I have been reading this Lang article about references and have a couple of questions:
1. The part about crossing diagonals to find quadratic surd-based references starts on page 24. At the bottom of page 33 Lang mentions "dividing fraction y by a factor b" to aid with this technique, and I haven't been able to make any sense of it. How exactly is this used? I may be unclear on exactly what is being multiplied by b, but I have failed to see the benefit or reasoning of this technique.
2. Do you guys know of any other articles or learning materials concerning this kind of more advanced reference-finding?
2 Referencing Questions
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Re: 2 Referencing Questions
The answer, from me, to question 2 would be no. Sorry.
Question 1, though...
As far as my understanding (based on a quick read of pages 24-33) are as follows;
You're looking for a way of finding some irrational number in the form of a+b√2
The table/Figure 24 on page 33 lets you find a few common surds and their reciprocals (despite them being labelled as "1-y" instead of "1/y").
However, if you have a fraction with a multiple of √2 more than 1, e.g. 3+7√2, they don't work as they are.
Assuming you've got the length of the side of your square = x(3+7√2) = w(1/y + 1/z)
And you take x=w, then you have 3+7√2 = 1/y + 1/z
In order to make things simpler, let's take 3+7√2 = ((10) + (7√2-7))
So, 1/y = 10 and 1/z = 7√2-7 = 7(√2-1)
y = 1/10, z = (1/(7(√2-1))) = (1/7)(1/(√2-1)) = (1/7)(√2+1)
I assume that this means you're supposed fold the diagonal to find the (√2+1) and then divide into 7ths to give you the (1/7)(√2+1) for z. Fold the other side into 10ths to get y and the intersection is where it should be.
Question 1, though...
As far as my understanding (based on a quick read of pages 24-33) are as follows;
You're looking for a way of finding some irrational number in the form of a+b√2
The table/Figure 24 on page 33 lets you find a few common surds and their reciprocals (despite them being labelled as "1-y" instead of "1/y").
However, if you have a fraction with a multiple of √2 more than 1, e.g. 3+7√2, they don't work as they are.
Assuming you've got the length of the side of your square = x(3+7√2) = w(1/y + 1/z)
And you take x=w, then you have 3+7√2 = 1/y + 1/z
In order to make things simpler, let's take 3+7√2 = ((10) + (7√2-7))
So, 1/y = 10 and 1/z = 7√2-7 = 7(√2-1)
y = 1/10, z = (1/(7(√2-1))) = (1/7)(1/(√2-1)) = (1/7)(√2+1)
I assume that this means you're supposed fold the diagonal to find the (√2+1) and then divide into 7ths to give you the (1/7)(√2+1) for z. Fold the other side into 10ths to get y and the intersection is where it should be.
Last edited by Falcifer on August 7th, 2012, 6:53 pm, edited 1 time in total.