I really don't understand how to do proofs on convergence at ALL.
I know you're supposed to use
$|x_i - x| < \epsilon$
but I have no idea how to apply this to this question:

Show that if $x$ is any real number then there is an infinite sequence of rational numbers converging to $x$.

Any hints or help will be greatly appreciated!


Are you familiar with the density of the rationals? If so,

For every $n\in \mathbb{N}$, let $x_n$ be a rational in the open ball $B(x,\frac{1}{n})$. By the density of the rationals, there is always one such a $x_n\in\mathbb{Q}$. Then for any $\epsilon>0$, there exists $N$ such that $\frac{1}{N}<\epsilon$ by the Archimedean property. Then, for all $n>N$, $x_n \in B(x,\frac{1}{N})\subset B(x,\epsilon),$ meaning $|x_n -x|< \epsilon$.


Assume $x>0$ and irrational. If $x$ is rational, define $x_n:=x$ for all $n$.

Let $a_n:=\left\lfloor x10^n\right\rfloor$ and define $b_n:= \left\lfloor 10^{n} x-10a_{n-1}\right\rfloor$. And finally $x_n:=\sum\limits_{k=0}^{n} \frac {b_k}{10^k}$. Basically I'm truncating $x$ to $n$ decimals at the $n$'th term in the sequence.

The breakdown:

We will write $x=b_0\cdot b_1 b_2 b_3 \ldots\;$, where $b_i \in\{0,1,2,\ldots,9\}$ for each $i \in \mathbb{N}$ and $b_0 \in \Bbb{N}_0$, with the convention that there does not exist an $N \in \Bbb{N}_0$ such that for all $i> N$ we have $b_i=9$. We do this to rule out ambiguities like $1=0.\dot{9}$

How you might write $x$ as a decimal is kinda the above method unless you know something special about the number. We want the $n$'th digit so we compute $a_{n-1}$ and $\lfloor 10^n x \rfloor$, and plug it into $b_n$ above. ($b_n$ also equal $\lfloor 10^n x\rfloor-10a_{n-1}$ because the second term is an integer less than or equal to the first term)

To show this works, suppose we have the decimal expansion as above.

$a_n=\left\lfloor x10^n\right\rfloor=\left\lfloor b_0 b_1 b_2 b_3 \ldots b_n\cdot b_{n+1}b_{n+2}\ldots\; \right\rfloor=b_0 b_1 b_2 b_3 \ldots b_n$ as a decimal number, not multiplied.

Then we have that

$\left\lfloor 10^{n} x-10a_{n-1}\right\rfloor=\left\lfloor b_0 b_1 b_2 b_3 \ldots b_n\cdot b_{n+1}b_{n+2}\ldots\;-b_0 b_1 b_2 b_3 \ldots b_{n-1} 0\right\rfloor=\left\lfloor b_n\cdot b_{n+1}b_{n+2}\ldots\;\right\rfloor$

which is just $b_n$. So we now know the $n$'th digit, so we more or less say what we started with, that $$x=\sum\limits_{k=0}^{\infty} \frac {b_k}{10^k}=b_0\cdot b_1 b_2 b_3 \ldots\;$$

and we just take the partial sums, which are finite sums of rationals and so are rational:

$$x_n:=\sum\limits_{k=0}^{n} \frac {b_k}{10^k}=b_0\cdot b_1 b_2 b_3 \ldots b_n\dot0\;$$

Also then $|x-x_n|<\large\frac{1}{10^n}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.