a) As there are irrational numbers in base two, then there exist irrational decimal numbers consisting of only two distinct digits, $p$, $q$, so $\frac{p}{9}$ and $\frac{q}{9}$ are two rational numbers that work. b) Similarly, we let find an irrational number in base $2$, replace all $0$'s with the the string of digits digit $\underbrace{p\dots p}_{2017\ p \textnormal{'s}}$, and $1$'s with a the same string of digits with a second digit $p$ appended on the end ($\underbrace{p\dots p}_{2017\ p \textnormal{'s}}q$). Letting our rational decimal be $0.\overline{p}$, at most $\frac{1}{2018}$ of the digits are different, so these decimals meet this requirement.

I will be a bit sketchy here. We can write $p=0.\overline{p_1p_2p_3 \ldots p_k}$ and $q=0.\overline{q_1q_2q_3 \ldots q_k}$. Consider the set of all numbers of the format $N=n_1n_2 \ldots n_k$ with $n_i \in \{p_i,q_i\}$. We can pick two of them if they are different, in fact we can just choose $P=p_1p_2p_3 \ldots p_k$ and $Q=q_1q_2q_3 \ldots q_k$. Then we can write any irrational number, say $\pi$, in base 2 and write a number I that way: if the $j$th digit of $\pi_2$ is 0, then copy the expansion of $P$, else copy the expansion of $Q$. As the expansion of $\pi_2$ isn't periodical, the same will occur with $I$. We can use the same idea to the second problem: use $P$ and $Q$ with periods of size $k$ differing in exactly one digit, and use the expansion of $\sqrt{\pi}$: $P$ if it is 0, $Q$ otherwise.The number of digits differing between $P$ and $I$ will be at most $\frac{1}{k}$ of the total.

Isn't it "a second digit $q$ appended on the end..."? programjames1 wrote: a) As there are irrational numbers in base two, then there exist irrational decimal numbers consisting of only two distinct digits, $p$, $q$, so $\frac{p}{9}$ and $\frac{q}{9}$ are two rational numbers that work. b) Similarly, we let find an irrational number in base $2$, replace all $0$'s with the the string of digits digit $\underbrace{p\dots p}_{2017\ p \textnormal{'s}}$, and $1$'s with a the same string of digits with a second digit $p$ appended on the end ($\underbrace{p\dots p}_{2017\ p \textnormal{'s}}q$). Letting our rational decimal be $0.\overline{p}$, at most $\frac{1}{2018}$ of the digits are different, so these decimals meet this requirement.

a) $p=0.3333...$ and $q=0.444...$ and $r=0.434334333433334....$ b) take a rational number $q=0.111...$ and take the $2017k_{th}$ indexes of $q$ and put the $k_{th}$ index of $r$ in part a

a) Take the following numbers: $p = 0.111...$ , $q = 0.222...$ and $r = 0.12112111211112...$. It's not hard to see that r is irrational because if it was rational, r would have a period. Let k be the size of this period. Notice that there is a string of 2k ones in decimal representation of r which implies that the period is equal to 1111...11, k times. It's an absurd because there exist 2 in the decimal representation of r. b) Now take $s= 0.9999...$. We will have an irrational number r that is almost like s but there are some strings of ones in it. You will create the irrational number by the following algorithm: Take the number $0.9$ and keep on adding nines to the end of the number until you may add one different digit( according to the conditions of the problem). Then you add 1 instead of 9. Now keep on adding nines until you may add 2 different consecutive digits and then you add 11 instead of 99. After that you keep on adding nines to the end of obtained number until you may add 3 different consecutive digits. Then you add 111 instead of 999 and keep on doing this procedure. The number would be like: $ 0.99...9199...9119...91119...911119....9111119....9... By the same argument of a) this number is irrational and we're done.

Is it true that given any rationals $p \neq q$ between 0 and 1, we can construct an irrational number $\alpha$ in the manner of (a)?

Th3Numb3rThr33 wrote: Is it true that given any rationals $p \neq q$ between 0 and 1, we can construct an irrational number $\alpha$ in the manner of (a)? No. For example: $p=0.1000000=\frac{1}{10}$ and $q=0.3000000=\frac{3}{10}$ Still interesting: Is it true that given any periodic rationals $p \neq q$ between 0 and 1, we can construct an irrational number $\alpha$ in the manner of (a)? With diferent period.

a) consider $\frac{1}{9}$, $\frac{2}{9}$ and $0.121121112\ldots$ b) consider $\frac{1}{9}$ and $0.\underbrace{1\ldots 1}_{2017\ 1 \textnormal{'s}}2\underbrace{1\ldots 1}_{2018\ 1 \textnormal{'s}}2\underbrace{1\ldots 1}_{2019\ 1 \textnormal{'s}}2\ldots$