Download LONG DIVISION AND HOW IT REVEALS THAT

LONG DIVISION AND HOW IT REVEALS THAT:
.9 6= 1,
The Existence of Subnumbers,
And The Infinitesimal
by
Eric Sanders
Edward Sanders
1
T ABLE OF CON T EN T S
Forward
Introduction
.9 6= 1
Explanations and Proofs
The Repetend Symbol
Error Analysis
Subnumbers
Infinitesimals
Conclusion
page 3
page 4
page 6
page 15
page 20
page 23
page 26
page 30
page 37
2
F ORW ARD
This paper is built on two unassailable truths. The first truth is that
two different numbers are not equal to each other. Therefore, .9 = 1 is not
accepted in this paper as ever having been correct. The second truth is
that the ”check”, the process of inserting potential solutions back into the
original problem or running a series of math operations backward to check for
validity, is a fundamentally valid way to determine if a potential solution is
a correct solution. And proponents of .9 = 1, who have unceasingly rejected
the results of checking the answers of certain long division problems that
generate repeating decimals, have never given any logical reasons or proofs
of any kind that explain why the ”check”, the most successful problem solving
strategy ever devised by mankind, is flawed or defective. If the check did not
work, we would have no economic system more developed than bartering,
and science and technology simply would not exist.
The conclusion that .9 = 1 has only ever been accepted by a minority
of the people exposed to it. The vast majority of people have vehemently
disagreed with .9 = 1, but, because the errors in the algebraic process that
converts a repeating decimal back into a fraction could not be found, this
majority have remained relatively quiet.
Those errors are unveiled in this paper which, also, further develops some
of the more interesting or speculative conclusions that can be drawn from
fixing those errors.
This paper is intended to spark debate and discussion. It is, also. intended to generate a collaborative effort among mathematicians, scientists,
and engineers worldwide to develop further the maths presented here.
All rules and definitions postulated in this paper are tentative only and,
merely, induced from observations of the limited number of examples presented. Changes are not only welcome but are expected.
The development of the maths in this paper is just in its infancy, so the
reader is encouraged to sit back and relax with a glass of a favorite beverage.
Just enjoy the ride.
Eric Sanders
3
IN T RODU CT ION
Long division performed routinely in the schools produces three different
types of answers. The first is the quotient combined with a remainder.
1 R8
9)17
−9
8
The second type of answer is to convert the remainder into a fraction.
1+
9)17
−9
8
8
9
= 1 89
The third type of answer is to continue the long division past the decimal
point and generate decimals in the quotient until the division terminates or
is rounded to an appropriate decimal place.
1.8 ≈ 2
9)17.0
−9
80
-72
8
This paper is built on performing long division to create an exact answer
to the above problem by combining the second and third types of answers.
This is achieved by including the traditionally omitted decimal points and
zeros.
1.8 +
9)17.0
−9
8.0
-7.2
.8
.8
9
4
Also, this paper uses the repeating bar in repeating decimals such as .8 in
a far more mathematically rigorous way. Specifically, it reveals and addresses
the difference in the number of digits repeated in different repeating decimals.
For example, the number of eights in 10 × .8 = 8.8 is the same as the number
of eights in .8, but the number of repeated eights in the 8.8 is one less than
the number of repeated eights in .8.
5
.9 6= 1
The fallacy of .9 = 1 originates in human error while performing long
division. Additional errors are committed when performing math on repeating decimals. These errors are in how we write answers and in how we use
the repeating symbol to write repeating decimals, not in the arithmetic itself.
8
9
.8888
9)8.0000
-72
80
-72
80
-72
80
-72
8
8
=
.8
9
Now, for the check.
9 × 89 = 9 × .8
8 = 9 × .8
This is a difficult check to visualize because multiplying starts with the
right side of a number, and there is no way to reach the end of an infinite
number of eights. But we can construct an answer.
9
9
9
9
9
×
×
×
×
×
.8 = 7.2
.88 = 7.92
.888 = 7.992
.8888 = 7.9992
.8 = 7.999...92
It is too cumbersome to write 7.999...92 all the time. Instead, it can
be written as 7.92. This is a novel use of the repeating symbol, but it will
work if there is a redefining of the symbol, and since new rules for repeating
decimals must be created, this paper will expand the current definition of
the repeating symbol which will give to it the capability to produce rigorous
6
mathematics.
DEFINITION: Let the repeating symbol equal an indefinite number of
repeating digits.
This is actually the dictionary definition but a clarification is needed
because too many people have substituted the word ”infinity” for the word
”indefinite” and believe that the repeating symbol represents an infinite number of repetitions and that is now the widely accepted informal definition.
Clarification: The number of repetitions can be less than as well
as equal to infinity.
And thinking that the repeating symbol is equal to an infinite number of
repetitions is what has led to misunderstandings, one of which is .9 = 1. This
way we can avoid what appears to be, at first glance, a number with more
than an infinite number of digits, ie. 7.92. The entire string may have an
infinite number of digits, but the number of repeating nines would be ∞ − 2
in length. Look back at the construction of 9 × .8 = 7.92. Notice how the
number of repeating eights is one more than the number of repeating nines.
If 7.92 has an ∞ − 2 number of repeating nines, then .8 must have an ∞ − 1
number of repeating eights. So now 7.999...992 = 7.92.
Back to the check.
9 × .8 = 7.92
or
.8 =
7.92
9
But now there is another problem. The repeating symbols in the .8 and
7.92 represent two different lengths of repeating digits. The construction of
7.92 from 9 × .8 demonstrates this. As will be shown later in this paper,
the use of two repeating symbols, each representing a different length of repeating digits, both in the same problem, is one of the two major errors that
has led to the false conclusion that .9 = 1. This error must be corrected by
equalizing the lengths represented by the repeating symbol. The repeating
symbol in .8 is one digit more in length than the repeating symbol in 7.92.
By removing an 8 from underneath the repeating symbol, .88, the length
represented by the repeating symbol is shortened by one and now matches
the length represented by the repeating symbol in 7.92. Placing the lone 8
on the right hand side of the repeating symbol instead of the left is arbitrary.
It could have been written .88. Choosing which one to use depends upon
which one makes doing math operations easier. Both will be used in the
7
same problem later in this paper. .88 = .88 So now...
9 × .88 = 7.92
But let’s return to the check. We were expecting the .8 × 9 to result in
an answer of 8 for our check, but, instead, the result was 7.92. The long division answer did not pass the check. The answer is wrong. There is an error.
And that error is an error of omission. Conventional long division allows us
to omit ”unnecessary” zeros and decimal points in order to streamline an
already lengthy process. All of the missing zeros and decimal points must be
put back into the problem along with a fractional remainder.
Let’s do the division of 89 again, but, this time, one step at a time and
checking each of those steps.
.8 +
9)8.0
-7.2
.8
.8
9
Let’s check the answer.
9×
.8
9
8
9
= .8 +
8
9
= 9 × .8 + 9 ×
.8
9
8 = 7.2 + .8
8=8
It passed the check.
Now for the second step of the long division of 89 .
.88 +
9)8.00
-7.2
.80
-.72
.08
.08
9
8
Let’s check the answer.
8
9
9×
= .88 +
8
9
.08
9
= 9 × .88 + 9 ×
.08
9
8 = 7.92 + .08
8=8
It passed the check.
Note in the second line of both checks that a 7.92 is beginning to be
generated. This is the result of trying to check 89 = .8 (from the first page of
this section). The .8 is just the quotient of the long division of 98 . Let’s take
this to its logical conclusion and perform the long division of 89 an infinite
number of steps.
.8888.......
9)8.0000
-7.2
.80
-.72
.080
-.072
.0080
-.0072
.0008
Note all of the remainders. The fraction 89 is not a terminating decimal
so a nonzero remainder is always generated. The remainders become smaller
but never zero because there is always another smaller remainder. This shows
that...
8
9
= .8 + a REMAINDER
Therefore, the quotient of just .8 is not an exact answer.
Omitting the remainder is a HUGE error. Look at the pattern of remainders. A .08 is being generated. So the long division answer is .8 + .08
.
9
8
9
= .8 +
.08
9
9
Therefore,
8
9
6= .8
Now for the check. But the repeating symbols represent different lengths.
They must be made the same length. And the length of the string of repeating zeros is one digit less than the length of the string of repeating eights.
So, an eight must be removed from underneath the repeating symbol The
traditional quotient of .8 becomes .88.
8
9
= .88 +
.08
9
Now, let’s do the check on our new answer.
9×
= 9 × .88 + 9 × .08
9
8 = 7.92 + .08
7.92
+.08
8.00
8=8
It passed the check.
8
9
The answer to the long division finally checks correctly, and that is because the remainder was included in the answer. The remainder is very small,
and that explains why the 7.92 is so close to the correct answer of 8, and
why many have mistakenly believed that two numbers so close together are
actually equal to each other, ie. .9 = 1.
But it has been thought that .9 could not be generated by long division
because there is no fraction that produces .9 as an answer, and, therefore,
no remainder to fill in the gap between .9 and 1.
But there is a fraction that, using long division, generates .9. And that
fracton is...
1
More specifically,
1
1
10
.9999........
1)1.0000
-.9
.10
-.09
.010
-.009
.0010
-.0009
.0001
1
1
= .9 + .01
1 = .9 + .01
1 6= .9
The remainder needed to make .9 equal to 1 is missing.
But .9 + .01 needs to be written correctly in order to perform math on
it. In its current state, it looks like .9 + .01 would equal .91.
.9
+.01
.91
But it does not. It equals 1. Look at all of the remainders for the long
division of 11 . The number of zeros to the left of the 1 in every remainder
is always one less than the number of nines in the quotient that generates
a specific remainder. This means that the repeating symbols in .9 + .01
represent two different lengths of strings of repeating numbers, just like it
did in the answer to 89 . Two different lengths of repeating decimals cannot
be arithmetically combined to produce a correct answer, as seen above. The
lengths must be made the same... by shortening the longer string. Two new
rules are required in order to fix this problem that has now appeared twice
in this paper.
Rule 1: The lengths of the strings of repeating decimals represented by
the repeating symbols in any problem or any line of a problem
must all be of the exact same length.
Rule 2: The repeating symbol will represent the length of the shortest
length of repeating decimals.
Now for the check to the long division of
11
1
1
1 = .9 + .01 can now be written correctly.
1 = .99 + .01
Once again, I chose to place the digit removed from the repeating string
of digits on the right hand side because it makes arithmetic operations easier
to see and perform.
1 = .99 + .01
.99
+.01
1.00
1=1
It passed the check.
Therefore, .9 6= 1
The mistake all along was using just the quotients, .8 and .9, from the
and .01 respectively.
long division problems and omitting the remainders, .08
9
This allowed for the false conclusion that .9 = 1. The quotients and their
remainders must travel together in order to produce exact answers. And the
remainders look suspiciously like the infinitesimal.
1 = .9 + .01
1 − .9 = .01
1 − .9 = .... is the classic definition for the infinitesimal. And that makes
.01 as well as .08
infinitesimals.
9
Now, let’s see what fractions .8 and .9 truly turn into. There is an algebraic process that can convert a decimal into a fraction, but it only works
correctly with repeating decimals if the above rules are followed. I call that
algebraic process the 10n-n process. Below is the 10n − n performed traditionally. Let n = .8.
10n = 8.8
−n −.8
9n = 8
n = 89
This process seems to convert .8 back into 89 but only because it breaks
rules 1 and 2. Let’s fix the errors. Multiplying .8 by 10 decreases the length
12
of the string of repeating eights to the right of the decimal point by one digit
less than the string of eights in .8. Let’s construct the answer.
.8
×10
0
8
8.0
.88
×10
00
88
8.80
.888 .8888
×10
×10
000
0000
888
8888
8.880 8.8880
.8
×10
8.80
Many believe that multiplying .8 by 10 shifts the string to the left by one
decimal place. It does not. The string is expanded by one decimal place to the
left. We omit the zero. We should not. The difference of the lengths of the
strings of repeating decimal eights in the .8 and the 8.8 must be addressed.
Let’s redo the 10n-n process with the .8 and include the missing zero while
applying rules 1 and 2. The repeating symbol in .8 must represent the exact
same length as the repeating symbol in 8.8 does. So, .8 can be rewritten as
.88 or as .88. Placing the 8 that cannot be included in the repeating string
of eights on either the right or left hand side depends upon which version
makes the math easier to perform. Starting with the traditional ”.8 ”, let n
= .88, or let n = .88.
10n = 10 × .88
10n = 8.80
-n
-.88
9n = 7.92
n = 7.92
9
Therefore, .8 =
Notice how both .88 and .88 are used in line 1
and in line 3, respectively.
7.92
9
This is the true fraction that .8 equals and note that it is the fraction
produced by the very first check performed in this paper. Let’s do the 10n-n
process with .9. First, the traditional way. Let n = .9.
10n = 9.9
−n −.9
9n = 9
n = 99
13
n=1
Now, let’s fix the errors and repeat the process again. Let n = .99, or let
n = .99.
10n = 9.90
−n −.99
9n = 8.91
n = 8.91
9
Therefore, .9 =
8.91
9
This is the true fraction that .9 equals. In addition, this implies that...
.9 6= 1
And this is clear proof that all repeating symbols in a problem must equal
the same number of repeating digits. Not doing so is analogous to adding
fractions without first finding a common denominator.
Clearly, 7.92 and 8.91 are not rational numbers. And since they are the
numerators of the two fractions, whose denominators are whole numbers,
that represent .8 and .9 respectively, then .8 and .9 are irrational as well.
This specific conclusion can be intuitively extended to all repeating decimals. None are rational numbers, and, so, the classification of repeating
decimals must be officially changed from rational to irrational.
Classification: Repeating decimals are irrational numbers.
Some mathematicians have pointed to the infinite sum of a geometric
a1
. But since the exact value of the limit
series as proof that .9 = 1, Sn = 1−r
depends on summing together an infinite number of terms, and infinity is
endless, the formula is being written incorrectly and needs to be redefined.
DEFINITION: Sn <
a1
1−r
These quotients and remainders that continuously increase or decrease in
magnitude because they are repeating decimals or have a repeating decimal
in them are difficult to visualize as numbers. They make more sense when
viewed as a being subnumbers, parts and pieces of complete numbers. Add
together the two subnumbers generated by one specific long division problem,
and the result is a complete number.
14
EXP LAN AT ION S AN D P ROOF S
The heart of the error in writing long division answers lies in convention.
Long division with a remainder in the answer is performed by students for
only a few years during grammar school. And the remainder is calculated
before long division generates decimals in the quotient.
7
4
1
4)7
-4
3
In later years, the long division is performed without remainders, resulting in terminating decimal answers or truncated/rounded answers.
1.75
4)7.00
-4
30
-28
20
or
.888 ≈ .89
9)8.000
-72
80
-72
80
-72
8
But long division is not traditionally performed into the the decimal tail
of a number, and, then, a remainder calculated.
15
.888 + R?
9)8.000
-72
80
-72
80
-72
8
The remainder can not be calculated correctly without the missing decimal points and zeros.
.888 + R .008
9)8.000
-7.2
.80
-.72
.080
-.072
.008
The fraction equivalent answer is .888 + .008
. But the remainder is neces9
sary in order to use the 10n-n process correctly. It has already been demonstrated that the 10n − n process can convert a decimal into an equivalent
fraction. What will happen if the full, correct answer to a long division problem (quotient plus remainder) is put into the 10n−n process. A general form
of the process shows this. Let n = a.
10n = 10a
−n −a
9n = 9a
n= a
What you put in, is what you get out. Let’s put the correct answer to
, let n
the division of 98 into the 10n-n process. Using the answer of .88 + .08
9
= .88 + 1 × .08
, or let n = .88 + 1 × .08
.
9
9
16
10n = 10(.88 + .08
)
9
10n = 8.80 + 10 × .08
9
−n −.88 −1 × .08
9
9n = 7.92 + 9 ×
9n = 7.92 + .08
9n = 7.92
+.08
8.00
9n = 8
n = 98
8
9
= .88 +
.08
,
9
.08
9
not .8
The 10n − n process does indeed work. Let’s try it with the long division
answer to 11 and see if 1 is the output. Let n = .99 + .01, or let n = .99 + .01.
10n = 9.90 + 10 × .01
.01
−n −.99 −
9n = 8.91 + 9 × .01
n = .99 + .01
.99
+.01
1.00
n=1
Therefore, 1 = .99 + .01, not .9.
There are a few ”proofs” that supposedly demonstrate that .9 = 1. The
first involves the fraction 91 .
.111 ...
9)1.000
-9
10
-9
10
-9
1
17
So, according to tradition,
1
9
= .1
The ”proof” goes as follows.
1
9
= .1
9 × 19 = 9 × .1
1 = .9
The astonishing part of all of this is that the above ”proof” is not even
a proof in the first place. It is a check. Specifically, it is the check to the
long division answer of 19 . And the answer of .1 fails to check. Let’s do this
all over again correctly including writing .1 as .11 to equalize its repeating
symbol with the repeating symbol in its remainder.
.111 ...
9)1.000
-.9
.10
-.09
.010
-.009
.001
1
9
= .11 +
.01
.
9
Let’s do the check.
= .11 + .01
9
9 × 91 = 9 × .11 + 9 ×
1 = .99 + .01
.90
+.01
1.00
1=1
1
9
.01
9
The check once again demonstrates that using just the quotient from
long division and omitting the remainder introduces significant error. There
is another ”proof’ that, also, shows this using 13 .
18
.333 ...
3)1.000
-9
10
-9
10
-9
1
So, according to tradition,
1
3
= .3
The ”proof” goes as follows.
1
+ 31 + 13 = .3 + .3 + .3
3
3
= .9
3
1 = .9
Let’s do the long division of
”proof”.
1
3
correctly and put that answer into this
.333 ...
3)1.000
- .9
.10
-.09
.010
-.009
.001
1
3
= .3 +
1
3
+
1
3
+
.01
3
1
3
3
3
= .33 +
.01
3
= .33 + .01
+ .33 +
3
= .99 + 3 × .01
3
1 = .99 + .01
.99
+.01
1.00
.01
3
+ .33 +
.01
3
1=1
The check is proof that omitting remainders causes significant error and
incorrect conclusions.
19
T HE REP ET EN D SY M BOL
The repeating, or repetend, symbol has been misunderstood. It was intended to represent an indefinite number of repeating decimals. This rigorous
definition was replaced, informally, with the less acurate concept that the repeating symbol is synonymous with an infinite number of repetends (which
can, unfortunately, create a string of decimals which contains a greater than
infinity number of digits, which is impossible to do, ie. .12). And this is
where all the conceptual errors begin. The repeating symbol in the quotient
of 8 ÷ 9 = .8 (or any other quotient that generates a repeating decimal)
misleads us into falsely concluding that the long division is finished because
we have gone out an ”infinity” worth of repetends and can go no further.
That causes us to ignore the remainders. In addition, the repeating symbols
in the quotients used in the 10n − n process are viewed as infinite in length
which led us into thinking that the string of repeated 8’s in 8.80 is the same
length as the string of 8’s in .8.
Merely written down by itself and not generated by long division, .8 has
a repeating symbol that represents an ∞ number of digits. Derived from the
division problem 8 ÷ 9, the repeating symbol in the remainder of the answer,
, represents a number of digits no greater than ∞-1.
.8 + .08
9
So far, the differences in the lengths represented by the repeating symbols
have been manageable since the difference inside the example problems has
always been just one. But later on in this paper the differences will become
significantly larger, and the length of digits represented by the repeating
symbol will actually change from line to line within a single problem. For
example, the repeating decimal .25 has a repeating symbol that can represent
a number of repetitions no greater than 12 ∞. Otherwise, the symbol would
represent a number of digits greater than ∞ which would violate the concept
of infinity in the first place. Repeating decimals with a single digit repetend
and not generated by division are just an infinitesimal fraction of all repeating
decinals. So, the idea that the repeating symbol equals an ∞ number of
repeating digits is just plain unrealistic and highly unusual when it does
happen.
The key is to see the repeating symbol as representing an indefinite number of repetitions that would create a string of digits no longer than ∞ decimal
places. The number of repetitions and the number of digits generated by the
repetitions need to be tracked throughout an entire problem. A qualifier
would be helpful. The symbol for the qualifier used for the remainderof this
paper is r̄.
Definition: Let r̄ symbolize the qualifier for repetends.
20
Definition: Let r̄a,b be the composite of symbol r̄ and variables a and b.
The r̄ is the character that symbolizes a repeating decimal’s
repetend. Information about a repeating decimal’s repetend
is contained in the a and b where a equals the length of a
repeating decimal’s repetend and where b represents the
number of repetitions generated by the repetend symbol in
that repeating decimal.
The values of a and b are chosen to ensure that the total number of digits
in the entire number does not exceed ∞ . Examples:
.8, r̄1,∞
.88 +
.08
,
9
r̄1,∞−1
.25, r̄2, 1 ∞
2
.2525 +
.0025
,
99
r̄2, 1 ∞−1
2
The 21 ∞ − 1 comes from the fact that the total number digits generated
by the repetend symbol is ∞ minus the lone 25 tail in .2525. The number of
repetions of the repetend, therefore, must be 21 the number of digits generated
by the repetend. 12 (∞ − 2) = 12 ∞ − 1
There is an interesting mathematical manipulation that comes with the
combining of two or more repeating decimals with unequal length repetends
when adding or subtracting.
.25
+ .3
r̄2, 1 ∞
2
r̄1,∞
The repetends are mismatched and, therefore, cannot be added together
correctly. The repetend lengths must equalized. If these repeating decimals
had been generated by long division, then it would be easy to see that for
every 25 generated, there would be a 33 generated. Therefore the .3 can be
rewritten as .33 and, then, added properly to .25.
.25
+ .33
.58
r̄2, 1 ∞
2
21
Another rule comes from all of this.
Rule 4: The length of a new repetend will be equal to the lowest common
multiple of the repetend lengths that generated the new repetend.
This rule only applies to addition and subtraction of repeating decimals.
The mathematics for multiplying repeating decimals is far trickier and
should be explored in its own paper. the tangents to this paper must be
brief. Here is another example.
.123
+ .12
r̄3, 1 ∞
3
r̄2, 1 ∞
2
Now equalize the lengths of the repetends.
.123123 r̄6, 1 ∞
6
+ .121212
.244335
The repetend variable can quantify the length of the repetend and the
number of digits generated by the repetend symbol for more than just individual repeating decimals. It can represent the characteristics for all repetends
of length a and number of digits generated b in any line of a problem or for an
entire problem, like in the addition problems above. This makes everything
clearer and equalized. For example:
n = .88 +
.08
9
r̄1,∞−1
Rule 5: Represent all repetends of length a and number of digits b
generated by the repetend symbol with only one repetend
variable to avoid mismatching errors. Multiple repetend symbols
can appear in a problem but only if each symbol represents a
different length of repetend from the others.
22
ERROR AN ALY SIS
This paper must now address a major point that, so far, has not been
discussed because the reader needed to accept its validity quickly so that it
would not distract from showing that .9 6= 1. That point is the unconventional
remainder created after calculating a decimal quotient. To my knowledge,
this has never been done before, at least not routinely.
For example, in the long division of 89 out to three decimal places, the
remainder is .008
.
9
The original use for the remainder was to allow grammar school children
the ability to calculate exact answers while avoiding decimals which they had
yet to be taught. When children are, finally, introduced to decimals, remainders are abandoned in favor of extending the quotient until it terminates or,
if termination is impossible, until an acceptable number of decimal places
has been calculated and the quotient, then, rounded. The decimals in this
new type of remainder mix decimals and fractions. Routinely placing such a
decimal/fraction into a long division answer would be unique.
Yet doing so opens the door to a more accurate strategy for calculating error for real world problems. The exact error in any division problem
between two whole numbers or terminating decimals can now be found.
In engineering and science applications, the error generated by division
can now be tracked exactly, using this new type of remainder. Remainders
can be added together which allows for an exact value of the accumulative
error in an application. Even the error introduced by rounding quotients can
be found exactly. For answers that round down, the error is the remainder.
For answers that round up the error is, actually, smalller than the remainder,
if the quotient had just been truncated at the acceptable decimal place.
For example, in the division of 94 to three decimal places, the anwer would
, so the remainder is the
be rounded to .444. The exact answer is .444 + .004
9
error.
In the division of 59 to three decimal places, the rounded answer is .556.
The exact answer is .555 + .005
. The truncation error is the remainder, but
9
the error from the rounding is .004
which is less than the remainder. So,
9
rounding is still very much an excellent strategy for controlling error.
So far, we have worked with decimals inside of a fraction like .008
. There is
9
a simpler and less confusing way to write the remainder. It is with a fraction
inside of a decimal, .00 89 . This is, indeed, a very unconventional number, but
it does work, and it makes it easier to write and understand the remainder.
For example, the answer to 8 ÷ 9 out to 3 decimal places is .888 + .008
. This
9
8
can be rewritten as .888 + .00 9 The remainder is the error when rounding
down takes place. The error in rounding up to .889 is .001 − .00 89 = .00 91 .
23
Performing math with these remainders is tricky. If the answer to .888 +
is written as .888 89 , then the 89 has been mistakenly moved from the
1,000th decimal place to the 10,000th decimal place. Here’s how to do it
correctly.
.00 89
.888 + .00 89 =
+ .00 89 =
.88 72
9
.88 80
9
The error is now in the correct decimal place but has been combined with
the last 8 in .888 to form an improper fraction. The error is hidden inside
the fraction and not obvious at all, an unwelcome side effect. There might be
some other way to write this such as .88.8 89 where the second decimal point
means that the mixed number actually occupies just one decimal place, in
this case the 1,000th place. But in both ways of writing the remainder combined with the quotient, the visual effect takes some getting used to. So, I
am just writing the quotient and remainder separately for now. For example:
1
9
1
+ 11
= .11
+ .01
9
= +.0909 + .0001
11
r̄1,∞−1
r̄2, 1 ∞−1
2
There are, clearly, repetends of different lenghts that are being added
together. The repetends must be equalized,
.1111 +
+.0909 +
.0001
9
.0001
11
r̄2, 1 ∞−1
2
Find a common denominator.
.1111 + .0011
99
+.0909 + .0009
99
.2020 +
1
9
+
1
11
=
.2020 +
.0020
99
20
99
Now do 20÷99 and the answer is:
.2020
99
Involving the repeating symbol in a decimal that contains a fraction just
makes everything worse, but it does make a major and very significant point
24
about infinity. The .88 89 that comes from adding together the .88 quotient
and 0 89 remainder (r̄1,∞−1 for the last three repeating decimals) from 8÷9
shows that the repeating symbol used in the traditional 8 ÷ 9 = .8 does
not even remotely represent an infinity worth of decimal digits. At most it
might represent 12 ∞ decimal places since the 98 generates its own ”infinite”
string of 8’s. But that second string of 8’s has another 89 as a remainder,
.8̄8̄ 98 . This repeating the repeating 8 now means that the repeating symbol
is at most equal to 31 ∞. This process of converting each remainder of 98
into another infinite string of 8’s followed by another and another can be
repeated for,... well,... an ∞ number of times. This then would create a
1
repeating symbol that at most is equal to ∞
∞ which either equals 1 or is
indeterminate. Therefore, thinking of the repeating symbol as creating an
infinite number of digits is incorrect and dangerous. I use the ∞ to represent
the number of digits in a repeating symbol because I have to, not because
I want want to. The ∞ symbol is just too convient to use. It is visually
intuitive and concise.
Thinking that we can reach infinity has always been a major issue and
error. Since we have a symbol for infinity, ∞, we have made the mistake of
treating ∞ as a number, and a number is a limit. But infinity is limitless.
A number is a specific point on a numberline. Infinity is that entire
numberline. It contains all the numbers in an endless sequence but is not a
number itself. Numbers and infinity, figuratively, represent, two incompatible
dimensions, the 0th and 1st dimensions, respectively. Therefore, combining
numbers and infinity is like trying to combine a distance with an area, or an
area with a volume. The units are not the same, so the answer is nonsensical.
Infinity is not a number, so it should not be used as one. But I know of
no other way of conveniently and concisely representing the indefinite length
of strings of repeating decimals without using it.
25
SU BN U M BERS
8
9
= .88 +
.08
9
, r̄1,∞−1
The 98 can be found on a numberline at an exact point. But .88 and .08
9
cannot. This is because, while still existing somewhere on the number line,
.88 increases in value, always approaching but never reaching 89 . The same is
only it decreases in value, always approaching but never reaching
true for .08
9
0. When added together, the two sum to an exact point.
It is as if long division has split a point into two pieces that cannot be
found exactly on a number line. Therefore, the two points are not points and,
therefore, cannot be considered to be numbers. They would be considered
to be subnumbers. An analogy is an atomic particle which can be split
into subatomic particles. Together, the subatomic particles make the atomic
particle.
make the number 89 . So, all repeating deciThe subnumbers .88 and .08
9
mals and their associated decimal/fraction remainders are subnumbers.
It appears that this phenomenon occurs when the denominator of a fraction contains a prime number that is not a factor of 10, our base system.
Otherwise, the fraction becomes a terminating decimal.
And the issue with infinity (subnumbers are infinite length strings of
digits) arises only because humans attempt to represent an abstract concept,
numbers and subnumbers, which have no dimensions, with two dimensional
symbols. The string of digits written down would stretch out infinitely, but
the subnumbers themselves have no length at all. So, the infinity issue does
not exist at all with regards to math, just with regards to symbology.
In the past, we have tried to manipulate and interpret the main subnumbers, the larger ever-increasing repeating decimals, alone without their
partner remainders. This has led to to the false conclusion of .9 = 1. Subnumbers must travel together with their associated partner subnumbers and
never should be removed from the context, the mathematical process, that
generates them.
Subnumbers necessitate the formation of two new definitions and one new
rule.
Definition: Subnumbers are non-terminating decimals whose magnitudes
either increase or decrease. Subnumbers are irrational.
Definition: Numbers are whole numbers, fractions, or terminating
decimals and are rational.
26
Continuing with this physics/chemistry analogy, prime numbers would be
considered the atoms. Products of prime numbers (atoms) could be considered molecular numbers (molecules). And products of two or more molecular
numbers could be considered compound numbers (compounds). For example, 2 and 3 are prime numbers (atoms). They multiply to 6 which would
now be called a molecular number (molecule). So to with 5 and 7 (prime
numbers) whcih multiply to 35 ( a molecular number). The two molecular
numbers 6 and 35 multiply to the compound number 210. This is an intriguing way to look at and name numbers.
Rule 6: Never interpret repeating symbols, and the lengths they represent, outside the context of the math that generates them.
To extend the logic in the first definition above, all irrational numbers
could be considered to be subnumbers. The subnumbers generated by long
division might be called additive subnumbers since the quotient, .8, and
remainder, .08
, add to a number.
9
Another type of subnumber would be the
√multiplicative subnumber. They
multiply to make a number. For example, 2 is a multiplicative subnumber.
√
2×
√
2=2
or (1.414213562...)(1.414213562...) = 2
√
The subnumber 2 is irrational, but multiply it by itself and the product is the rational number 2. Roots of any kind that do not result in a
number which is an integer, fraction, or terminating decimal (rational) are
subnumbers (irrational)
√ that can multiply with themselves to form a number.
Another example is 3 7 = 1.912931183....
(1.912931183...)(1.912931183...)(1.912931183...)
=7
√
√
√
3
3
3
or
7× 7× 7=7
Another type of subnumber is e. It could be thought of as an exponential
subnumber while its inverse ln(x) (with its cousin log(x)) could be thought
of as logarhithmic subnumbers. Together, the exponential and logarithmic
subnumbers make a number. For example e = 2.718281828... and ln2 =
.6931471806... are irrational but put them together and they make 2.
2.718281828....6931471806... = 2
or
eln2 = 2
It appears that all irrationals are subnumbers that come with partner
27
subnumbers that together make a number. Even π, which is irrational, has a
partner subnumber. π is an infinite sum of fractions with odd denominators,
each one of which produces a subnumber remainder. Summing all the fractions creates π. Summing all the remainders would create one subnumber
remainder partner. The number that π plus the remainders represent is a
rational fraction whose numerator and denominator continually increase in
size. This increasing is what gives π its irrational nature.
Even imaginary numbers can be thought of as subnumbers. In polynomials, they travel with partners (conjugates) and, both, add and multiply
back into a number. This makes them bizarre, but they fit the description
for subnumbers.
i × i = -1
i × -i = 1
(4 + 3i) + (4 - 3i) = 8
(4 + 3i)(4 - 3i) 25
Imaginaries are both additive and multiplicative subnumbers at the same
time. While as real subnumbers are the result of splitting a real number (or
point), imaginary subnumbers seem to be the result of splitting nothing into
two pieces. Not the splitting of 0 which is ”something” but the splitting of
nothing. Zero can be split.
0
1
.111...
1)0.000
-.1
-.10
-.01
-.110
-.001
-.111
0
1
= .111... + −.111...
0
1
=0
The interesting thing about this particular long division problem is that
the remainder is not a string of repeating zeros followed by a number or numbers like .01 or .08
which we have seen before. Instead, it is just the negative
9
of the quotient, and that makes sense. So what could those remainders with
28
the repeating strings of zeros be, because they do not make sense. They
could be the elusive but hypothesized... infinitesimal.
29
IN F IN IT ESIM ALS
So far, this paper has concentrated on main subnumbers like .8 and .9
and the need to pair them with their partner remainders. But the remainders
are just as interesting and need to be explored. One of the definitions of the
infinitesimal is 1 - .9. For centuries no one had an answer to this problem,
so many mathematicians just assumed that the answer was 0 which means
that .9 eventually grows large enough in value to equal 1. But, now, there is
a numerical answer to the problem.
1 = .99 + .01 , r̄1,∞−1
1 − .99 = .01
This makes .01 an infinitesimal. But, by the same logic, .0 89 is an infinitesimal as well.
8
9
= .88 + .0 98 , r̄1.∞−1
8
9
− 88 = .0 89
And .0 98 is smaller than .01. Extending the logic, since there are an infinite
number of repeating decimals that are generated by an infinite number of
long division problems, then there are an infinite number of their partner
remainders which means that there is an infiite number of ... infinitesimals.
Kind of ironic, isn’t it? An infinity of infinitely small numbers (which we
should be calling subnumbers).
What happens when a main subnumber is added to an infinitesimal subnumber? There is one on page 2 of this paper. It is the result of the check.
9 × .88 = 7.92 , r̄1,∞−2
7.92 = 7.9 + .02
And that is why the answer to the check has been ignored for all of these
centuries. It is a truly bizarre answer, a combination of the misunderstood
repeating decimal main subnumber and the unrecognized infinitesimal. But
now we know that it is completely legitimate.
There is another definition for the infinitesimal. 0 < x < n1 The infinitesimal is the x when n approaches ∞. The n1 approaches 0 but never
1
equals it. What number could possibly fit between 0 and ∞
? Not a number,
30
but a subnumber could. Specifically, .01.
In fraction form, .01 = 101∞ . Let’s apply this infinitesimal of .01 into a
real world problem; the derivative. I shall use the derivative derived from
the infinitesimal approach used by Isaac Newton and others. This approach
involves division, potentially by zero. Newton and others got around this
problem by hypothesizing the existence of a number so small that it could
not be measured. When added to a number, the number’s value would not
change, yet, not being zero, you could divide by it. This number was called
the infinitesimal. Now, there is a value for the infinitesimal which, algebraically, is labeled ∆x, ∆y, etc.
Definition: An infinitesimal is any subnumber that starts, on the left side,
with a repeating zero (.0)
Examples are .01, .0 89 , and .0201.
y = x2
y + ∆y = (x + ∆x)2
y + ∆y = x2 + 2x∆x + ∆x2
∆y = 2x∆x + ∆x2
∆y = ∆x(2x + ∆x)
∆y
= 2x + ∆x
∆x
Let’s now find the value of the derivative at x = 1 and ∆x = .01 , r̄1,∞−1 .
∆y
∆x
= 2(1) + .01
∆y
∆x
= 2.01 , r̄1,∞−2
The 2.01 is the true value for the derivative of y = x2 at x = 1 using the
infinitesimal that fits the conventional definitions for the infinitesimal.
Let’s look at line 4 of the above construction of the derivative. ∆y (a differential) can be calculated if x and ∆x are defined. This is not the derivative,
just an intermediate step, but it leads to some interesting math that further
expands the rules for dealing with repeating decimals. Let x = 1 and ∆x =
.01 , r̄1,∞−1 .
∆y = 2x∆x + ∆x2
∆y = 2(1)(.01) + (.01)2
31
∆y = .02
+ ?
(.01)2 = ((.01)(.01) = ?
This is where the math becomes interesting. Some may calculate the
answer .0 0 1. But this answer leads to the paradox of an infinite string of
zeroes followed by another infinite string of zeros and, then, followed by a 1.
A construction of the answer shows otherwise.
(.1)2 = .1 × .1 = .01
(.01)2 = .01 × .01 = .00 01
(.001)2 = .001 × .001 = .00 00 01
(.0001)2 = .0001 × .0001 = .00 00 00 01
(.01)2 = .0001 , r̄1,∞−1 , r̄2, 1 ∞−1
2
The repeating symbol in the .0001, r̄2, 1 ∞−1 represents a length that is
2
two times the length represented by the repeating symbol in .02, r̄1, 1 ∞−1 .
2
The number of zeros is always odd and the number of pairs of zeros in the
answer is equal to the number of single zeros being repeated in the number
being squared. For cubing, (.01)3 , it is the number of triplets that equals the
number of singles with a .001 as the exposed tail. (.01)3 = .000001, r̄1, 1 ∞−1 ,
3
r̄3, 1 ∞−1 .
3
Rule 7: An infinitesimal, with a single digit end, raised to a power results
in the lengthening of the number of repeated digits by a factor
equal to the power. The tail digit is treated as a decimal raised
to the power.
Example: (.01)4 = (.0.1)4 = (.0)4 (.1)4 =
.0000.0001 =
.00000001 r̄4, 1 ∞−1
4
Putting the same digit multiple times under one repeating symbol might
be a way to represent a string of numbers that is a multiple length of the
original string’s length. But such a subnumber would have to be broken
apart in order to add it to other subnumbers because the repeating symbols
represent different lengths of repeating digits, and that would violate Rule 1
and Rule 2.
Multiplying the grammar school way shows us how to construct and to
32
visualize the answer.
.1
.01
×.1 ×.01
1
01
.01
00
.0001
.001
×.001
001
000
000
.000001
It is clear that there is always a missing zero at the far left side of the
answer when the columns are added. So, the subnumber could be written
as .0001 r̄2, 1 ∞−1 . The middle two zeros with the one repeating symbol over
2
them can be split apart to make the repeating symbol represent the same
length string of digits that the repeating symbol in .02 does, .0 0 0 1. But
since every digit is a zero except for the very last one, the lone zero and the
repeating zeros can be arranged in what ever way is needed to facillitate the
math. For our purposes, the subnumber should be written as .0001 r̄1, 1 ∞−1 .
2
Why cannot the .0001, r̄2, 1 ∞−1 , and the .000001, r̄3, 1 ∞−1 , and, now, the
2
3
.0001, r̄1, 1 ∞−1 , be written just as .01? Why isn’t (.01)2 = .01, r̄1,∞−1 ?
2
Since the lengths of the strings represented by the repeating symbol are so
close to infinity, what is the difference of a couple of digits? It would still be
just infinity. Even a multiple of infinity is just infinity. This is what mathematicians have been doing for years. The reason why is that the answer
would come out wrong because those ”couple of digits” have a significant
impact on the exact length of the string of digits, and the conventional approach of letting infinity ”absorb” numbers added or multiplied to it only
works in limit problems. What we are performing is arithmetic and algebra,
so the traditional approaches and rules do not apply because our problems
require exactness. Limits are just approximations. To demonstrate, let’s put
both alternative forms into the differential.
Let ∆x = .01, and let (.01)2 = .0001 , r̄1, 1 ∞−1 .
2
∆y = 2x∆x + ∆x2
∆y = 2(1)(.01) + (.01)2
∆y = .02 + .0001
And now they can be added.
.0001
.02
.0201
∆y = .0201
33
Let’s now let x = 1 and ∆x = .01 and make (.01)2 = .01 , r̄1,∞−1 .
∆y = 2x∆x + ∆x2
∆y = 2(1)(.01) + (.01)2
∆y = .02 + .01
∆y = .03
This is not even close to the answer above. But is .0201 even the correct
answer in the first place?
A critical and important point needs to be made. The number of digits in
.0001 appears to be infinity. The repeating symbol represents a length of ∞ 2 zeros. So, the repeating symbol represents a length of ∞−2
number of pairs
2
of zeros. The number of digits in the .02, also, appears to be infinity. But
how can that be? Remember the definition of the repeating symbol uses the
word indefinite, not infinite. And that is part of the problem. The number
of digits in .02 does not have to be anywhere cole to infinity. The answer
∆y = .0201 suggests a solution to this paradox.
The key to understanding all of this is to perceive the mistake we have
been making all along. We have been viewing repeating decimals in a static
way and always interpreting the number after it is finally built with an infinite
number of digits. But you cannot do that. Since an infinite number of digits
can never be reached, the length of the string is always increasing. And it
never stops. So, the best we can do is take a snap shot of the string at any
instant in time. And the .0001 builds (lengthens) twice as fast as the .02
does. So, adding them together to make .0201 is completely valid. Each of
which can be rewritten
the repeating symbols in .0201 has a length of ∞−2
2
1
as 2 ∞ − 1.
The 21 and the −1 cannot be absorbed into the ∞. So, we need two new
rules to differentiate the use of ∞ in problems that contain repeating decimals and in problems that use limits.
Rule 8: Define ∞ to be the longest string of repeating decimals within
the context of the problem. All other lengths will be, by
definition, less than ∞.
Rule 9: When using the repetend symbol, r̄a,b , all coefficients, constants,
and exponents connected to an ∞ must be preserved. They
cannot be absorbed into the ∞ and disappear.
So, .02 does not have an infinite number of digits. If you follow Rules 8
and 9, the .0001 has the greatest infinite number of digits within the context
34
of the entire problem. So, .02 has a 21 ∞ − 1 number of digits.
In other words, it is not the length of the string of digits that is important.
It is the rate of the lengthening of the string of digits that is important. I
used to call repeating decimals ”dynamic” numbers for just this very reason.
Their magnituteds constantly change. Numbers are ”static”. Their magnitudes never change. Subnumbers are ”dynamic”. And the missing perception
of this has led to errors. Let’s construct .0201.
.02
+.0001
.0201
.002
+.000001
.002001
.0002
+.00000001
.00020001
.02
+.0001
.0201
Without construction, this is difficult to see.
Rule 10: Construct an answer to deduce and reveal the dynamic
nature of the subnumber.
Examples of ”constructing” an answer appear on pages 2, 7, 26, 27, and
29.
The .0201 is an infinitesimal with an infinitesimal of it as a tail. This
might seem bizarre, but it simply cannot be ignored. With ∆x = .01 and
∆y
derived earlier with x = 1 and
∆y = .0201, we can verify the value of the ∆x
∆x = .01.
∆y
∆x
=
.0201
.01
= 2.01 , r̄1,∞−2
∆y
This is exactly the same answer for ∆x
from earlier in the paper. If we
had used the ∆y = .03 that was derived from the alternative (.01)2 = .01,
∆y
03
the answer would have been ∆x
= . .01
= 3 which does not equal the answer
to the derivative from earlier nor does it equal the answer for the traditional
limit definition of the derivative.
Our derivative process ignores the ∆x = .01 because the process is just
an approximation.
y = x2
y 0 = 2x
35
The lone ∆x’s are just the infinitesimals which become so small they can
be ignored. And that is what we do. But now we have an actual subnumber
to ignore, not just an abstract concept.
36
CON CLU SION
·
·
·
·
·
·
·
repeating decimals are irrational
.9 6= 1
subnumbers must join the pantheon of types of numbers
one subset of subnumbers is repeating decimals
another subset of subnumbers is the infinitesimal
long division is a number splitter
all irrational numbers are subnumbers
This has already been stated, but it bears repeating. The errors with and
misconceptions of repeating decimals lie with the symbols we use to represent
them. All infinities appear in writing the subnumbers. There are no infinities
in the subnumbers themselves.
Long division has largely been ignored by mathematicians and consigned
to grammar shool textbooks to be used soley for the purpose of converting
a fraction into a mixed number or into a decimal.
Division is, in actuality, the most powerful of all algorithms. It can split
a number. And into immeasurably small pieces. Yet, those infinitesimal
pieces, when ignored, can lead to false conclusions such as .9 = 1 which kept
subnumbers and, specifically, infinitesimals hidden beneath layer upon layer
of errors.
As Shelock Holmes in THE SIGNS OF FOUR states, ”An exception disproves the rule.” And .9 was that exception.
.9 6= 1
Contact: [email protected]
37