Suspiciously precise floats, or, how I got Claude's real limits
1–7 of 7 posts
Re: Suspiciously precise floats, or, how I got Claude's real limits
#2Neat analysis.
Although noticing the repeated pattern of a multiple of 9 in the fraction 0.16327272727272726 naturally suggests multiplying by 11, and then we get the much simpler value 1.796, at which point it's much easier to continue. I wouldn't have broken out a general analysis method for this, although it's neat to know that they exist.
Re: Suspiciously precise floats, or, how I got Claude's real limits
#3Neat analysis. Although noticing the repeated pattern of a multiple of 9 in the fraction 0.16327272727272726 naturally suggests multiplying by 11, and then we get the much simpler value 1.796, at which point it's much easier to continue. I wouldn't have broken out a general analysis method for this, although it's neat to know that they exist.
I think the standard way to convert repeating decimals or decimals that appear to have a certain repeating pattern to fractions is to take the first repeating period and divide by 0.999.. with the number of 9s matching the period length.
0.163272727.. = 0.163+0.00027/0.99 = 163/1000+27/99000 = 449/2750
Re: Suspiciously precise floats, or, how I got Claude's real limits
#4Neat analysis. Although noticing the repeated pattern of a multiple of 9 in the fraction 0.16327272727272726 naturally suggests multiplying by 11, and then we get the much simpler value 1.796, at which point it's much easier to continue. I wouldn't have broken out a general analysis method for this, although it's neat to know that they exist.
> naturally suggests multiplying by 11
Is this a named concept that I can learn about?
Re: Suspiciously precise floats, or, how I got Claude's real limits
#5Neat analysis. Although noticing the repeated pattern of a multiple of 9 in the fraction 0.16327272727272726 naturally suggests multiplying by 11, and then we get the much simpler value 1.796, at which point it's much easier to continue. I wouldn't have broken out a general analysis method for this, although it's neat to know that they exist.
> naturally suggests multiplying by 11 Is this a named concept that I can learn about?
Not sure if this helps but check it out
https://mathcentral.uregina.ca/QQ/database/QQ.09.07/h/jack1....
Re: Suspiciously precise floats, or, how I got Claude's real limits
#6Neat analysis. Although noticing the repeated pattern of a multiple of 9 in the fraction 0.16327272727272726 naturally suggests multiplying by 11, and then we get the much simpler value 1.796, at which point it's much easier to continue. I wouldn't have broken out a general analysis method for this, although it's neat to know that they exist.
> naturally suggests multiplying by 11 Is this a named concept that I can learn about?
The repeating pattern has 2 digits, and is a multiple of 9. 9 times 11 is 99, so that multiplication gives a "repeating pattern of a multiple of 99" — equivalently, a multiple of a repeating pattern of 99. And I assume you're familiar with the concept of .99999... equaling 1.
Re: Suspiciously precise floats, or, how I got Claude's real limits
#7Neat analysis. Although noticing the repeated pattern of a multiple of 9 in the fraction 0.16327272727272726 naturally suggests multiplying by 11, and then we get the much simpler value 1.796, at which point it's much easier to continue. I wouldn't have broken out a general analysis method for this, although it's neat to know that they exist.
I think the standard way to convert repeating decimals or decimals that appear to have a certain repeating pattern to fractions is to take the first repeating period and divide by 0.999.. with the number of 9s matching the period length. 0.163272727.. = 0.163+0.00027/0.99 = 163/1000+27/99000 = 449/2750
(This works because x/9 = 0.xxxx..., xy/99 = 0.xyxyxy... and so on). And that is true intuitively because when you long divide in order to get a repeating pattern you need the remainder to be the same as what you started with. I.e if you long divide
0.n
-----
a| b.0
Or going the other way, if d = 10^n - 1 then [10 a = a (mod d)] so your remainders never change. And then note that
a * 10^n = a * (10^n - 1) + a