1 is 0.9999999999999............

[BdS]

What percentage is 1|3 of the pie? it is impossible to have 3 identical pieces via percentage... each slice (s) must be s1 = 33.333... % s2 = 33.333... % s3 =33.333... %
s1 + s2 + s3 = 99.999... %
where is the crumb that is 0.000...r1 % of the 100% pie? one slice would have to be 33.333...4 to make it 100% and make the slices not equal...

What percentage is 20 sec of a minute?
20 / 60 * 100 = 33.333... %
33.333... % * 3 = 99.999... %
same 0.000...r1 % missing

What percentage is 120 degrees of 360 degrees?
120 / 360 * 100 = 33.333... %
33.333... % * 3 = 99.999... %
same 0.000...r1 % missing

 

[Motor Daddy]

What percentage is 1|3 of the pie? it is impossible to have 3 identical pieces via percentage... each slice (s) must be s1 = 33.333... % s2 = 33.333... % s3 =33.333... %
s1 + s2 + s3 = 99.999... %
where is the crumb that is 0.000...r1 % of the 100% pie? one slice would have to be 33.333...4 to make it 100% and make the slices not equal...

What percentage is 20 sec of a minute?
20 / 60 * 100 = 33.333... %
33.333... % * 3 = 99.999... %
same 0.000...r1 % missing

What percentage is 120 degrees of 360 degrees?
120 / 360 * 100 = 33.333... %
33.333... % * 3 = 99.999... %
same 0.000...r1 % missing

Nice post.

 

[someguy1]

3. You are repeating the same nonsense as Undefined. The center point can be shared by all slices. It has zero measure , so it can be divided amongst any number of slices.
Besides, if you are intent into falling into the philosophical nonsense, you should have made the same argument about the endpoints of the arcs defined at point 1.

First, as far as dividing a circle into 3 equal parts, clearly each arc contains one of its endpoints and not the other. That's the only way to have a disjoint partition of the circle into three equal parts.

If we're dividing a disc, the question is what to do about the center. If you say you can include the center with each part, that violates disjointness. The whole idea is to partition the given object. That means to subdivide it into parts such that the intersection of any two parts is empty, and the union of the parts is the original object. You can't share the center point.

Now you are correct that you could arbitrarily assign the center to any one of the parts since a point has zero measure. That's correct. However it violates the idea of congruence, which means that each of the equal parts can be mapped to any other via a rigid motion, or isometry. Isometry captures the essence of the ancient idea of congruence. If one part of the disc includes the center and the other two parts don't, then you do indeed have three parts of equal measure, but NOT three congruent parts. [And note that each of the parts includes one of its boundary lines and not the other]

So it's accurate to say that you can partition a disc into three parts of equal measure; but not three congruent parts.

 

[someguy1]

1= 1 kilogram

I have no idea what you mean by that. 1 does not equal 1 kilogram. 1 is a unitless number.

If you allow 1 = 1kg then it's also true that 1 = 1 apple and therefore an apple = a kilogram. Of course that's nonsense.

 

[Quantum Quack]

I have no idea what you mean by that. 1 does not equal 1 kilogram. 1 is a unitless number.

If you allow 1 = 1kg then it's also true that 1 = 1 apple and therefore an apple = a kilogram. Of course that's nonsense.

Oh I think you know what I mean, however the following diagram concerning the proof in question explains it better.
I posted this earlier about 5 pages back in this thread and discussed it with Pete...you may have missed it...

by extending the proof by displaying all the hidden and implied values helps to clear up the context.

1 = 1 what? sort of analysis/confirmation/validation exercise.
Any way I am sure you can work it out for yourself...
As far as I am concerned, Dinosaur and Billy T are both correct and the proof is a dead issue.

The Faculty of Science/School of Mathematics at Monash University Melbourne should find it most interesting...

0.999..=1 can not be calculated and can only be defined using a process of limits placed upon infinity. [ via calculus ]

@Tach... Now that's one for your text books..

 

[Monimonika]

What percentage is 1|3 of the pie? it is impossible to have 3 identical pieces via percentage... each slice (s) must be s1 = 33.333... % s2 = 33.333... % s3 =33.333... %
s1 + s2 + s3 = 99.999... %
where is the crumb that is 0.000...r1 % of the 100% pie? one slice would have to be 33.333...4 to make it 100% and make the slices not equal...

What percentage is 20 sec of a minute?
20 / 60 * 100 = 33.333... %
33.333... % * 3 = 99.999... %
same 0.000...r1 % missing

What percentage is 120 degrees of 360 degrees?
120 / 360 * 100 = 33.333... %
33.333... % * 3 = 99.999... %
same 0.000...r1 % missing

So it all boils down to:

What percentage is 1 of 3?
1 / 3 * 100 = 33.333... %
33.333... % * 3 = 99.999... %
same 0.000...r1 % missing

In other words, both BdS and Motor Daddy think it is impossible to divide 3 (which is the result of adding three 1s) by 3 because percentage-wise there will always be a "0.000...r1 %" missing.

Let me make this clearer. BdS and Motor Daddy both believe that it is IMPOSSIBLE for 3/3 to equal 1. Since they both insist that anything that is 3 times 33.333....% of 100% must always be 0.000...r% less than 100%, it also applies when the 100% is 3.

I'll say it again (just because it's so absolutely hilarious and saddening at the same time):

BdS and Motor Daddy both believe that it is IMPOSSIBLE for 3/3 to equal 1.

 

[Billy T]

First, as far as dividing a circle into 3 equal parts, clearly each arc contains one of its endpoints and not the other. That's the only way to have a disjoint partition of the circle into three equal parts.

If we're dividing a disc, the question is what to do about the center. If you say you can include the center with each part, that violates disjointness. The whole idea is to partition the given object. That means to subdivide it into parts such that the intersection of any two parts is empty, and the union of the parts is the original object. You can't share the center point.

Now you are correct that you could arbitrarily assign the center to any one of the parts since a point has zero measure. That's correct. However it violates the idea of congruence, which means that each of the equal parts can be mapped to any other via a rigid motion, or isometry. Isometry captures the essence of the ancient idea of congruence. If one part of the disc includes the center and the other two parts don't, then you do indeed have three parts of equal measure, but NOT three congruent parts. [And note that each of the parts includes one of its boundary lines and not the other]

So it's accurate to say that you can partition a disc into three parts of equal measure; but not three congruent parts.

This is at best confused* by the unsupported assertion that there cannot be a large number of zero measure points at any defined XY location (say with edges of the table helping to define locations with the origin at the center of the pie initially where the X-axis & Y-axis cross.**) still there at x=0 & y=0 when each of the three, now separated, pie slices has a point at the apex. Or, more correctly: there is a point at the location of the apex of each pie slice, always. I. e. the slices can all be both congruent and of equal measure. At least until you can both clearly state and prove the postulated assertions :
(1) Disjointness cannot be violated.
AND
(2) How the idea of congruence is violated.
AND
(3) Cleary define what is "disjointness." I understand congruence as: That for every point of "A" there exist at least one point at the same relative location as all the points of "B."

* I. e. you seem to think that a moved object takes its prior location points with it as it moves. It does not. There is still a zero measure point where any part of it was before. Points are not movable, but locations. Location is the ONLY property of a point, not its apple flavor.

** in addition to the "center of the pie" initially at (0,0) there is also an "x-axis" point at (0,0) and an "y-axis" point at (0,0) all with their own different definitions (three just given, but million of more could be).

BTW, my computer's spell checker is complaining that there is no such thing as "disjointness" defined in English.

 

[BdS]

Let me make this clearer...

3/3 = 1

P = 1/3 = 0.333...
M = 1/3 = 0.333...
S = 1/3 = 0.333...

P + M + S = 3/3 = 0.999...

is 3/3 = 0.999... ?

<-- when I put this here, it means Im 99.999...r1 % confused...

 

[Motor Daddy]

BdS and Motor Daddy both believe that it is IMPOSSIBLE for 3/3 to equal 1.

I'll say it again because you lived up to it: You are not the sharpest tool in the shed.

3/3 is three divided by three, or 300%/3=100% Check your work: 100*3=300! Yup, good to go! That is what you are talking about.
1/3 is one divided by three, or 100%/3=33.333...%. Check my work: 33.333...*3=99.999... Wait, something is wrong! That is what I am talking about.

 

[Tach]

First, as far as dividing a circle into 3 equal parts, clearly each arc contains one of its endpoints and not the other. That's the only way to have a disjoint partition of the circle into three equal parts.

If we're dividing a disc, the question is what to do about the center. If you say you can include the center with each part, that violates disjointness. The whole idea is to partition the given object. That means to subdivide it into parts such that the intersection of any two parts is empty, and the union of the parts is the original object. You can't share the center point.

You are mixing up the mathematical model (the division of the circumference and/or disk with the ruler and compass) with the physical realization (the "cutting" of the disk with some implement). This is the root of your error.

Now you are correct that you could arbitrarily assign the center to any one of the parts since a point has zero measure. That's correct. However it violates the idea of congruence, which means that each of the equal parts can be mapped to any other via a rigid motion, or isometry. Isometry captures the essence of the ancient idea of congruence. If one part of the disc includes the center and the other two parts don't, then you do indeed have three parts of equal measure, but NOT three congruent parts. [And note that each of the parts includes one of its boundary lines and not the other]

Nope, in the mathematical modeling the center is divided by all the slices, I was very clear on that. Since it has zero measure (it is a point) it can be divided among any number of slices.

So it's accurate to say that you can partition a disc into three parts of equal measure; but not three congruent parts.

No, it is not. See above, the reasons for your confusion.

 

[Billy T]

You are mixing up the mathematical model (the division of the circumference and/or disk with the ruler and compass) with the physical realization (the "cutting" of the disk with some implement). This is the root of your error. ... No, it is not. See above, the reasons for your confusion.

I agree and explained that too in post 967, also saying he was confused and asking him to define his terms (hoping that he would remove his confusion in the process).

 

[Tach]

BdS and Motor Daddy both believe that it is IMPOSSIBLE for 3/3 to equal 1.

They are just trolling you.

 

[Billy T]

I think, Post 941 proves valid* and gives the ultimate in compact, easy to calculate, Room Assignment Algorithms, RAA, or solution to Pete's Puzzle (Infinite number of buses, B equal to a positive integer, all full with an infinite number of passengers, P**, wanting the vacant even number rooms in Hilbert's infinite hotel.) which is:
When P is < or = B, passenger B*P's room is 2[(B-1)^2 +P], otherwise it is 2[(B-1)^2 + B + P].
Note that P is not larger than B when this RAA is applied and passenger with P = B, B*P, (or B*B or P*P or P*B as all four point to same guy) gets his room by the "pre-otherwise" part.

* No rooms doubly assigned and none left empty.
** Note B is a unique integer, but there are an infinite number of passengers with the same "P number" (one on each bus).

Quick summary of proof idea: There are an infinite number of squares also named by N = B, with area of N^2. The "boundary group," also called B, of square N makes its area greater than that of square (N-1) by 2N+1 new unit area squares and has 2B+1 new passenger (one P per unit area) wanting the next 2N+1, not yet assigned even rooms.
The P< or = B part of the above RAA gives rooms to B+1 of the 2B+1 "boundary passengers" being added to square B-1 to make it square B (or N as they name the same square next larger than square N-1), which has (N-1)^2 area or Ps (as they are equal numbers) who already have been assigned rooms. Read post941 for more clear / proven discussion of the RAA.

 

[someguy1]

Nope, in the mathematical modeling the center is divided by all the slices, I was very clear on that. Since it has zero measure (it is a point) it can be divided among any number of slices.

That's a very interesting way of looking at points.

So if I'm understanding you correctly, I can demonstrate a disjoint partition of the unit interval

I = {x in R : 0 <= x and x <= 1] as follows:

I define two sets X and Y like this:

X = {x in I : 0 <= x and x <= 1/2]

and

Y = X = {x in I : 1/2 <= x]

In other words I partition the unit interval into the two closed intervals [0, 1/2] and [1/2, 1].

Then if someone says to me: Well, that's not a disjoint partition because 1/2 is in both parts; I can answer, according to your logic: "Since 1/2 is a point with zero measure, it can exist in both partitions yet there's no overlap, because the point 1/2 is "split"".

I must stand in awe of your new concepts. You have just overthrown 200 years of real analysis and general topology. I salute you.

 

[origin]

by extending the proof by displaying all the hidden and implied values helps to clear up the context.
..

I have deleted this abortion that I wrote. The strange notation threw me for a loop apparently.[shrug]

 

[rpenner]

I agree with origin. If $$a \neq 0$$ and $$b \neq 0$$ then pretty much by definition $$\frac{a \cdot b}{a \cdot b} = \frac{\not a \cdot \not b}{\not a \cdot \not b} = 1$$. Likewise saying $$x = 1 (0.999...) $$ is the same as saying $$x = 0.999...$$ so the right side is a confused mismash of bad ideas and pseudomathematics. Because you eliminate the variable x in the right side your conclusion is merely a repetition of your premise and so you have "proved" nothing.

While on the left side, you have failed to understand the proof you are trying to quote because statements don't obviously follow from each other. A more correct proof is:

$$\begin{array}{rrclcl}
1. & x & = & 0.999... & \quad \quad \quad & \textrm{Premise: x is a number, ... means the 9's go on forever}
2. & 10 x & = & 9.999... & & \textrm{Property of Decimal fractions, even infinite ones, is a shift of the decimal point right is a multiplication by 10}
3. & 10 x & = & 9 + 0.999... & & \textrm{Property of Decimal fractions is that they are absolutely convergent sums and may be partitioned arbitrarily without changing the total}
4. & 10x - x & = & 9 + 0.999... - 0.999... & & \textrm{Equals subtracted from equals are equals, here we subtract line 1 from line 3}
5. & (10 - 1) x & = & 9 + 0.999... - 0.999... & & \textrm{Distributive property of multiplication over subtraction}
6. & 9 x & = & 9 + 0.999... - 0.999... & & \textrm{Replacing 10 - 1 with an equal expression}
7. & 9 x & = & 9 & & \textrm{Cancellation of additive inverses}
8. & \frac{1}{9} 9 x & = & \frac{1}{9} 9 & & \textrm{Multiplication of both sides by the same number}
9. & x & = & \frac{1}{9} 9 & & \textrm{Cancellation of multiplicative inverses}
10. & x & = & 1 & & \textrm{Cancellation of multiplicative inverses}
\end{array}$$

 

[Tach]

In other words I partition the unit interval into the two closed intervals [0, 1/2] and [1/2, 1].

This is a perfectly ok partition, I don't understand why you have so much trouble with it. It must be due to your continuing confusion between mathematical models and reality, noted previously. The point located at 0.5 is just shared by the two intervals. You can have either partition:

1. [0,1/2) , (1/2, 1]
or
2. [0,1/2] , [1/2, 1]

Both of them are equally valid. Your concern about what happens with x=1/2 is totally unfounded and appears to be deeply rooted in your co-mingling mathematical models with the reality they model.

I must stand in awe of your new concepts. You have just overthrown 200 years of real analysis and general topology. I salute you.

This is the third time you fail in understanding a simple concept. <shrug>

 

[Motor Daddy]

I agree with origin. If $$a \neq 0$$ and $$b \neq 0$$ then pretty much by definition $$\frac{a \cdot b}{a \cdot b} = \frac{\not a \cdot \not b}{\not a \cdot \not b} = 1$$. Likewise saying $$x = 1 (0.999...) $$ is the same as saying $$x = 0.999...$$ so the right side is a confused mismash of bad ideas and pseudomathematics. Because you eliminate the variable x in the right side your conclusion is merely a repetition of your premise and so you have "proved" nothing.

While on the left side, you have failed to understand the proof you are trying to quote because statements don't obviously follow from each other. A more correct proof is:

$$\begin{array}{rrclcl}
1. & x & = & 0.999... & \quad \quad \quad & \textrm{Premise: x is a number, ... means the 9's go on forever}
2. & 10 x & = & 9.999... & & \textrm{Property of Decimal fractions, even infinite ones, is a shift of the decimal point right is a multiplication by 10}
3. & 10 x & = & 9 + 0.999... & & \textrm{Property of Decimal fractions is that they are absolutely convergent sums and may be partitioned arbitrarily without changing the total}
4. & 10x - x & = & 9 + 0.999... - 0.999... & & \textrm{Equals subtracted from equals are equals, here we subtract line 1 from line 3}
5. & (10 - 1) x & = & 9 + 0.999... - 0.999... & & \textrm{Distributive property of multiplication over subtraction}
6. & 9 x & = & 9 + 0.999... - 0.999... & & \textrm{Replacing 10 - 1 with an equal expression}
7. & 9 x & = & 9 & & \textrm{Cancellation of additive inverses}
8. & \frac{1}{9} 9 x & = & \frac{1}{9} 9 & & \textrm{Multiplication of both sides by the same number}
9. & x & = & \frac{1}{9} 9 & & \textrm{Cancellation of multiplicative inverses}
10. & x & = & 1 & & \textrm{Cancellation of multiplicative inverses}
\end{array}$$

Let's just cut to the chase in #2:

1. x=.999...
2. 9x=??

 

[Tach]

Let's just cut to the chase in #2:

1. x=.999...
2. 9x=??

9x=9

Of course, you will never accept that and you will continue to troll on the subject for the rest of your life.

 

[rpenner]

$$x = 9 \times \sum_{k=1}^{\infty} \frac{1}{10^k}
\\ \begin{eqnarray} 9x & = & 9 \times 9 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 81 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 80 \times \sum_{k=1}^{\infty} \frac{1}{10^k} & + & 1 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 \times \sum_{k=1}^{\infty} \frac{1}{10^{k-1}} & + & 1 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 \times \sum_{k=0}^{\infty} \frac{1}{10^k} & + & 1 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 \quad + \quad 8 \times \sum_{k=1}^{\infty} \frac{1}{10^k} & + & 1 \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 \quad + \quad \left( 8 + 1\right) \times \sum_{k=1}^{\infty} \frac{1}{10^k} \\ & = & 8 + x \end{eqnarray}$$