# Thread: Find an "interesting" Cauchy sequence.

1. ## Find an "interesting" Cauchy sequence.

For a rational number $\displaystyle x \neq 0,$ write it as $\displaystyle 2^k\frac{m}{n},$ where $\displaystyle m$ and $\displaystyle n$ are integers that have no common factors, and let $\displaystyle |x|_{2} = 2^{-k}.$ Define $\displaystyle |0|_{2} = 0.$ Let $\displaystyle d(x,y) = |x-y|_{2}.$

Prove $\displaystyle d(x,z) \leq d(x,y) + d(y,z)$ for all rational $\displaystyle x, y, z.$

Now since $\displaystyle d(x,y) = d(y,x)$ and $\displaystyle d(x,y) \geq 0$ with equality only if $\displaystyle x = y.$ Write an interesting Cauchy sequence in this metric.

2. Originally Posted by cgiulz
For a rational number $\displaystyle x \neq 0,$ write it as $\displaystyle 2^k\frac{m}{n},$ where $\displaystyle m$ and $\displaystyle n$ are integers that have no common factors, and let $\displaystyle |x|_{2} = 2^{-k}.$ Define $\displaystyle |0|_{2} = 0.$ Let $\displaystyle d(x,y) = |x-y|_{2}.$

Prove $\displaystyle d(x,z) \leq d(x,y) + d(y,z)$ for all rational $\displaystyle x, y, z.$

Now since $\displaystyle d(x,y) = d(y,x)$ and $\displaystyle d(x,y) \geq 0$ with equality only if $\displaystyle x = y.$ Write an interesting Cauchy sequence in this metric.

For positive integers m,n we have that m > n <==> 2^(-m) < 2^(-n) , so

|x + y|_2 <= max{|x|_2 , |y|_2} <= |x|_2 + |y|_2 and from here the

triangle inequality follows.

As for an interesting Cauchy sequence: what about the sequence

{2^n} = {2, 4, 8, 16,...} ?

Tonio