Or lets simplify it and call it just value (v)
v/inf = 0
0 · inf = v
Yea, seems untrue
2 years ago
Anonymous
>uses inf/inf
nope
2 years ago
Anonymous
no, since inf/inf and 0*inf are both undefined.
Can't get from 1/inf = 0 to 0 · inf = 1 using algebraic manipulation without bumping into inf/inf
2 years ago
Anonymous
>Can't get from 1/inf = 0 to 0 · inf = 1 using algebraic manipulation without bumping into inf/inf
If i have 666/69 = 9,652..
then i can 9,652.. x 69 = 666 >Simple as
2 years ago
Anonymous
So what you really mean is 666/69 = 9,653
2 years ago
Anonymous
>If i have 666/69 = 9,652.. >then i can 9,652.. x 69 = 666
now do it using algebraic manipulation, moron
can't avoid 69/69
maybe I should get a tape measure in base 3, that way I can cut the 12 inches in equal parts 1.2 inches.
this shit ain't hard. There's nothing special about numbers not divisible by 2 or 5(Or 1 divided by powers of 2 or 5 multiplied), it's just our base being 2*5. In base 12 the problem number would be other prime, but it's never been in the real thing.
The number line is fractal it seems in this sense,
1 = infinity.
1 can never be reached, if starting from 0, likewise .0002 can never be reached, because .000000000001...
But the decimals from 0 to 1 are the same problem as the infinity from 1 to 99999999...
Unless we consider that the number line need not be a type rope to 2d walk on from left to right, but a 3d object to look at, and maybe as the quantities go up left to right, in the depth of your looking ahead, the infinite decimals at each point between them head off infinitly past the horizon.
Is it as simple as 10 cannot be cut into 3 equal parts using base ten number system?
This problem would be different if we were asking about how many times 3 goes into 9, but what perfect numbers are involved with 13 divided by 7.
The very idea is the uneveness
Is the frustration that this .999... Situation is as close to possibly a number flushly going into another though not?
This thread is now about infinite decimals with signed digits.
We will work in base ten but extend the set of digits to include negative digits:
{-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Negative digits will be denoted with a bar of the digit, like so:
[math]{bar{9}, bar{8}, bar{7}, bar{6}, bar{5}, bar{4}, bar{3}, bar{2}, bar{1}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}[/math]
You can write [math]bar{0}[/math] instead of 0 if you want; they mean the same thing, but we will usually not need to.
Now you can't actually write an infinite decimal. When we speak of infinite decimals, what we mean is a process which generates a stream of digits which continues for as long as we want it to. We can take these streams and perform operations on them to make new streams, and this is how we will do arithmetic with infinite decimals.
It is critical that we understand the meaning of an initial portion of an infinite decimal, since this is all we will ever actually have.
Consider an infinite decimal starting with the digits "3.14". What are the possible values for this number?
The largest possible value is if all the remaining digits are 9, that is, 3.14[9, repeating] = 3.15.
The smallest possible value is if all the remaining digits are [math]bar{9}[/math], that is, 3.14[[math]bar{9}[/math], repeating] = 3.13.
As you can see, there is always an uncertainty of [math]pm 1[/math] in the place value of the last digit.
Let's try adding two numbers. We start at the left-hand side since it is clearly impossible to start at the right.
3.??? + 1.???
If we add the digits 3 and 1, we get 4. But this result has an uncertainty of [math]pm 2[/math], so we cannot include the digit 4 in the output stream at this point. We will need to request more digits of the addends.
After requesting more digits, we have:
3.1??? + 1.9???
We add the 1 and 9 in the tenths places together, obtaining 10 tenths. We must carry a 1 to the ones place, changing the 4 there to a 5.
Our provisional result is 5.0 and our uncertainty on this result is [math]pm 0.2[/math]. There is too much uncertainty to output 5.0, but we can output 5, since we know the true answer is between 4.8 and 5.2 inclusive, and thus between 4 and 6 inclusive. We send the digit 5 to the output stream. This digit is now final and cannot be changed by future calculations.
To proceed further, we must request more digits of the addends. We are given:
3.14??? + 1.95???
Adding the digits in the hundredths places, we obtain 9. Our provisional result is now 5.09 with an uncertainty of [math]pm 0.02[/math]. In other words, we know the true answer is between 5.07 and 5.11 inclusive. Since we do not know that the answer is between 4.9 and 5.1 inclusive, we cannot output a "0" for the tenths place. But we can adjust our answer by subtracting 10 from the hundredths place and adding 1 to the tenths place, obtaining [math]5.1bar{1}[/math]. Since we do know the answer is between 5.0 and 5.2 inclusive, we can output a "1" in the tenths place. The part of the answer that is now final is 5.1.
Let's go further.
3.141??? + 1.958???
Our new provisional result is [math]5.1bar{1}9 pm 0.002[/math]. Since this result does not allow us to output [math]bar{1}[/math] as a final digit, we change it to [math]5.10bar{1} pm 0.002[/math]. We can now output a "0" in the hundredths place, and the part of our answer that is now final is 5.10.
Further:
3.1415??? + 1.9584???
Our new provisional result is [math]5.10bar{1}9 pm 0.0002[/math]. We change it to [math]5.100bar{1} pm 0.0002[/math] and output a "0" in the thousandths place. The final part of the answer is now 5.100.
Further:
3.14151??? + 1.95841???
Our new provisional result is [math]5.100bar{1}2 pm 0.0002[/math]. We output a "[math]bar{1}[/math]" in the 1/10,000s place. The final part of the answer is now [math]5.100bar{1}[/math].
One thing to note is that if we're adding the numbers algorithmically, we don't get to use our knowledge of how the pattern continues, only the information supplied by the digits themselves.
first stage:
0.??? + 0.??? = 0 [math]pm[/math] 2
0.??? + 0.??? = 0 [math]pm[/math] 2
not enough info to output a digit yet
next:
0.6??? + 0.3??? = 0.9 [math]pm[/math] 0.2 = [math]1.bar{1} pm[/math] 0.2
0.9??? + 0.0??? = 0.9 [math]pm[/math] 0.2 = [math]1.bar{1} pm[/math] 0.2
we can now output a "1" for the ones place, final part of answer so far: 1
0.66??? + 0.33??? = [math]1.bar{1}9 pm[/math] 0.02 = [math]1.0bar{1} pm[/math] 0.02
0.99??? + 0.00??? = [math]1.bar{1}9 pm[/math] 0.02 = [math]1.0bar{1} pm[/math] 0.02
we can now output a "0" for the tenths place, final part of answer so far: 1.0
0.666??? + 0.333??? = [math]1.0bar{1}9 pm[/math] 0.002 = [math]1.00bar{1} pm[/math] 0.002
0.999??? + 0.000??? = [math]1.0bar{1}9 pm[/math] 0.002 = [math]1.00bar{1} pm[/math] 0.002
we can now output a "0" for the hundredths place, final part of answer so far: 1.00
This pattern will continue. For both problems we compute 1.[0, repeating].
What's that N1 represent, explain that please
There's an aleph0 of digits in the binary representation of 0.9999999
Therefore 1-0.999999 = 1/2^aleph0
Or 1/aleph1
In the surreal numbers there is indeed an infinitesimal between 0.999... and 1.
Only if ... falls short of absolutely infinite.
Transfer principle.
What you mean?
under R, 1/inf=0
Can you rig 1/inf=0 into multiplication to see if it stands
no, since inf/inf and 0*inf are both undefined
if 1/inf = 0 then 0 · inf = 1
Or lets simplify it and call it just value (v)
v/inf = 0
0 · inf = v
Yea, seems untrue
>uses inf/inf
nope
no, since inf/inf and 0*inf are both undefined.
Can't get from 1/inf = 0 to 0 · inf = 1 using algebraic manipulation without bumping into inf/inf
>Can't get from 1/inf = 0 to 0 · inf = 1 using algebraic manipulation without bumping into inf/inf
If i have 666/69 = 9,652..
then i can 9,652.. x 69 = 666
>Simple as
So what you really mean is 666/69 = 9,653
>If i have 666/69 = 9,652..
>then i can 9,652.. x 69 = 666
now do it using algebraic manipulation, moron
can't avoid 69/69
You can successfully divide one thing into 3 equal parts easily
Numbers aren’t actually real
#
>You can successfully divide one thing into 3 equal parts easily
Let's say a cake was 10 inches long, and you wanted to cut it into 3 even pieces;
You take out your tape measure; what are the 2 numbers you would make your cuts on?
maybe I should get a tape measure in base 3, that way I can cut the 12 inches in equal parts 1.2 inches.
this shit ain't hard. There's nothing special about numbers not divisible by 2 or 5(Or 1 divided by powers of 2 or 5 multiplied), it's just our base being 2*5. In base 12 the problem number would be other prime, but it's never been in the real thing.
>You take out your tape measure; what are the 2 numbers you would make your cuts on?
Cut number 1 and Cut number 2
wiseguy
anon i don't think a tape measure is appropriate for cutting a cake
not a sensical thing to write
benis ins bagine
The number line is fractal it seems in this sense,
1 = infinity.
1 can never be reached, if starting from 0, likewise .0002 can never be reached, because .000000000001...
But the decimals from 0 to 1 are the same problem as the infinity from 1 to 99999999...
Unless we consider that the number line need not be a type rope to 2d walk on from left to right, but a 3d object to look at, and maybe as the quantities go up left to right, in the depth of your looking ahead, the infinite decimals at each point between them head off infinitly past the horizon.
Is it as simple as 10 cannot be cut into 3 equal parts using base ten number system?
This problem would be different if we were asking about how many times 3 goes into 9, but what perfect numbers are involved with 13 divided by 7.
The very idea is the uneveness
Is the frustration that this .999... Situation is as close to possibly a number flushly going into another though not?
is math about measuring or pretend
Is it possible for a number to be unable to be divided into 3 exactly equal parts?
Or the possible infinities of decimals always let's you find the equal parts?
no
This thread is now about infinite decimals with signed digits.
We will work in base ten but extend the set of digits to include negative digits:
{-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Negative digits will be denoted with a bar of the digit, like so:
[math]{bar{9}, bar{8}, bar{7}, bar{6}, bar{5}, bar{4}, bar{3}, bar{2}, bar{1}, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}[/math]
You can write [math]bar{0}[/math] instead of 0 if you want; they mean the same thing, but we will usually not need to.
Now you can't actually write an infinite decimal. When we speak of infinite decimals, what we mean is a process which generates a stream of digits which continues for as long as we want it to. We can take these streams and perform operations on them to make new streams, and this is how we will do arithmetic with infinite decimals.
It is critical that we understand the meaning of an initial portion of an infinite decimal, since this is all we will ever actually have.
Consider an infinite decimal starting with the digits "3.14". What are the possible values for this number?
The largest possible value is if all the remaining digits are 9, that is, 3.14[9, repeating] = 3.15.
The smallest possible value is if all the remaining digits are [math]bar{9}[/math], that is, 3.14[[math]bar{9}[/math], repeating] = 3.13.
As you can see, there is always an uncertainty of [math]pm 1[/math] in the place value of the last digit.
Let's try adding two numbers. We start at the left-hand side since it is clearly impossible to start at the right.
3.??? + 1.???
If we add the digits 3 and 1, we get 4. But this result has an uncertainty of [math]pm 2[/math], so we cannot include the digit 4 in the output stream at this point. We will need to request more digits of the addends.
After requesting more digits, we have:
3.1??? + 1.9???
We add the 1 and 9 in the tenths places together, obtaining 10 tenths. We must carry a 1 to the ones place, changing the 4 there to a 5.
Our provisional result is 5.0 and our uncertainty on this result is [math]pm 0.2[/math]. There is too much uncertainty to output 5.0, but we can output 5, since we know the true answer is between 4.8 and 5.2 inclusive, and thus between 4 and 6 inclusive. We send the digit 5 to the output stream. This digit is now final and cannot be changed by future calculations.
To proceed further, we must request more digits of the addends. We are given:
3.14??? + 1.95???
Adding the digits in the hundredths places, we obtain 9. Our provisional result is now 5.09 with an uncertainty of [math]pm 0.02[/math]. In other words, we know the true answer is between 5.07 and 5.11 inclusive. Since we do not know that the answer is between 4.9 and 5.1 inclusive, we cannot output a "0" for the tenths place. But we can adjust our answer by subtracting 10 from the hundredths place and adding 1 to the tenths place, obtaining [math]5.1bar{1}[/math]. Since we do know the answer is between 5.0 and 5.2 inclusive, we can output a "1" in the tenths place. The part of the answer that is now final is 5.1.
Let's go further.
3.141??? + 1.958???
Our new provisional result is [math]5.1bar{1}9 pm 0.002[/math]. Since this result does not allow us to output [math]bar{1}[/math] as a final digit, we change it to [math]5.10bar{1} pm 0.002[/math]. We can now output a "0" in the hundredths place, and the part of our answer that is now final is 5.10.
Further:
3.1415??? + 1.9584???
Our new provisional result is [math]5.10bar{1}9 pm 0.0002[/math]. We change it to [math]5.100bar{1} pm 0.0002[/math] and output a "0" in the thousandths place. The final part of the answer is now 5.100.
Further:
3.14151??? + 1.95841???
Our new provisional result is [math]5.100bar{1}2 pm 0.0002[/math]. We output a "[math]bar{1}[/math]" in the 1/10,000s place. The final part of the answer is now [math]5.100bar{1}[/math].
Exercises for the reader:
Compute 0.[6, repeating] + 0.[3, repeating].
Compute 0.[9, repeating] + 0.[0, repeating].
1,1
0.[9, repeating]
0.[9, repeating]
One thing to note is that if we're adding the numbers algorithmically, we don't get to use our knowledge of how the pattern continues, only the information supplied by the digits themselves.
first stage:
0.??? + 0.??? = 0 [math]pm[/math] 2
0.??? + 0.??? = 0 [math]pm[/math] 2
not enough info to output a digit yet
next:
0.6??? + 0.3??? = 0.9 [math]pm[/math] 0.2 = [math]1.bar{1} pm[/math] 0.2
0.9??? + 0.0??? = 0.9 [math]pm[/math] 0.2 = [math]1.bar{1} pm[/math] 0.2
we can now output a "1" for the ones place, final part of answer so far: 1
0.66??? + 0.33??? = [math]1.bar{1}9 pm[/math] 0.02 = [math]1.0bar{1} pm[/math] 0.02
0.99??? + 0.00??? = [math]1.bar{1}9 pm[/math] 0.02 = [math]1.0bar{1} pm[/math] 0.02
we can now output a "0" for the tenths place, final part of answer so far: 1.0
0.666??? + 0.333??? = [math]1.0bar{1}9 pm[/math] 0.002 = [math]1.00bar{1} pm[/math] 0.002
0.999??? + 0.000??? = [math]1.0bar{1}9 pm[/math] 0.002 = [math]1.00bar{1} pm[/math] 0.002
we can now output a "0" for the hundredths place, final part of answer so far: 1.00
This pattern will continue. For both problems we compute 1.[0, repeating].