Courant and Fritz, Constructing the real numbers

[GodfreyHW]

In chapter 1, page 10, real numbers are found by confining them to an interval that shrinks to "zero" length. Basically, if [MATH]x[/MATH] is between [MATH]c[/MATH] and [MATH]c+1[/MATH], then we can divide that interval into ten subintervals, and we can, then, have [MATH]c+\frac{1}{10}c_1\leq x\leq c+\frac{1}{10}c_1+\frac{1}{10}[/MATH], where [MATH]c_1[/MATH] is a digit from zero to nine.

Repeating this process, and making [MATH]n\to\infty[/MATH] subdivisions, we'll eventually get [MATH]x=c+0.c_1c_2c_3...[/MATH]
I am confused by this, though:

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1) How is he constructing that [MATH]I_{n+1}[/MATH] interval?
If I suppose [MATH]c_0-1\leq x[/MATH], then I can do [MATH]c_0-1+\frac{1}{10}c_1+...+\frac{1}{10^n}c_n\leq x[/MATH], and if I say that [MATH]n\to\infty[/MATH], then I can also write [MATH]\underbrace{(c_0-1+\frac{1}{10}c_1+...+\frac{1}{10^n}c_n)}_x-\frac{1}{10^{n+1}}\leq x[/MATH], so that [MATH]\frac{1}{10^{n+1}}[/MATH] is like an infinitesimal.
Would this be correct for [MATH]\left[x-\frac{1}{10^{n+1}},x\right][/MATH]? It should be the same with [MATH]x\leq c_0+1[/MATH] by subtracting [MATH]\frac{c_i}{10^i}[/MATH]s and adding [MATH]\frac{1}{10^{n+1}}[/MATH].

2) How is he getting the second representation with [MATH]c_n-1[/MATH]? (typo?)
I see that, maybe, we'd do [MATH]c_0\leq x[/MATH], and get it to [MATH]c_0+\frac{1}{10}c_1+...+\frac{1}{10^n}c_n\leq x[/MATH], but we would have [MATH]c_{i>0}=9[/MATH].

Thank you for your time.

 

[JeffM]

It does seem to me that the text is obscure.

They are writing generally about a class of numbers. As I tell my students in first year elementary algebra, if a general statement about numbers seems mysterious, restate it in terms of a specific number. In this case, they start by dealing with integers. In the next to the last sentence, they suddenly introduce a new class of numbers, namely those rational numbers that can be expressed as an integer over a power of 10. (Perhaps this makes sense in context, but it comes out of the blue for me.) So I would take this in two stages, first for an integer and second for a rational in the proper class, say 6 and then for 3/500 = 6/1000 = 6/10^3.

If x is 6, we could choose c_0 as 6 or as 6 - 1 = 5. If we choose c_0 = 6, our first interval is

[MATH][6,\ 7] \implies \left [6 + \dfrac{0}{10},\ 7 - \dfrac{9}{10} \right] \equiv [6,\ 6.1] \implies\\ \left [6 + \dfrac{0}{10} + \dfrac{0}{100},\ 7 - \dfrac{9}{10} - \dfrac{9}{100} \right] \equiv [6.0,\ 6.01] \text { and so on.}[/MATH]If we choose c_0 = 5, our first interval is

[MATH][5,\ 6] \implies \left [5 + \dfrac{9}{10},\ 6 - \dfrac{0}{10} \right] \equiv [5.9,\ 6] \rightarrow \\ \left [5 + \dfrac{9}{10} + \dfrac{9}{100},\ 6 - \dfrac{0}{10} - \dfrac{0}{100} \right] \equiv [5.99,\ 6] \text { and so on.}[/MATH] So far, I think everything is straightforward. What I think happens is that in the middle of the paragraph, they stop talking about integers and start talking about this special class of rationals that they do not mention until the end of the paragraph.

See if the text makes sense if you follow through with the example of 3/500 = 6/1000. If it does, try a more complex rational of the same type, perhaps 37/125. Hope this helps.

 

[GodfreyHW]

It does seem to me that the text is obscure.

They are writing generally about a class of numbers. As I tell my students in first year elementary algebra, if a general statement about numbers seems mysterious, restate it in terms of a specific number. In this case, they start by dealing with integers. In the next to the last sentence, they suddenly introduce a new class of numbers, namely those rational numbers that can be expressed as an integer over a power of 10. (Perhaps this makes sense in context, but it comes out of the blue for me.) So I would take this in two stages, first for an integer and second for a rational in the proper class, say 6 and then for 3/500 = 6/1000 = 6/10^3.

If x is 6, we could choose c_0 as 6 or as 6 - 1 = 5. If we choose c_0 = 6, our first interval is

[MATH][6,\ 7] \implies \left [6 + \dfrac{0}{10},\ 7 - \dfrac{9}{10} \right] \equiv [6,\ 6.1] \implies\\ \left [6 + \dfrac{0}{10} + \dfrac{0}{100},\ 7 - \dfrac{9}{10} - \dfrac{9}{100} \right] \equiv [6.0,\ 6.01] \text { and so on.}[/MATH]If we choose c_0 = 5, our first interval is

[MATH][5,\ 6] \implies \left [5 + \dfrac{9}{10},\ 6 - \dfrac{0}{10} \right] \equiv [5.9,\ 6] \rightarrow \\ \left [5 + \dfrac{9}{10} + \dfrac{9}{100},\ 6 - \dfrac{0}{10} - \dfrac{0}{100} \right] \equiv [5.99,\ 6] \text { and so on.}[/MATH]So far, I think everything is straightforward. What I think happens is that in the middle of the paragraph, they stop talking about integers and start talking about this special class of rationals that they do not mention until the end of the paragraph.

See if the text makes sense if you follow through with the example of 3/500 = 6/1000. If it does, try a more complex rational of the same type, perhaps 37/125. Hope this helps.

Thank you for your answer!
For both examples, I'd just consider putting [MATH]x=6[/MATH] and [MATH]x=37\times8=296[/MATH], do the intervals thing until I get the sought after expansion, then divide by [MATH]1000[/MATH] (coincidentally the same number for both cases).
I want to add something, though. I wanted to know how he derived the subdivisions' expression from the inequalities, but you have just applied it.
I think that what I have proposed is correct, but the starting points should be [MATH]x\leq c_0+1[/MATH] and [MATH]c_0\leq x[/MATH].

 

[JeffM]

Glad to have helped. I still find the text obscure. Somewhere in the middle, I am 99% convinced that they stop talking about integers and, without any indication, start talking about this special class of rationals. I suspect that there is some nuance in there as a result, but I have insufficient time to figure out where and what it it is. (Maybe I should say insufficient interest; analysis does not charm me.)