Help me understand

Quote from: John Davis on December 07, 2017, 07:21:01 PM

Globe math. Just great. 1 + 2 + 3 + 4 + 5 ...= -1/12.

No. That equals 15, silly. Again, you are confusing a finite series with an infinite one.

So you filled in the dots, eh?

Nothing on this page has much to do  with proving the shape of the planet.   It's clearly spherical using very simple geometry.

Well you were talking about the difference of 1 and 0.999...999 so the distinction was important.

Again, if you think 1 is different to 0.9 recurring, the distinction is clear.
If you don't, it makes no difference.
Ether way, all you are doing is stalling.
Now answer the question:
Do you agree that multiple and division and inverse functions of each other?
That is if you have a*b=c, then c/b=a?

I have to admit, its almost impossible to parse this globularist barble and I am an expert in mathematics. What exactly are you asking?!

This has nothing to do with the globe. If you were an expert in math you would understand quite simply.

In general, if you have a function f(x), and in inverse function to that, e.g. g(x), then g(f(x))=x.
That is, if you take a number, put it into a function, and then put that result into the inverse function, you get your original number back.

Multiplication and division is a simple example of this.
If you take a function f(x)=b*x, then the inverse function g(x)=x/b.
Thus g(f(x))=f(x)/b=(b*x)/b=x.

You can also get that from the relation I gave, but I will spell it out a bit more:
Assuming "a", "b" and "c" are all finite, non-zero, real numbers; assuming the relation a*b=c holds, then does the relation a=c/b also hold?

Does that make it easy enough for an "expert" like you to understand?

Quote from: boydster on December 07, 2017, 07:34:50 PM

Quote from: John Davis on December 07, 2017, 07:21:01 PM

Globe math. Just great. 1 + 2 + 3 + 4 + 5 ...= -1/12.

No. That equals 15, silly. Again, you are confusing a finite series with an infinite one.

So you filled in the dots, eh?

That's some disputeone-level bullshit right there. Don't misquote me just because you made a mistake and can't admit it.

Now substitute in 1 = 0.999...9999. This leaves you with -1/12 - ( 0.0000...01)

But, 1 ≠ 0.999...9999 no matter how many decimal places you take it to.
So your substitution is invalid - try again!

And you call yourself a mathematician - piffle!

Quote from: John Davis on December 07, 2017, 07:18:42 PM

Now substitute in 1 = 0.999...9999. This leaves you with -1/12 - ( 0.0000...01)

But, 1 ≠ 0.999...9999 no matter how many decimal places you take it to.
So your substitution is invalid - try again!

And you call yourself a mathematician - piffle!

1 = 0.9999...9999 as much so as 1 = 0.9999999...

Quote from: John Davis on December 07, 2017, 07:35:49 PM

Quote from: boydster on December 07, 2017, 07:34:50 PM

Quote from: John Davis on December 07, 2017, 07:21:01 PM

Globe math. Just great. 1 + 2 + 3 + 4 + 5 ...= -1/12.

No. That equals 15, silly. Again, you are confusing a finite series with an infinite one.

So you filled in the dots, eh?

That's some disputeone-level bullshit right there. Don't misquote me just because you made a mistake and can't admit it.

Misquote you? I clearly labelled where I inserted the dots.

You added the dots after, then asked if I had filled them in somehow, when they didn't exist in your original statement.

What number comes between 0.999... and 1?

Why is that relevant? There need not be a number between two numbers for them both to be distinct numbers and not equal each other.

If there is no number between them, then they are not different, distinct numbers.

The series of repeating 9s needs to be finite (as you no doubt know, as you have been clear to illustrate it with your notation) for there to be a difference.

Why is that relevant? There need not be a number between two numbers for them both to be distinct numbers and not equal each other.

9*1 = 9
9/9 = 1
9*1/9 = 1

1/9 = 0.1111....
9*0.1111....=0.999....

0.9999....=1

Can we be done with this stupid tangent now?

9*1 = 9
9/9 = 1
9*1/9 = 1

1/9 = 0.1111....
9*0.1111....=0.999....

0.9999....=1

Here's your problem:
9*0.1111... = 0.9999... != 1

Ah, so FE math states that 9/9 != 1

Have you met Danang? You guys will get along famously.

If there is no number between them, then they are not different, distinct numbers.

The series of repeating 9s needs to be finite (as you no doubt know, as you have been clear to illustrate it with your notation) for there to be a difference.

So let me understand this correctly.

We have 3 numbers a,b and c. If a = b, but b != c, that makes neither a nor c a number, in spite of if a < c !?

Quote from: realNarcberry on December 07, 2017, 01:52:22 PM

I just want to make sure I'm solving the right problem, so I need to understand which numbers you mean

I have already provided them.

If you think these numbers are not the same, and I have asked you what is 1*9, why ask about other numbers.

If they were not the same number then what you are doing amounts to something like this exchange:

Me: "What is 1*9?"
You: "1 or 57?"
Me: "1"
You: "9 or 32?"

Notice how your questions are just pathetic attempts at stalling??

I have provided the numbers, 1 and 9.
What is the product of 1 and 9?
Can you answer that simple question, or must you continually avoid it because you know you are full of shit?

I think he's referring to the fact that 1 is easily shown to be equal to .999... in a few steps.

He's trying to, at least.

Quote from: boydster on December 08, 2017, 08:46:22 AM

If there is no number between them, then they are not different, distinct numbers.

The series of repeating 9s needs to be finite (as you no doubt know, as you have been clear to illustrate it with your notation) for there to be a difference.

So let me understand this correctly.

We have 3 numbers a,b and c. If a = b, but b != c, that makes neither a nor c a number, in spite of if a < c !?

If a < c, then what is c - a?  If there is no number between c and a, then c - a = 0 and a = c.  If a = c then they are not different numbers.

Quote from: ItsRoundIPromise on December 08, 2017, 08:48:33 AM

9*1 = 9
9/9 = 1
9*1/9 = 1

1/9 = 0.1111....
9*0.1111....=0.999....

0.9999....=1

Here's your problem:
9*0.1111... = 0.9999... != 1

Except that makes no sense.

9 * 0.1111....= 0.9999...  We agree on this.

9* 1/9 = 1 We agree on this.

1/9 = 0.1111.... We agree on this.

If a * b = x
and a * c = y
and b = c
then x = y

Plug the numbers in and see what happens.

Plug the numbers in and see what happens.

Every calculator is going to have this rounding error.

Quote from: boydster on December 08, 2017, 08:46:22 AM

If there is no number between them, then they are not different, distinct numbers.

The series of repeating 9s needs to be finite (as you no doubt know, as you have been clear to illustrate it with your notation) for there to be a difference.

So let me understand this correctly.

We have 3 numbers a,b and c. If a = b, but b != c, that makes neither a nor c a number, in spite of if a < c !?

Much like before, you are misrepresenting what was stated and creating an argument against a strawman. And I know you know that.

Quote from: ItsRoundIPromise on December 08, 2017, 10:54:46 AM

Plug the numbers in and see what happens.

Every calculator is going to have this rounding error.

It's not a rounding error and you don't need a calculator.  I'll spell it out for you.

If a * b = x                                               a = 9   b= 0.1111....     a * b = 0.9999...    0.9999....= x
and a * c = y                                            a = 9    c = 1/9              a * c = 1               1 = y
and b = c                                                 b = c               0.1111...= 1/9   
then x = y                                                x = y               0.9999... = 1

No calculator necessary, and no rounding.  A simple algebraic substitution and basic arithmetic.  Can we finally be done with this?  It's not something anyone with an ounce of reasoning should have to question for more than a moment or two.

Quote from: realNarcberry on December 08, 2017, 10:56:11 AM

Quote from: ItsRoundIPromise on December 08, 2017, 10:54:46 AM

Plug the numbers in and see what happens.

Every calculator is going to have this rounding error.

It's not a rounding error and you don't need a calculator.  I'll spell it out for you.

If a * b = x                                               a = 9   b= 0.1111....     a * b = 0.9999...    0.9999....= x
and a * c = y                                            a = 9    c = 1/9              a * c = 1               1 = y
and b = c                                                 b = c               0.1111...= 1/9   
then x = y                                                x = y               0.9999... = 1

No calculator necessary, and no rounding.  A simple algebraic substitution and basic arithmetic.  Can we finally be done with this?  It's not something anyone with an ounce of reasoning should have to question for more than a moment or two.

This shows nothing. You're just complicating your last, failed, example.

Why is that relevant? There need not be a number between two numbers for them both to be distinct numbers and not equal each other.

Yes there is.
The number line is a continuum.
Between any 2 numbers there exists another.
In fact, between any 2 numbers there exists a number in the middle, the average.
If you have the numbers a and b, you can find the average (c) by the formula:
c=(a+b)/2.
If these numbers are the same, the average will be that number, if they are different there will be another number between.

So lets try:
c=(1+0.999...)/2
=1.999.../2
=0.999...

Which was the first number (not one of them as this shows they are the same number).

Quote from: ItsRoundIPromise on December 08, 2017, 08:48:33 AM

9*1 = 9
9/9 = 1
9*1/9 = 1

1/9 = 0.1111....
9*0.1111....=0.999....

0.9999....=1

Here's your problem:
9*0.1111... = 0.9999... != 1

Nope. No problem there.

The simple fact is 1/9=0.999...
The simple fact is 9*1=9
The simple fact is 9/9=1
The simple fact is 9*1/9=(9*1)/9=9/9=1
The simple fact is 9*1/9=9*(1/9)=9*0.111...=0.999...

Thus the simple fact is 1=0.999...

If you wish to disagree, point out what the error is.
You can't just object to the conclusion as it is based entirely upon the above simple facts.
As such, you need to object to one of them.
Which do you object to?

Quote from: boydster on December 08, 2017, 08:46:22 AM

If there is no number between them, then they are not different, distinct numbers.

The series of repeating 9s needs to be finite (as you no doubt know, as you have been clear to illustrate it with your notation) for there to be a difference.

So let me understand this correctly.

We have 3 numbers a,b and c. If a = b, but b != c, that makes neither a nor c a number, in spite of if a < c !?

How is that anything like what he said?

You have 3 numbers, 1, 1 (0.999...) and 2.
1=1, and 1!=2.
a and c are still numbers.

Or did you get confused at distinct numbers?

If a and b are equal, they are not distinct. They both are the same number and thus represent a single distinct number.

I stated that pretty poorly. To say:
-- if there is no number between two numbers .: those numbers are equal
Is to say:
-- all numbers are equal, and there are no distinct numbers at all.

I'll have to check whenever I get time to spend in my library, but I believe I can source this with Ludwig Wittgenstein.

I stated that pretty poorly. To say:
-- if there is no number between two numbers .: those numbers are equal
Is to say:
-- all numbers are equal, and there are no distinct numbers at all.

I'll have to check whenever I get time to spend in my library, but I believe I can source this with Ludwig Wittgenstein.

You're not making sense.  3 and 4 are between 2 and 5.  None of those numbers are equal and they are distinct numbers.

3.11111112 is between 3.11111111 and 3.11111113.  They aren't equal and they are distinct numbers. 

There is no number between 5 and 5 because they are equal, and not distinct from one another.

Ok, so given these two numbers:

a = 0.000000.....00000001 (as our convention seems to allow)
b = 0.000000.....00000002

This would lead us, given your statements, to believe a=b as there is no number between them. This then means, as one can easily see, that no numbers exist and all numbers are equal.

Ok, so given these two numbers:

a = 0.000000.....00000001 (as our convention seems to allow)
b = 0.000000.....00000002

This would lead us, given your statements, to believe a=b as there is no number between them. This then means, as one can easily see, that no numbers exist and all numbers are equal.

One of the many numbers between them is:

x = 0.000000.....000000015

Does that clear it up for you?

Quote from: John Davis on December 08, 2017, 01:08:03 PM

Ok, so given these two numbers:

a = 0.000000.....00000001 (as our convention seems to allow)
b = 0.000000.....00000002

This would lead us, given your statements, to believe a=b as there is no number between them. This then means, as one can easily see, that no numbers exist and all numbers are equal.

One of the many numbers between them is:

x = 0.000000.....000000015

Does that clear it up for you?

That is certainly not between them. That is larger than both of them.