .999... is not a number, anyfinite number of digits is. 1/0 is not defined, 10* .999.. is notdefined.
I am not a mathematician, I am anengineer who has done numerical analysis and computations.
We say that (limit x→ 0 : 1/x) is infinity because 1/x grows beyond any limits. Usually formulated as: for any value y there is a value e such that 1/e > y
That is: for ANY value y you can find a value x that makes 1/x greater than that y.
That is why it is said to have no limit and is infinit.
Why? Do you also mean that pi isnt a number?
Pi has also a infinite number of decimals.
Nah. An engineer should know his tools. You obviously doesnt know real numbers.
RPN has to do with it, in that it's in the name of the specific calculator app (RPN Scientific) on my phone where the programmers were intelligent enough to not cause a truncation error when the user specified an exact rational value. Dividing one by three displays '.33' and multiplying that result by ten displays '3.33'. Multiplying that result by three displays '10'. Dividing one by nine displays '.11', multiplying by three displays '.33', and multiplying by three again displays '1'.
Computers certainly implement floats and doubles, but it's ultimately up to the programmer to specify how to do the math and precision errors when working with precise values (as opposed to say the reals) only crop up due to imprecise thinking. Limitations of the IEEE floating point standard don't dictate what mathematical operations are valid ones - and I'm fully capable of calculating 10 * 8.999... in my head or on a piece of paper. The IEEE standard does not define algebra.
.999... and .333... specifies a number, not a process.
Hee Hee.
Now you are just playing games. Read what I wrote. PI can not be represented as a finite number, to use it you must pcick a finite truncation. When I do quick mental approximations I use 3. in Scilab the constant PI is 3.1415927
lim x -> 0 1/x is infinity or goes to infinity as a limit.
lim x -> x -> inf 1/x is zero or is approached as a limit.
In both cases 0 and inf are limits that are asymptotically approached but never reached.
lim x -> inf [5/(1 + (1/x))] is 5. We take it as 5 computationally knowing it is not exactly 5 but an asymptote.
A basic equation for a simple operational amplifier circuit..
Gain = [1/((1/A) + Ri/Rf))] where A is amp open loop gain and Ri,Rf are resistors. We say in the limit as A -> inf gain goes to Rf/Ri.
How you say it in words doesn't change anything. The soothing beauty of math vs philosophy.
In your calculator infinity is an overflow condition in a variable.
RPN has to do with it, in that it's in the name of the specific calculator app (RPN Scientific) on my phone where the programmers were intelligent enough to not cause a truncation error when the user specified an exact rational value. Dividing one by three displays '.33' and multiplying that result by ten displays '3.33'. Multiplying that result by three displays '10'. Dividing one by nine displays '.11', multiplying by three displays '.33', and multiplying by three again displays '1'.
Computers certainly implement floats and doubles, but it's ultimately up to the programmer to specify how to do the math and precision errors when working with precise values (as opposed to say the reals) only crop up due to imprecise thinking. Limitations of the IEEE floating point standard don't dictate what mathematical operations are valid ones - and I'm fully capable of calculating 10 * 8.999... in my head or on a piece of paper. The IEEE standard does not define algebra.
.999... and .333... specifies a number, not a process.
I have been using HP RPN calculators since th 80s.
. http://www8.hp.com/us/en/products/calculators/product-detail.html?oid=7029051#!tab=features
RPN uses a a stack to order the sequence of calculations, RPN in itself does not affect truncation or accuracy. The actual floating point operations are the same PPN or otherwise.
It has to do with algebra and finite arithmetic. Regardless of calculator accuracy will change based on order of operations. I'd have to get my book out for examples, I refer you to Knuth's Semi Numerical Algorithms.
A calculator does not know whether a number is a repeating decimal or an integer. It uses rules for rounding and truncation.
1/3 displays .33? hmmmm.
From Scilab. It is riounding up.
->x = 1/3
x =
0.3333333
-->y = .3 * x
y =
0.1
Direct entry of .333333. Same result.
-->x = .3333333
x =
0.3333333 --- 7 digits entered by hand
-->y = x * .3
y =
0.1
Entering less than an overflow and theresult is .99999. In your calculator entering 1/3 forces a variableoverflow and a rounding. In writing numerical algorithms or evendoing a series of sequential hand calculations with a calculator youhave to have guard digits to prevent the error you are demonstrating. Your calculator does not know if you entered .33333333or if it was result of 1/3. As I showed if I enter .333333 to the pointof overflow and multiply by .3 I get 1 which is Incorrect.
I wet through all this 30 years agowhen I learned numerical methods. It was the reason why I got myfirst PC. I started out writing my own math routine for work....andfun.
-->x = .333333
x =
0.333333 ---6 digits entered by hand
-->y = .3 * x
y =
0.0999999
Last edited:
beero1000
Veteran Member
Jesus Christ. I leave for a day and it gets so much worse. 
If this is the logic underlying peoples' number concept, then it's no wonder I was having so much trouble getting them to grasp the paradox. Nearly all of mathematics is vitally dependent on precise number concepts. Disallowing completeness basically means you reject the real numbers, and that means you essentially reject all of calculus and more.
Part of me wants to create a tutorial thread on convergence and number systems, and the rest of me thinks that it won't help.
If this is the logic underlying peoples' number concept, then it's no wonder I was having so much trouble getting them to grasp the paradox. Nearly all of mathematics is vitally dependent on precise number concepts. Disallowing completeness basically means you reject the real numbers, and that means you essentially reject all of calculus and more. Part of me wants to create a tutorial thread on convergence and number systems, and the rest of me thinks that it won't help.
Pi is less than 4, and greater than 3. Thus it is a finite number.
What you talk about is a truncated floating point representation.
Wrong. Using "limit" means "the asymptotical value" so lim x -> inf [5/(1 + (1/x))] is exactly 5.
A basic equation for a simple operational amplifier circuit..
The binary representation of numbers as IEEE real numbers is not in any way a more real representation than algebraic representations.
Rounding errors a little more clearly.
First, 6 decimal places and a correct answer.
-->1.333333*.3
ans =
0.3999999
Second, 7 decimal places and a rounding error..
-->1.3333333 * .3
ans =
0.4000000
Entering 1/3 in your calculator is the second condition above.
First, 6 decimal places and a correct answer.
-->1.333333*.3
ans =
0.3999999
Second, 7 decimal places and a rounding error..
-->1.3333333 * .3
ans =
0.4000000
Entering 1/3 in your calculator is the second condition above.
Jesus Christ. I leave for a day and it gets so much worse.
Part of me wants to create a tutorial thread on convergence and number systems, and the rest of me thinks that it won't help.
Math reduces to counting, addition, and subtraction with multiplication as multiple additions and division multiple subtractions.
You still have not shown how you get 42 balls in the vase or refuted the simulations I have shown.
Please do that before presuming to lecture on numerical analysis.
PI can not be expressed as a finite number, it is an irrational number.
So .333333... is correctly called an irrational number not an infinity. Still does not change the actual math.
We take the asymptote to be 5.0 knowing in reality it is not exactly 5 for any finite value of x. The debate is philosophical as 5 would be used regardless.
beero1000
Veteran Member
Jesus Christ. I leave for a day and it gets so much worse.
Part of me wants to create a tutorial thread on convergence and number systems, and the rest of me thinks that it won't help.
Math reduces to counting, addition, and subtraction with multiplication as multiple additions and division multiple subtractions.
You still have not shown how you get 42 balls in the vase or refuted the simulations I have shown.
Please do that before presuming to lecture on numerical analysis.
Oh, I wouldn't presume to lecture you. Like I said, I don't really think it would help. If you ever want to learn though, I'd be happy to teach you.
:laughing-smiley-014
So have I. I also spent the lion's share of my professional life as a programmer. You didn't by chance work on the Patriot missile system, did you?
Steve, c'mon down man. We're all worried about you.
Archimedes
Member
I'd definitely be interested in such a thread.Part of me wants to create a tutorial thread on convergence and number systems, and the rest of me thinks that it won't help.
Archimedes
Member
No. 0.33333.... is correctly called a rational number as it can be expressed as the ratio of one integer to another, i.e. 1/3