bonanova 77 Report post Posted July 5, 2012 Title is fav quote by Kipling, on equality. How about you on the same subject, but mathematically, not philosophically: Can two numbers x and y written in decimal expansion differ in every decimal place, yet be equal? Share this post Link to post Share on other sites

0 ujjagrawal 5 Report post Posted July 5, 2012 0.9999999999999999999999... 1.0000000000000000000000... almost equal Share this post Link to post Share on other sites

0 envy 2 Report post Posted July 5, 2012 0.999...... 1.000...... are not "almost equal" ... they are EQUAL proof 1)) 1/3 = 0.333...... 3*1/3 = 0.333*3 but also 3*1/3 = 1 0.999.... = 1 proof 2)) let x = 0.999..... 10x = 9.999.... subtract the two equations.... 9x = 9 ... or x=1, but assumption 1, x=0.9999 ..... thus 0.999.... = 1 ....or if you like, 1.0000..... 2 Share this post Link to post Share on other sites

0 ujjagrawal 5 Report post Posted July 5, 2012 0.999...... 1.000...... are not "almost equal" ... they are EQUAL proof 1)) 1/3 = 0.333...... 3*1/3 = 0.333*3 but also 3*1/3 = 1 0.999.... = 1 proof 2)) let x = 0.999..... 10x = 9.999.... subtract the two equations.... 9x = 9 ... or x=1, but assumption 1, x=0.9999 ..... thus 0.999.... = 1 ....or if you like, 1.0000..... Thanks for the proofs... Even though I will stick too... they are not "EQUAL" but almost equal ... the difference may be infinitesimally small but it exists... Share this post Link to post Share on other sites

0 witzar 18 Report post Posted July 5, 2012 the difference may be infinitesimally small but it exists... 1. The difference of two real numbers numbers always exists and is also a real number. 2. A real number cannot be "infinitesimally small". 3. Please don't answer "thanks for the explanation, but I'll stick to... " 1 Share this post Link to post Share on other sites

0 dpalmer 1 Report post Posted July 5, 2012 I can think of one example: 1.000... and 0.9999... are equal but differ in every decimal place! This is the only one that comes to mind for me. 1 Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted July 5, 2012 1. The difference of two real numbers numbers always exists and is also a real number. 2. A real number cannot be "infinitesimally small". 3. Please don't answer "thanks for the explanation, but I'll stick to... " To what class of numbers do those that are infinitesimally small belong? Share this post Link to post Share on other sites

0 jim 1 Report post Posted July 5, 2012 In standard mathematics 1.000... and 0.999... are considered to be two different ways to express exactly the same number. Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted July 5, 2012 0.9999999999999999999999... 1.0000000000000000000000... almost equal Let x = 0.999... y = 1.000... and z = y - x. [1] describe your understanding of the notation "..." [2] describe z. Share this post Link to post Share on other sites

0 jim 1 Report post Posted July 5, 2012 There are infiniesinally small real numbers in the non-standard real number system of Abraham Robinson which he used in his non-standard analysis. That system allows a form of calculus where the derivative is the ratio of two infinitesimals. Share this post Link to post Share on other sites

0 jim 1 Report post Posted July 5, 2012 If x=0.999... and y=1.000... then x, if it exists, is a number that differs from a finite sequence of 0.99...99 by an amount that is smaller than any positive number if we carry the expansion out far enough. For example, it will be less than one millionth if we go out 7 decimal places or more. The difference y-x, if it exists, is an approximation for 0.00...001 that has an error that can be made smaller than any positive nuimber if we go to a large enough number of decimal places This is the definition in standard math. If you wanted to define two real numbers as equal only if the rational approximations used to define the numbers were eventually identical you could do so and get two distinct numbers infinitesimally close together--but that is not the definition that is used in standard math. Share this post Link to post Share on other sites

0 Yoruichi-san 116 Report post Posted July 5, 2012 This is getting surreal... ...or perhaps you're trying to push us to our limits ;P (On a totally unrelated and being a smart@z note in answer to the OP...scientific notation, anyone? ) Share this post Link to post Share on other sites

0 ujjagrawal 5 Report post Posted July 6, 2012 Let x = 0.999... y = 1.000... and z = y - x. [1] describe your understanding of the notation "..." [2] describe z. In arithmetic, recurring decimal notation "..." is a way to represent rational numbers (e.g. 1/3= 0.3333...) because of ability to write repeating number infinite times. 1 being a whole number need not to be represented using recurring decimal notation like 0.9999... instead can be simply represented as 1.0 if required to be in decimal form. Recurring notation "..." was just used in my answer to represent repeating 9's and 0's in order to do a decimal place by place comparison. Take this example, two parallel lines meets in infinity that means they never meet. 1.0000.... 0.9999.... are equal to each other only when we goes until infinity. Like there is always some distance between two parallel lines, there is always some difference between the above two numbers. Yes, it is least important to us most of the times (like here) and hence we are accustomed to ignore it. Also, I have no issues accepting the two numbers as equal if it solves this problem for me. Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted July 6, 2012 [1] 1.0000.... 0.9999.... are equal to each other only when we goes until infinity. [2] there is always some difference between the above two numbers. [1] And so we have it ... two numbers written in decimal, differing in every decimal place, nevertheless equal. Why? Because the "..." notation is a symbol that says, in your words, "go to infinity." That's the answer to the OP. [2] There is never any difference between the above two numbers. You, as jim, refer to 0.99... as if it were one of the terms of the sequence 0.9, 0.99, 0.999, 0.9999 .... It is not one of the terms of that sequence; it is the limit of that sequence. As such, it does not become something, it does not change, it is not approximately anything. It is -- they are both -- precisely, exactly, and forever identical with, unity. 1 Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted July 6, 2012 If x=0.999... and y=1.000... then x, if it exists, is a number that differs from a finite sequence of 0.99...99 by an amount that is smaller than any positive number if we carry the expansion out far enough. For example, it will be less than one millionth if we go out 7 decimal places or more. The difference y-x, if it exists, is an approximation for 0.00...001 that has an error that can be made smaller than any positive nuimber if we go to a large enough number of decimal places This is the definition in standard math. If you wanted to define two real numbers as equal only if the rational approximations used to define the numbers were eventually identical you could do so and get two distinct numbers infinitesimally close together--but that is not the definition that is used in standard math. You discuss the concept of limit. But you alternate your focus between the limit and the entity that possesses the limit. And your conclusion suffers from that blurred distinction. The terms of a convergent sequence [partial sums of a convergent infinite series are one example] need not ever equal, and usually are always different from, their limit. That is a very different statement from saying the limit does not exist, or is in any way imprecisely known. Saying that in your terminology, it must be recognized that the limit does not "eventually" become some value. To say that 0.99... differs even infinitesimally from 1.00... Is to deny the convergence of the sequence 0.9, 0.99, 0.999, ... Mathematically, yes, in standard mathematics, .99... is defined as the limit [not one of the terms] of the sequence a_0 = .9, a_1 = .99, a_2 = .999, a_3 = .9999, ... if it exists. Since that limit does exist, and is equal to 1, .99... == 1. Share this post Link to post Share on other sites

0 EventHorizon 7 Report post Posted July 6, 2012 First, an excerpt from the wikipedia page for "decimal." Consider those rational numbers which have only the factors 2 and 5 in the denominator, i.e., which can be written as p/(2^{a}5^{b}). In this case there is a terminating decimal representation. For instance, 1/1 = 1, 1/2 = 0.5, 3/5 = 0.6, 3/25 = 0.12 and 1306/1250 = 1.0448. Such numbers are the only real numbers which do not have a unique decimal representation, as they can also be written as a representation that has a recurring 9, for instance 1 = 0.99999…, 1/2 = 0.499999…, etc. The number 0 = 0/1 is special in that it has no representation with recurring 9. This leaves the irrational numbers. They also have unique infinite decimal representations, and can be characterised as the numbers whose decimal representations neither terminate nor recur. So in general the decimal representation is unique, if one excludes representations that end in a recurring 9. The same trichotomy holds for other base-n positional numeral systems:Terminating representation: rational where the denominator divides some n^{k}Recurring representation: other rationalNon-terminating, non-recurring representation: irrational A version of this even holds for irrational-base numeration systems, such as golden mean base representation. And another excerpt from the page for "0.999..." Every nonzero, terminating decimal has an equal twin representation with trailing 9s, such as 8.32 and 8.31999... The terminating decimal representation is almost always preferred, contributing to the misconception that it is the only representation. The same phenomenon occurs in all other bases or in any similar representation of the real numbers. The equality of 0.999... and 1 is closely related to the absence of nonzero infinitesimals in the real number system, the most commonly used system in mathematical analysis. Some alternative number systems, such as the hyperreals, do contain nonzero infinitesimals. In most such number systems, the standard interpretation of the expression 0.999... makes it equal to 1, but in some of these number systems, the symbol "0.999..." admits other interpretations that contain infinitely many 9s while falling infinitesimally short of 1. So since decimal places are the digits after the decimal point, it seems there is one pair of equal decimal expansion representations for each positive integer and each negative integer. And this is exactly the set of all the numbers that fit the OP. Any other terminating decimal expansion will have the pair of representations equal at one or more decimal places. It is interesting that 0 fails in this regard. The method of decimal representation we use has no way of approaching 0 with repeating 9's due to the fact that adding value in a decimal place always increases the distance from 0. If you change the OP to require a different digit in each place (not just decimal place, and can't simply not have a value explicitely put in a place (ie, both 0's in hundreds place of 34 and 67)), you find that at least one number needs to be infinite. You obviously cannot have an infinite number equal to a finite one. But what about two infinite numbers? Sure, if you subtract the decimal representations of two infinite numbers created by an infinite decimal expansion, you could end up with a finite decimal expansion. But it seems it isn't possible in the case where the digits are different at every position. If you try the "trailing 9's" approach you end up with ....99999.9999... = ...00000.0000..., but the number on the right is 0, while the other is infinite. But are all infinite numbers created by this decimal expansion method equal anyway? For instance, is ...22222.2222... = ...11111.1111... ? You can keep subtracting 1 from both and never reach a negative number (or even a finite number). I'm thinking every infinite number created using this method of decimal expansion is equal (and equal to aleph-naught), and so the subtraction of the decimal expansions doesn't really apply since all are the same number/concept... and the difference is simply undefined. Sure, there are higher degrees of infinity, but they are essentially 2^(one of these infinite number) or more in "value." Hyperreals sound interesting. I can't, at the moment, understand how there can be a non-zero infinitesimal. I'll quickly look that up. It seems that they handle it essentially like the different degrees of infinity, and just call all the infinitesimals nonzero. It seems to me like they are kept as two separate concepts, and when you order values you order them first by its "standard part" and then by the infinitesimals included. Anyone have thoughts on any of this or disagree vehemently on a point? Share this post Link to post Share on other sites

0 Yoruichi-san 116 Report post Posted July 6, 2012 First, an excerpt from the wikipedia page for "decimal." And another excerpt from the page for "0.999..." So since decimal places are the digits after the decimal point, it seems there is one pair of equal decimal expansion representations for each positive integer and each negative integer. And this is exactly the set of all the numbers that fit the OP. Any other terminating decimal expansion will have the pair of representations equal at one or more decimal places. It is interesting that 0 fails in this regard. The method of decimal representation we use has no way of approaching 0 with repeating 9's due to the fact that adding value in a decimal place always increases the distance from 0. If you change the OP to require a different digit in each place (not just decimal place, and can't simply not have a value explicitely put in a place (ie, both 0's in hundreds place of 34 and 67)), you find that at least one number needs to be infinite. You obviously cannot have an infinite number equal to a finite one. But what about two infinite numbers? Sure, if you subtract the decimal representations of two infinite numbers created by an infinite decimal expansion, you could end up with a finite decimal expansion. But it seems it isn't possible in the case where the digits are different at every position. If you try the "trailing 9's" approach you end up with ....99999.9999... = ...00000.0000..., but the number on the right is 0, while the other is infinite. But are all infinite numbers created by this decimal expansion method equal anyway? For instance, is ...22222.2222... = ...11111.1111... ? You can keep subtracting 1 from both and never reach a negative number (or even a finite number). I'm thinking every infinite number created using this method of decimal expansion is equal (and equal to aleph-naught), and so the subtraction of the decimal expansions doesn't really apply since all are the same number/concept... and the difference is simply undefined. Sure, there are higher degrees of infinity, but they are essentially 2^(one of these infinite number) or more in "value." Hyperreals sound interesting. I can't, at the moment, understand how there can be a non-zero infinitesimal. I'll quickly look that up. It seems that they handle it essentially like the different degrees of infinity, and just call all the infinitesimals nonzero. It seems to me like they are kept as two separate concepts, and when you order values you order them first by its "standard part" and then by the infinitesimals included. Anyone have thoughts on any of this or disagree vehemently on a point? You should also take a look at surreal numbers, which is somewhat related. It's a notation that involves taking the numbers "between" other numbers and is a very interesting way of looking at infinitesimals . Share this post Link to post Share on other sites

Title is fav quote by Kipling, on equality.

How about you on the same subject, but mathematically, not philosophically:

Can two numbers

xandywritten in decimal expansion differ in every decimal place, yet be equal?## Share this post

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