We know C=F, -40. What is |K|=|F|?

Originally posted by sonhouse
I thought I said that then I thought you said you included the infinite series which is why I'm confused.

Included, yes. Written down, not all of them. 🙂

Originally posted by SwissGambit
Included, yes. Written down, not all of them. 🙂

but how does the equation know you haven't written the whole thing down?

Maybe thinking of writing down the infinite series will help.

Take a subtraction of two numbers with repeating decimals:

1.2686868....
-0.1686868....
---------------
What is the result?

Does it change if I write down a bunch more 68's for each one?

Originally posted by SwissGambit
Maybe thinking of writing down the infinite series will help.

Take a subtraction of two numbers with repeating decimals:

1.2686868....
-0.1686868....
---------------
What is the result?

Does it change if I write down a bunch more 68's for each one?

Wouldn't the second number be 0.126868686....?, wait, I get 1.1 with your numbers.

how did you eliminate the digit 2?

I did your original math on the PC calculator and it came out:

164167692.9749999999999571429

Are you just rounding off to the 975?

Originally posted by sonhouse
Wouldn't the second number be 0.126868686....?, wait, I get 1.1 with your numbers.

how did you eliminate the digit 2?

I did your original math on the PC calculator and it came out:

164167692.9749999999999571429

Are you just rounding off to the 975?

I'm working on that stuff. Do you know the history of that trick?

Originally posted by sonhouse
Wouldn't the second number be 0.126868686....?, wait, I get 1.1 with your numbers.

how did you eliminate the digit 2?

I did your original math on the PC calculator and it came out:

164167692.9749999999999571429

Are you just rounding off to the 975?

Yes, it's 1.1 exactly! That's the point. The repeating digits are eliminated by the subtraction. This example was not related to the 'trick' I am describing. It was just to show that repeating digits can be eliminated via subtraction.

-----

No, I didn't round anything off. You are getting 164167692.9749999999999571429 because you aren't using the same number of decimal digits for both numbers when you punch them into the calculator.

Originally posted by sonhouse
I'm working on that stuff. Do you know the history of that trick?

I learned it in grade school math. I don't know who originally thought of it.

Originally posted by sonhouse
I'm working on that stuff. Do you know the history of that trick?

I think its a standard method for working out the sum of a geometric series.

e.g.
Let S = 1/3 + 1/9 + 1/27 + ...
then 3S = 1+ 1/3 + 1/9 + 1/27 + ...

3S - S = 1
therefore S = 1/2

with recurring decimals you just multiply up by the appropriate power of 10

e.g.
Let S = 0.33333333
then 10S = 3.3333333333

10S-S = 3
therefore S=3/9=1/3

Originally posted by wolfgang59
I think its a standard method for working out the sum of a geometric series.

e.g.
Let S = 1/3 + 1/9 + 1/27 + ...
then 3S = 1+ 1/3 + 1/9 + 1/27 + ...

3S - S = 1
therefore S = 1/2

with recurring decimals you just multiply up by the appropriate power of 10

e.g.
Let S = 0.33333333
then 10S = 3.3333333333

10S-S = 3
therefore S=3/9=1/3

How can that be a standard when it does not give valid results? If I go S=.33 and multiply it by 10, and go 10S-S , therefore 3/9 or 3 when 3 most distinctly does not =0.33? The infinite series 0.333333333......ends up equaling 3 sure, but .33 misses being 3 by 0.003333333333...., a real and positive number. So how can anyone use that formula?

Originally posted by sonhouse
How can that be a standard when it does not give valid results? If I go S=.33 and multiply it by 10, and go 10S-S , therefore 3/9 or 3 when 3 most distinctly does not =0.33? The infinite series 0.333333333......ends up equaling 3 sure, but .33 misses being 3 by 0.003333333333...., a real and positive number. So how can anyone use that formula?

You are still confused. You should not 'go S=.33' but rather 'S=.33...' - the repeating decimals are not thrown out. Of course that would lead to an invalid result.

When you enter things into a calculator, you obviously can't punch in an infinite number of 3's, so you must realize that the repeating digits will cancel out and thus enter the subtraction such that that happens.

For example, you can enter:
3.33 - 0.33 or
3.3333 - 0.3333 or
3.3 - 0.3 etc.
but not 3.33 - 0.333, because the 3's at the ends will not cancel out.

A generic version may.. or may not.. be helpful.

The numbers that that deals with are of the form

x = a + b (1 + c + c^2 + c^3 + c^4 + ...) = a + b + bc + bc^2 + bc^3 + ...

For example, 1.283434343434.. = 1.28 + 0.0034 x (1 + 0.01^1 + 0.01^2 + ...)
with a = 1.28, b = 0.0034, c = 0.01.

if you multiply the whole thing with c, you get

cx = c (a + b (1 + c + c^2 + c^3 + c^4 + ...) = ac + bc + bc^2 + bc^3 + ...

Now subtract x from cx.

cx - x = (a + b + bc + bc^2 + bc^3 + ... ) - (ac + bc + bc^2 + bc^3 + ... )
= a + b - ac

because everything else is both added and subtracted, so it cancels out.

cx - x = a + b - ac
x (c - 1) = a + b - ac
x = ( a + b - ac ) / (c - 1)

So, every fraction that has a repeating sequence can be converted to the form of having one integer divided by another. For example;


x = 1.28343434 ...
0.01 x = 0.128343434 ...
x - 0.01 x = (1.2834 + 0.000034343434....) - (0.0128 + 0.000034343434....)
0.99 x = 1.2834 - 0.0128 + (0.000034343434.... - 0.000034343434....)
0.99 x = 1.2706 + 0
x = 1.2706 / 0.99 = 12706 / 9900

Originally posted by talzamir
A generic version may.. or may not.. be helpful.

The numbers that that deals with are of the form

x = a + b (1 + c + c^2 + c^3 + c^4 + ...) = a + b + bc + bc^2 + bc^3 + ...

For example, 1.283434343434.. = 1.28 + 0.0034 x (1 + 0.01^1 + 0.01^2 + ...)
with a = 1.28, b = 0.0034, c = 0.01.

if you multiply the whole thing with c, you get

cx = c ( ...[text shortened]... (0.000034343434.... - 0.000034343434....)
0.99 x = 1.2706 + 0
x = 1.2706 / 0.99 = 12706 / 9900

Thanks for the tip. Interestingly enough, I left my scientific calculator at home ( have to make several kinds of calculations including times in seconds of several different kinds of events requiring the calculation of various duty cycles like air compressor ones I did this morning) and there is a Staples a few blocks from work so stopped in and got another one, they are only about 20 bucks for the high end scientific calculators, as opposed to the low end programmables, anyway, the one I got was the Casio Fx-115 PLUS, and reading through the manual I came across a new entry I hadn't seen before in this level of calculator: Recurring Decimal Calculations.

It seems they allow for doing the 0.33.... thing you talked about as opposed to the 0.33 I was writing down, where you were pointing out the difference. So now you can input say, 2.23147321473214732... as 2.23 with the added 14732 with a line over it designating that part as a recurring decimal.

Funny how I started this line of questions and ended up actually learning something new and finding out a calculator can now do those kind of calculations directly without having to input an infinite number of recurrences!

In the next few pages of the manual it goes in to greater detail and I found this interesting fraction they used as an illustration:

123/999=41/333=0.123 bar (don't know how to draw the line over the top of the 123 part here at RHP so I call it bar.

1234/9999=0.1234 bar

12345/99999=0.12345 bar

123456/999999=0.123456 bar.

My goodness, there is no need to go all decimal, and then struggle to to get back to fractions!

(x-273.15)*1.8+32=-x

1.8x - 1.8*273.15 + 32 = -x

1.8*273.15 - 32 = 2.8x

now turn those to fractions
18/10 * 27315/100 - 32 = 28x/10

18 * 27315/100 - 320 = 28x

x = 18*27315 /(100*28) - 320/28

make the bases the same and add the fractions

x = (18*27315 - 320*100)/(100*28)

x = 459670/(100*28)

cancel out factor of 10

x = 45967/280

this is the simplest form because 45967 is not divisible by 2, 5 or 7 which are the factors of 280, however we can convert it to a mixed fraction if we like

x = 164+47/280

this would be an exact number (164.47) in base 280, the 47 there represents the 47th unit in that base and the 164 is the 164th unit.

Originally posted by iamatiger
My goodness, there is no need to go all decimal, and then struggle to to get back to fractions!

(x-273.15)*1.8+32=-x

1.8x - 1.8*273.15 + 32 = -x

1.8*273.15 - 32 = 2.8x

now turn those to fractions
18/10 * 27315/100 - 32 = 28x/10

18 * 27315/100 - 320 = 28x

x = 18*27315 /(100*28) - 320/28

make the bases the same and add the fractions
...[text shortened]... ) in base 280, the 47 there represents the 47th unit in that base and the 164 is the 164th unit.

Who's struggling? My computer reduced that fraction in less than a second. 😉

Originally posted by iamatiger
My goodness, there is no need to go all decimal, and then struggle to to get back to fractions!

(x-273.15)*1.8+32=-x

1.8x - 1.8*273.15 + 32 = -x

1.8*273.15 - 32 = 2.8x

now turn those to fractions
18/10 * 27315/100 - 32 = 28x/10

18 * 27315/100 - 320 = 28x

x = 18*27315 /(100*28) - 320/28

make the bases the same and add the fractions
...[text shortened]... ) in base 280, the 47 there represents the 47th unit in that base and the 164 is the 164th unit.

I already had found that easier path. See post 4 on page 1:

(x-273.15)*1.8+32=-x
1.8x -491.67+32=-x
-459.67=-2.8x
x=164.1678571

Go back to:

-459.67=-2.8x

Multiply both sides by 100:

-45967=-280x
x=45967/280

But I was showing sonhouse a general method for converting ANY number with a repeating decimal to a fraction. I just happened to use that number as an example.