Of course, reversing the order of a proof constitutes reordering it. This is rather fundamental. It appears that you don’t have any clear idea about the rules of logic or what constitutes a proof. Pointing this out does not constitute misinterpretation, “twisting things” to suit someone’s purposes, or being unfair in any way. As I have said before this forum is for the purpose of obtaining factual information. You can’t simply make up your own rules.
Once again (third time’s a charm) I am instructing you to stick strictly to factual information. Any further non-factual comments here or complaints about being forced to actually demonstrate your statements may jeopardize your posting privileges. You’re not going to be given an unlimited pass to post whatever you please.
Seems that your “refutation” is to simply delete some steps in the proof and scream “TAUTOLOGY”!
That’s not the way formal logic works, though. You have to leave the steps intact and point out a flaw in the reasoning that led from one step to the next. And no, deleting steps at random does not count as finding a flaw in the reasoning between those steps.
For instance, I could say that this is a bit clearer (at least to me)
x = 0.333…
if we multiply both sides by 10 we get
10x = 3.333…
if we add 3 to both sides on 1) we get
3 + x = 3.333…
2 and 3, taken together give us
10x = 3 + x
we substract x from both sides, arriving at
9x = 3
we divide by 9 on both sides
x = 3/9
we simplify 3/9
x = 1/3
substituting x from step 1 we arrive at
0.333… = 1/3
To disprove the reasoning above, you’d have to attack the validity of the step and/or its mathematical representation on the next line.
As a former mathematics student, I think there are three steps which could be attacked in the proof:
This assumes that .333… is a real number, i.e., that the series 3/10 + 3/100 + 3/1000 + … has a limit. It’s fairly easy to do so, but you do need a clear definition of a limit. In any case, I think this is the step that 7777777 has an objection to.
and 3) assume that ordinary arithmetic operations work for the decimal fraction .333… . They do for all decimal fractions (which again is fairly easy to prove), but they don’t for all infinite series.
However, I don’t think 7777777 knows enough about mathematical analysis to see these potential problems – which are only potential problems, not real problems.
Let me for the sake of reasoning play de7il’s advocate here. I don’t think bunch of sevens has the clarity of thought to pursue this line of reasoning, but if I understand anything that he has attempted to postulate, the following is a [del]logical[/del] sensible consequence.
The objection would come in line 2 which would be rewritten:
10x=3.3333…330 where the Zth decimal place holds a 3 and the infinityth decimal place is zero.
Line 3 then does not follow.
What is the most efficient way of addressing this fallacy?
Z must be a natural number in this case. If the Zth decimal place in 3.33… holds a 3 and the Z+1th decimal place in 3.33… holds a 0, then it is trivial to go back to 0.333… and show (by just pointing at the digit) that the Z+1th (and Z+2th, and Z+3th, and Z+WHATEVER NUMBER YOU WANT TO INPUT HEREth) is not zero. This works for any Z. No matter what you select Z as, you will find that the space after Z contained a 3 before the multiplication and therefore still does. This whole concept is just so fundamentally asinine that it hurts my brain.
I agree with you entirely – it is asinine.
However, this (meaning the argument you just presented) is precisely the line of reasoning that 7777777 rejects.
Hamster King laid down a challenge – a reasonable one at that. I was trying to envisage what the most probably response from 7777777 would be. (He did not respond as I expected.) So, my argument is purely hypothetical. Pretend I am a bunch of sevens. Shoot me down in colourful flames.
BPC is giving a proof by induction. It is a mark of math cranks that they do not use induction and do not understand induction when it is presented to them. They therefore cannot be shot down by induction because they simply ignore the bullets.
It is literally like the kid’s game of shooting people with their fingers. One kid makes a direct hit and yells “I got ya” and the other insists “You missed” despite evidence obvious to everyone around.
What confounds you is that adults are continuing to play.
As j_sum1 is noting, we could, if we wanted to play de7il’s advocate, consistently say “x = 0.3333… means x is given by Z many 3s after the decimal point. 10x, then, would be 3 + 0.333… with only Z - 1 many 3s after the decimal point. This would not be the same as 3 + x, which would have Z many 3s after the decimal point. We would in fact find that 10x - x = 3 - 10[sup]-Z[/sup], which would be less than 3. x would come out to 1/3 - 10[sup]-Z[/sup]/9, which would be less than 1/3.”
Along the same lines but more general, we might say “x = 0.3333… means x lies within (0.3, 0.4), (0.33, 0.34), (0.333, 0.334), etc. 10x, then, would indeed lie within (3, 4), (0.3, 0.4), (0.33, 0.34), etc., as would 3 + x. But just knowing this information would not allow us to determine the relationship between 10x and 3 + x; whether they are equal or one is larger than the other. We will similarly not be able to determine the relationship between 9x and 3, though we will know that 9x lies within (2.9, 3.1), (2.99, 3.01), (2.999, 3.001), etc. And similarly with the relationship between 3x and 1”.
The answer to all of these, of course, is “That’s not what I mean.”
.0000123456…/8 < .000002456… ?
.0000123456… < 8 x .000002456… ?
.0000123456… < .000019648… (iteration via numerical analysis will not be able to DISPROVE this statement)
Thanks, checking, meantime, please comment on this:
Lorentz Length Contraction
Ln = Lo sqrt [1 - (v^2/c^2)]
c = 299,792,458.000… (notice the zeros and the ellipsis indicating that the value is
defined with infinite precision)
=> c^2 = 89875517873681764.000… (again notice the ellipsis, the square of the speed of light is ALSO known
EXACTLY)
Check Lorentz Equation for .999… substitution for an object moving at 250,000,000
Ln = Lo sqrt [.999… - (62500000000000000/89875517873681764) ]
Ln = Lo sqrt [.999… - .69540628…]
Whatever the difference is inside the brackets it will be impossible to take the square root of it
and so the the .999… substitution makes the Lorentz length contraction formula useless.
No it won’t. Why do you say that? Do you foresee the same problem using 1 instead of .999…? You’ll get the same decimal difference inside the brackets either way.
No, it is impossible to take the square root of an infinite precision decimal real number to obtain a magnitude that can be used to convert the length. The speed of the object would be known with finite precision and so 1 minus a finite precision number results in another finite precision number. .999… minus a finite precision number results in an infinite precision number.
For example. if V^2/C^2 was .24712
Than .999… - .24712 would be .75287999…
(remember you CANNOT back substitute 1 in here because it DOESN’T exist conceptually per your equivalency)
What is the square root of .75287999… ?
I would need a magnitude to be able to multiply the length
Not simply a tautology of the square root of .75287999… is the square root of .75287999…
