... If and when you have a rational, logical and common sense, relatively simple explanation please share with me. I've not seen that from you ...r6
I and others, have PROVEN 0.9999... = 1/1 Perhaps mine is the most simple as does not use infinity or ANY of the general math operations as I define or derive every step used.
Here is link to full, old proof, I gave at
http://www.sciforums.com/showthread...ophy-of-Math&p=3136195&viewfull=1#post3136195
but the essence of it is:
... I first illustrate, several of the infinite numbers of examples, of true statements concerning terminating or infinitely Repeating Decimals. (And then a procedure for finding the Rational Fraction that equals to ANY given Repeating Decimal, RD):
1/3 =0.333333.... and 1/1 = 0.99999.... are rational fraction numbers with a "repeat length" of 1 in their equivalent decimal versions.
12/99 = 0.12121212... and 19/99 = 0.1919191919... and 34/99 = 0.343434... are rational numbers with a "repeat length" of 2 in their equivalent decimal versions.
In general, any integer less than 99 divided by (and not a factor of) 99 will produce a decimal repeating with length 2. Some of the factors will too. For example 3/99 = 03/99 = 0.03030303... does; but not 11 or 33. I.e. 11/99 =0.111... and 33/99 = 0.3333333333... In ALL cases with 99 as the denominator, the decimal repeats two numbers in blocks of two) is still true, as well as in "blocks" of one for some cases as bold type helps you see.
Likewise any integer less than 999 divided by 999 will be a decimal fraction with repeat length not more than 3 and always will repeat in blocks of 3, but for somecases, like 333 /999 = 1/3 the least long repeat block is less than 3. Check with your calculator if you like. Etc. For example, 678 /999 = 0.678,678,678, .... and that is slightly larger than 678 /1000, which equals 0.678 and should given you a hint of the proof to come.
However, any integer divided by a factor of the number base (1, 2 & 5 for base 10) or any product of these factors (like 4, 16, 2^n, 5 or 5^m, {2^n x 5^m} ) will terminate, not repeat. For example 17 /(1x4x5) = 0.85
The proof I and others have given that 1 = 9/9 = 0.99999.... is just particular case of the fact ALL rational fractions like a/b or a/9 (both a & b being integers and a < b) are equal to an infinitely repeating decimal (if they are not a finite decimal when b is a factor or product of factors of the base).
For example, the general proof of this goes like:
Rational Decimal, RD = 0.abcdefg abcdefg abcdefg .... Where each letter is one from the set (0,1,2...8,9) and the spaces are just to make it easier to see the repeat length in this case is 7.
Now for this repeat length 7 case, moving the decimal point 7 spaces to the right is not a multiply operation, but a notional change with the same effect on meaning as multiplying RD by 10,000,000. I. e. 10,000,000 RD = a,bcd,efg . abcdefg abcdefg ... Is a 2nd equation with comas for easy reading the integer part.
Now, after noting (10,000,000 - 1) = 9,999,999 and subtracting the first equation from the second, we have:
a,bcd,efg = 9,999,999 x RD. Note 9,999,999 certainly is not zero so we can divide by it to get: The Rational Fraction, RF = RD = a,bcd,efg / 9,999,999 I. e. the rational fraction of two integers exactly equal to the infinitely long repeating (with repeat length =7 in this case) decimal, RD.
Now lets become less general and consider just one of the repeat length = 7 cases. I. e. have a=b=c=d=e=f=g = 9 and recall RD was DEFINED as 0.abcdefg...so is now in this less general RD = 0.9,999,999,... and from green part of line above, The Rational Fraction which equals RD is 9,999,999 / 9,999,999, which reduces to the fraction 1/1 which is unity as the numerator is identical with the non-zero denominator. I.e. the least numerator rational fraction equal to 0.999,999... is 1/1.
By exactly the same procedure the RD = 0.123,123,123,.... is a case with repeat length of 3, can be shown be equal to the RF = 41/333 (as your calculator will show, as best as it can, if used to divide 41 by 333.
To prove this one moves the decimal point of the RD 3 places to the right get: RD' which is 1000 times larger than RD. I.e. RD' = 123.123123123... as the 2nd equation and then subtract the first from it to get after division by (1000 -1): RD = 123 /999, which reduces to 41/333 as the RF = 0.123123123...
The objectors to 1 =0.99999 need not only to give extraordinary proof for their objection but also need to explain why ONLY ONE of the other infinite number of successes of this procedure fails to produce the rational fraction that is exactly equal to the infinite Repeating Decimal , RD.
The objectors tend to fall into two classes: idiots and those not fully understanding the meaning of the "Bases, places and decimal point" notational system, so I will explain that below (with numbered logical steps from the starting definition for easy reference if you think there is any error in the logic or resulting proof than 1/1 = 0.999...):
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A shorter summary of the logical development first:
I started with definition of "line segment" quickly specialized to one between 0 and 1, to establish a scale. Then joined two of these unit length line segments to make line segment of length twice as long and introduced the name "2" both as the name of the not-zero, new end of the longer new line segment and as the sum of two integers. I said 1 + 1 = 2 and then that 2 + 1 = 3. etc. up to 8 + 1 = 9
I did thus define addition of integers, and noted there was no largest one N as N+1 was larger. I also noted we have no symbol for what 9 +1 is equal too.* These two facts forced the introduction of a line (the "number line") with no ends to that 1D item list and the introduction of the place and decimal systems (with B as base) so I could tell that 9 + 1 = 10 in the system in common use by people, and begin to discuss the meaning of the "Base, place, and decimal point" notation systems.
* The hexadecimal (base =16) notation system does have a symbol for 9+1. It is "a" I.e. in that base system, 9+1 = a.
Note my proof, now well founded on logical extension of the defined "line segment" (with length 1), does NOT use ANY of the normal GENERAL math operation (+ , - , x or /). It needs no algorithms for these operation, except adding and subtracting lengths on the number line. I do write the notation commonly used for fractions, but even there only with integers for a & b in a/b but never describe how the indicated division would be done as the proof has no need of that.
Further more, even when in the proof I do use my limited well defined "subtraction" it is only using powers of 10 (like 1000, when the repeat length of the RD is 3) and one step back of magnitude unity towards 0 on the number line. That is the proof does need "integer subtraction" or reduction by 1. For example fact that 1000 - 1 = 999 is a location on the number line one unit closer to 0 than 1000 is. It does also use the algebraic fact that (x+a) - (y+a) = x-y. I.e. two points on the number line that are both greater than x or y by the same amount (a in this case), have length difference or separation of x-y where x > y as I never use or need negative numbers in the proof.
PS: Not only do I avoid use of all of general math's algorithms for operations including any limiting process, but I have no need to try to defined "infinity" in my proof! The closest to that I come is to use a line with no ends, called the number line, and do all movement on it by unity steps for some starting point on the number line. (My additions and subtractions are not general algorithms, but deal only with integer steps and the subtractions I use in the proof ALWAYS have positive results like 99, or 999 or 9999 etc. but not 999...)
Now the full text of the logical development:
(1) A “line segment” is a math defined term, which is 1 dimensional (1D etc here after) with two ends. They can be called “a” & “b” or “0” & “1” for more convenience as then one can say “length' of the line segment is 1.
(2) If a second equal to the 0 to 1 line segment is joined to the end called 1, the total length is twice the length of the first alone. This can conceptually be done with out end as there is no “largest number,” N. (N+1 is larger.)
To cope with this fact and yet retain the idea that any well defined number can correspond to one and only one point on a “line,” we define a line to be 1D and like a line segment, but with no ends. Zero has a special roll to play. By common convention, all the numbers of / on the line to the left of 0 are “negative” and to the right are positive, but I don't need to say more about negative numbers, so won't.
(3) All number system notations use a “base” I'll refer to as “B.” The most common in public use has ten different value symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, & 9 and can be used to describe the length of the longer line segments produced by adding, end to end, more of the 0 to 1 line segments, but then two ends of line segments cease to exit. The pair of line segment of point (2) joined “end to end” have ends called, 0 & 2 and has length two, etc. as “2” = 1+1; 3 = 2 +1; etc. but when we get to 9 + 1 we lack a symbol to call it (except in higher than base ten systems, such as the b =16 system used in many computers. Then 9+1 = a; and a+1 = b etc. up to d + 1 = e, when this “hexadecimal” bases system also runs out of additional symbols. In general the number of symbols, #, is same as the base. # = B.
(4) Man used his fingers to count so and that is why B = ten is so commonly used. (Mental math would be easier if humans had twelve fingers, as division by any factor of the base is relatively easy and twelve has many more factors than ten does. Even dividing a base twelve number by 8 is easy as divide by 2 and then by 4, both factors of the base.) But I digress. Fact that transistors have two well defined states (conducting or not) has made B=2 a very commonly used number system for electronic machine made calculation.
(5) All base systems have the same meaning for 0 & 1 but that is all the symbols the “binary system" has, “places with values” in the notational system were invented. I'll count up to base ten 8 in binary to illustrate:
0; 1;
10;
11; 100; 101;
110; 1
11; 10
00. The value of each place is a power of base ten 's 2. i.e. the green ones or even the green 0 last binary number I listed (the one with value 8 in the B =ten system) all are in a “notation place”
one to the left of the first place and so have value, if not occupied by 0, of base ten's 2, which is 2^
1. likewise the red 1 in the second place to the left of the first place has value of base ten's 2 squared or 2^2 = 4 of base ten. Etc. I. e. a 1 in the fifth place to left of the first place, has value of base ten's 2 raised to the fifth power. All base notation systems work this same way. The value of places, if not occupied by 0, is B^n where in is an integer telling how many places to the left of the first place the space is, if occupied. For example in the hexadecimal base system 1b is the base ten number 16 + 2 = 18 but I am already using the B = ten system when writing this “18.” i.e. the 8 is in what I have been calling the “first place” and the 1 of the 18 is in the first place to the left of the “first place.” So that one, in any base system has value = B^1 and the next to left of it has value = B^2, etc.
(6) We often mark with a decimal point where the “first place is. For example the hexadecimal number 1b is more clearly written as 1b. And the 1 as stated early is worth B^1 = 16^1 = 16 in the base ten system. The decimal point is essential if we want to write numbers less than 1 as they are written to the right of the decimal point. In the hexadecimal system base ten fraction 1/16 is written 0.1 or is not worried the reader will fail to see the decimal point, just as .1 and a value sixteen times smaller is written as 0.01 etc.
(7) Now with this general background, lets speak mainly of the base ten system:
Any of the ten symbol can be in the first place to the right or left of the decimal point. If for example we have 12. than mean twelve. 1.2 means 1 + 1/5 and 012 means 1/10 + 2/100 and this is BUILT INTO THE MEANING OF THE NOTATION SYSTEM.
It is true that if I multiply 1.2 by 10, I get 12, but exactly that same effect can be achieved by
notational change ONLY. Just move the decimal point found in 1.2 to the other side of the 2. I. e. write 12.
Likewise 1.23456789 can be made ten times larger by move of decimal point one space to the right to get: 12.3456789 or a ten thousand (which is 10^4) times larger by move of decimal point 4 spaces to the right. I. e. 12345.6789 is ten times larger than 1.23456789 is. This is NOT an operation of multiplication, which I have not even defined any general algorithm for, but BUILT INTO THE MEANING OF THE base, place & point NOTATION SYSTEM.
For example as 16x16 = 256 then if I wanted to make the hexadecimal number 1b. larger by 256 times I just write it as 1b00. That hexadecimal number 1b. Had value in base 10 of 18 an if in base 10 I want the number which is 256 times larger I must multiply (or use my calculator) to get 4608.
Likewise I can increase the hexadecimal number 1b by 16^4 times easily. It is 1b0000 but to do that to base 10's 18 I will use my calculator to get: 1179648, but this only hints at the fact large values sre shorter to write in larger base number systems. Consider hexadecimal number ed. In decimal notation that is 16x15 + 13 = 253. I. e. take three places, not just two to write and even more to write in a binary system.
As I noted early:
Challenge to those who don't think 1/1 = 1 = 0.99999 exactly, tell what is wrong with the general procedure I gave for finding the rational fraction equal to any repeating decimal. Or the logical foundation I describe for it above.
