Otherwise the hits are gonna keep on coming, brud. There's plenty more cookin' where that came from.
Part 1: in which his claim of 0.333... = 0.3 + 0.333.../10 implies 0.999... = 1
Part 2: 0.333... = 0.3 + 0.333.../10 implies 0.999... = 1 - 1/10^(n) + 0.999.../10^(n,) directly contradicting his position that 0.999... = 1 - 1/10^n ⭐ REMOVED from /r/infinitenines
Today, we'll go into how his claim of 0.333... = 0.3 + 0.333.../10 directly contradicts his concept of references.
Here is a recent example:
> x = 0.999...9 , which reference set.
>
> 10x = 9.999...0
>
> 9x = 8.999...1
>
> x = 0.999...9 aka 0.999...
His concept can be explained as 0.999... = 1 - 1/10^(n) for integer n -- or rather that it is represented by this sequence, and that all operations on such a number must actually take place on the sequence of partial sums in this way.
He uses "references" to argue that 10 x 0.999... = 10 - 1/10^(n-1), and hence, the sequences must be aligned to share a common "reference".
It's very simple, Bob. First, let's do things in his "referenced" way, using his notation. Let's multiply 0.333... by 10 and use the same logic he does in explaining references above.
10 x 0.333... = 3.333...0
9 x 0.333... = 2.999...7
This all seems a bit spooky? You can always "understand" him this way: when he says 0.333... he means the sequence 1/3 x [1 - 1/10^(n)]) = (0.3, 0.33, 0.333, ...).
In other words, his argument is actually...
0.333... = [1 - 1/10^(n)] / 3
10 x 0.333... = [10 - 1/10^(n-1)] / 3
9 x 0.333... = [9 - 9/10^(n)] / 3 = 3 - 3/10^(n)
But if we apply this concept to the statement 0.333... = 0.3 + 0.333.../10, it all BREAKS DOWN:
0.333... = 0.3 + 0.333.../10
[1 - 1/10^(n)] / 3 = 3/10 + [1 - 1/10^(n)] / 30
[1 - 1/10^(n) - 1/10 - 1/10^(n+1)] / 3 = 3/10
[9/10 - 9/10^(n)] / 3 = 3/10
3/10 - 3/10^(n) = 3/10
- 3/10^(n) = 0
Or, in HIS language,
10 x 0.333... = 3 + 0.033...3
3.333...0 - 0.033...3 = 3
2.999...7 = 3
0.2999..7 = 0.3
-0.0000...3 = 0
Is that true, Bob? He would say NO. In fact, it means the ONLY way that his system of "references" can coexist with 0.333... = 0.3 + 0.333.../10 is to imply that...
Either way, you're cooked again, brud! 🍝
Next time: how he FAILS to account for BASIC FACTORING.
Thanks for playing.
Yesterday I demonstrated that one of his claims is devastating to his position:
He said
>0.333... = 0.3 + (1/10) * 0.333...
And I demonstrated that this directly leads to 9 x 0.999... = 9, or 0.999... = 1.
His rebuttal was to link to one of his many claims that 0.999... = 1 - 1/10^n.
In fact, his OWN argument allows us to DIRECTLY ATTACK and DISPROVE that claim.
We can make a stronger, more general construction like the one in his 0.333... post above:
0.999... = 1 - 1/10^n + 0.999... / 10^n
This formula holds for ANY integer n > 0, LIMITLESS or NOT. It nevertheless constructs 0.999... from a string of n initial 9s; and a string of n leading 0s followed by infinite 9s. Combined, these are still a cohesive whole, an unbroken string of 9s, just as 0.999... requires.
As he so often says, 1/10^(n) is never 0. Well, neither is 0.999... / 10^(n) .
This means his oft-repeated claim that 0.999... = 1 - 1/10^(n) is CONTRADICTED by the logical consequences of HIS OWN WORDS.
Thanks for playing, brud. Hope you're not too... steamed. 🍝
Next time: his OWN LOGIC has COMPLETED UNDERMINED his "concept" of references.
Yesterday I demonstrated that one of his claims is devastating to his position:
He said
>0.333... = 0.3 + (1/10) * 0.333...
And I demonstrated that this directly leads to 9 x 0.999... = 9, or 0.999... = 1.
His rebuttal was to link to one of his many claims that 0.999... = 1 - 1/10^n.
In fact, his OWN argument allows us to DIRECTLY ATTACK and DISPROVE that claim.
We can make a stronger, more general construction like the one in his 0.333... post above:
0.999... = 1 - 1/10^n + 0.999... / 10^n
This formula holds for ANY integer n > 0, LIMITLESS or NOT. It nevertheless constructs 0.999... from a string of n initial 9s; and a string of n leading 0s followed by infinite 9s. Combined, these are still a cohesive whole, an unbroken string of 9s, just as 0.999... requires.
As he so often says, 1/10^(n) is never 0. Well, neither is 0.999... / 10^(n) .
This means his oft-repeated claim that 0.999... = 1 - 1/10^(n) is CONTRADICTED by the logical consequences of HIS OWN WORDS.
Thanks for playing, brud. Hope you're not too... steamed. 🍝
Next time: his OWN LOGIC has COMPLETED UNDERMINED his "concept" of references.
Let's focus on Exhibit A.
>0.333... = 0.3 + (1/10) * 0.333...
>
>0.333... = 0.3 + 0.033...
That's a neat trick. Let's multiply by 3.
0.999... = 0.9 + 0.999... x (1/10)
Let's move one of those terms to the left.
0.999... - 0.999... x (1/10) = 0.9
(1 - 1/10) x 0.999... = 9/10
(9/10) x 0.999... = 9/10
Wait a minute! If we multiply by 10 (and remember 10 x 9/10 is... divide negation), that's just...
9 x 0.999... = 9
0.999... = 1
His OWN ARGUMENT has PROVED the 9 x 0.999... = 9 argument he has so strenuously tried to refute!
You're cooked, brud. 🍝
Next time: how this method completely negates his concept that 0.999... = 1 - 1/10^(n) for n "limitless".
Thanks for playing.
0.333... = 0.3 + (1/10) * 0.333...
1/3 = 0.333...3 + (0.000...1 / 3)
0.333...333... "indeed is" 0.333...
0.333... = 0.3 + [ 1 - 1/10^n ]/30
Hint: this means for ANY n, LIMITLESS or NOT...
0.333... = (1/3) x ( 1 - 1/10^n ) + (0.333...) x ( 1/10^n )
1 - 1/10^n is (0.9, 0.99, 0.999, ...) so 1/3 x ( 1 - 1/10^n ) is (0.3, 0.33, 0.333, ...).
Conversely, 0.333... x ( 1/10^n ) is (0.0333..., 0.00333..., ...).
Combined TOGETHER, these terms reconstruct a full infinite string of 3s.
Now, can you SPOT his ROOKIE ERROR that makes his whole position FALL APART?
> 1/2 + 1/4 + 1/8 + ... > > = 1 - 1/2^n with n integer starting at n = 1, then n continually upped. > > Now, 1/2 - 1/4 + 1/8 - ... > > = 1/2 - 1/4 + (- 1/4 + 1/4) + 1/8 - 1/16 + (-1/16 + 1/16) + > > 1/32 - 1/64 + (-1/64 + 1/64) + ... > > = 1/2 + 1/4 + 1/8 + ... - ( 1/2 + 1/8 + 1/32 + ... ) > > keep going brud.
Look what he does. He inserts and removes terms freely, like that (-1/4 + 1/4).
Well, we can do that too, can't we?
1 = 10/10
10/10 + (-1/10 + 1/10) = 9/10 + 1/10
9/10 + 1/10 + (-1/100 + 1/100) = 9/10 + 9/100 + 1/100
And we can do this infinitely. He does. He adds infinitely many terms that are, when wrapped in parentheses, equal to 0. So will I.
1 = 9/10 + 9/100 + 9/1000 + 9/10000 + ... = 0.999...
Thanks for playing.
The "rookie error" makers continue to insist they are THE SAME, but even HE won't say which one is greater!
>> We can all agree that the function f(n) = 1 - 10^-n represents the infinite sequence {0.9, 0.99, 0.999, ...}
> I just needed to read the first part. You messed up big time.
> 0.999... = 0.9 + 0.09 + 0.009 + ... = 1 - 1/10^n for the case n integer pushed to limitless, and the summation starts at n = 1.
>> I see justification of certain numbers, but not definition of real number itself. How do you construct real numbers in general? > > Basically numbers that can be written in decimal form. Rational and irrational numbers. > > Eg. 0.9 > > 0.999... > > 0.999...9 > > And we can put dots on top of symbols to denote recurring patterns.
> Where did I say 100...0 is a real number? It is a number. Just as 0.000...1 is a number. Just as 0.999... is a number (which is not one).
> When you multiply 0.999... by 10 to get 9.999..., the technical issue is, even though you do see .999... after the decimal point, the .999... from 0.999... is NOT THE SAME .999... from 9.999... > > The two lots of .999... are 'out of sync' by one sequence slot. > > Eg. ignore the 'nine' values for the moment, and take a look at 0.abcdef... > > Multiply by 10, and get a.bcdef... > > This means the original sequence after the decimal point was abcdef... But after the multiplication by 10, the sequence after the decimal point is .bcdef... > > So even though the values are all nines, you're taking the difference between two different sequences (ie. out of sync by one sequence slot). So with this difference, it is a difference between two DIFFERENT infinite sequences, and it is not simply going to be zero.
>> So sometimes they are the same and sometimes they aren’t? > > Yes. Correct. That's my assumption. I mention it here.
You know what they say about when you assume.
> The set {0.9, 0.99, 0.999, etc} where the 'etc' is an incarnation of 0.999... itself, ALREADY spans the entire nines space of 0.999...
>> Yes there is an endless amount of them, but that endless amount already exists. > > Wrong on your part brud. Already exists doesn't cut the mustard at all.