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## Solutions to TRTR Ch. 3 (Kinds of number in the physical world) 3.1-3.4


	
*NOTE**: these are my personal solutions. I have made them as logically sound as I can but they may still contain mistakes. Also there might be alternative ways of solving a problem. I will try to point those out whenever I am aware.


### 3.1 (easy)

The first two expansions have been thoroughly described in the text.

The integers that appear in the continued fraction expansion of $\pi$ form the sequence [A001203](https://oeis.org/A001203) in OEIS. The OEIS page did not provide any other context where this sequence occurs.

### 3.2 (medium)

The first expansion satisfies $x-1 = [2+(x-1)]^{-1}$ which is equivalent to the quadratic equation $x^2-2=0$. Its two solutions are $x=\pm\sqrt{2}$. Clearly the expansion represents a positive number, so it must be $\sqrt{2}$.

The second expansion satisfies $(x-5)^{-1}-3 = \{\{[(x-5)^{-1}-3]+2\}^{-1}+1\}^{-1}.$ We can first solve for $y=(x-5)^{-1}-3$, which satisfies $y^2+2y-2=0$. This gives us $y=-1\pm\sqrt{3}$, hence $x=(y+3)^{-1}+5=7\mp\sqrt{3}$. Meanwhile the expansion is clearly less than 6, therefore it must be $7-\sqrt{3}$.


	
*NOTE**: There is a construction in knot theory that is closely related to (finite) confinued fraction known as "Conway's square dance".


### 3.3 (medium)

The Eudoxan criterion says $Ma > Nb$ and $Mc < Nd$ for positive integers $M$ and $N$ and positive numbers $a$, $b$, $c$ and $d$. We can immediately deduce $a/b > N/M > c/d$.

### 3.4 (hard)

Given $a:b$ and $c:d$, we are going to define their sum and product using the Eudoxian method.

$(a:b)+(c:d)$ is a number that is greater than $\frac{A'D'+B'C'}{B'D'}$ and smaller than $\frac{A''D''+B''C''}{B''D''}$ for any positive integers $A'$, $B'$, $A''$, etc. satisfying $\frac{A'}{B'}<a:b<\frac{A''}{B''}$, $\frac{C'}{D'}<c:d<\frac{C''}{D''}$.

$(a:b)\times(c:d)$ is a number that is greater than $\frac{A'C'}{B'D'}$ and smaller than $\frac{A''C''}{B''D''}$ for any positive integers $A'$, $B'$, $A''$, etc. satisfying $\frac{A'}{B'}<a:b<\frac{A''}{B''}$, $\frac{C'}{D'}<c:d<\frac{C''}{D''}$.

Because rational numbers are "dense" in the reals, this uniquely nails down $(a:b)+(c:d)$ and $(a:b)\times(c:d)$.