The procedure given below works for ALL Repeating Decimals (all RDs) to produce the exactly equivalent rational fraction, but some stubborn and /or ignorant people posting here make the extraordinary claim that it does not when the RD = 0.9,999,999,999.... and offer zero proof of their claim when extraordinary proof is required!
i.e. People who think that 0.9999... is not 1.0 are not only wrong, but irrational (pun intended) when postulating, with no proof, that the validity of algebraic rules fails in the proofs showing that 0.999... = 1. They make the extraordinary claim that general truths, true in an infinity of different cases, that ALL repeating or terminating decimals all have equivalent rational fractions like a/b where a & b are positive integers, with a < b are not true in one special case (the case when a = b) without giving ANY, much less the extraordinary proof, required for their extraordinary claims!
I first illustrate, several of the infinite numbers of examples, of true statements concerning terminating or infinitely Repeating Decimals (I.e. about the rational fractions equal to RDs):
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.
and 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... but even in these cases the obvious pattern (decimal repeats the two numerator numbers) is still true. I.e. the two digit numerator repeats in blocks of two.
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, multiplying RD by 10,000,000 moves the decimal point 7 spaces to the right. 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 the integer:
a,bcd,efg = 9,999,999 x RD. Note there are no infinitely long numbers here and 9,999,999 certainly is not zero so we can divide by it to get: RD = a,bcd,efg / 9,999,999 the rational fraction of integers exactly equal to the infinitely long repeating (with repeat length =7) 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, RD = 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,.... a case with repeat length of 3, can be shown to be equal to 123/ 999, which happens to reduce to 41/333 as your calculator will show as best as it can.
To prove this one multiplies this repeat length 3 RD by 1000 to get the 2nd equation and then subtract the first from it to get: RD = 123 /999.
In general when the repeat length is "n" one always multiplies by 1 followed by n zeros (an integer power of 10) the DR defining equation and then subtracts the RD defining equation from the results of the multiplication to eliminate all infinitely long number strings and get an easy to solve equation for the DR now as a rational fraction.
Then objectors to 1 =0.99999 need not only to give extraordinary proof for their objection but also need to explain why 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.
i.e. People who think that 0.9999... is not 1.0 are not only wrong, but irrational (pun intended) when postulating, with no proof, that the validity of algebraic rules fails in the proofs showing that 0.999... = 1. They make the extraordinary claim that general truths, true in an infinity of different cases, that ALL repeating or terminating decimals all have equivalent rational fractions like a/b where a & b are positive integers, with a < b are not true in one special case (the case when a = b) without giving ANY, much less the extraordinary proof, required for their extraordinary claims!
I first illustrate, several of the infinite numbers of examples, of true statements concerning terminating or infinitely Repeating Decimals (I.e. about the rational fractions equal to RDs):
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.
and 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... but even in these cases the obvious pattern (decimal repeats the two numerator numbers) is still true. I.e. the two digit numerator repeats in blocks of two.
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, multiplying RD by 10,000,000 moves the decimal point 7 spaces to the right. 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 the integer:
a,bcd,efg = 9,999,999 x RD. Note there are no infinitely long numbers here and 9,999,999 certainly is not zero so we can divide by it to get: RD = a,bcd,efg / 9,999,999 the rational fraction of integers exactly equal to the infinitely long repeating (with repeat length =7) 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, RD = 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,.... a case with repeat length of 3, can be shown to be equal to 123/ 999, which happens to reduce to 41/333 as your calculator will show as best as it can.
To prove this one multiplies this repeat length 3 RD by 1000 to get the 2nd equation and then subtract the first from it to get: RD = 123 /999.
In general when the repeat length is "n" one always multiplies by 1 followed by n zeros (an integer power of 10) the DR defining equation and then subtracts the RD defining equation from the results of the multiplication to eliminate all infinitely long number strings and get an easy to solve equation for the DR now as a rational fraction.
Then objectors to 1 =0.99999 need not only to give extraordinary proof for their objection but also need to explain why 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.