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EDIT: embarrassingly this is not correct as it stands (see comments below). I’m thinking about how to fix it.

I don't think this should be hard. Think base 3. Something is irrational if and only if it does not have a ultimately periodic base 3 expansion. Something's in $C$ if and only if it has no 1's in its base 3 expansion.

Here's an explicit $t$ that I think does the job: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}. $$ In base 3 it's $$ t=0.10011110000000011111111111111110000000000000000000000000000000011\ldots $$ Since $t$ does not have an ultimately periodic base 3 expansion, it's irrational.

Let's call the coordinate ranges where $t$ has 0's "0-blocks" (i.e. from $2^k$ to $2^{k+1}-1$ for $k$ odd) and the coordinate ranges where $t$ has 1's "1-blocks". Now if you form $x+t$ for any $x\in C$, $x+t$ has arbitrarily long blocks with no 1's in the base 3 expansion (corresponding to the blocks of 0's in $t$) - if you're unlucky, the last digit of $x+t$ in a 0-block might be a 1, but there are no others.

Hence if assume for a contradiction that $x+t$ is rational (and so has ultimately periodic base 3 expansion), the repeating block must contain only 0's and 2's.

Now consider the expansion of $x+t$ on the 1-blocks. A calculation shows that the only way to avoid having 1's in the interior of the blocks is for $x$ to have 2's in those blocks and $x+t$ to consist of 0's in the 1-blocks.

Now since $x+t$ has arbitrarily long blocks of 0's, we deduce that the periodic block must consist of all 0's -- that is $x+t$ is a triadic rational.

Since $1-t$ has ternary expansion $$ 1-t=0.122111122222222111111111111111122222222\ldots $$ it's easy to check that there's no $x\in C$ with the property that $x+t$ is a triadic rational.

EDIT: This is modified from the original version which was incorrect, as pointed out by @KajetanJaniak.

I don't think this should be hard. Think base 3. Something is irrational if and only if it does not have a ultimately periodic base 3 expansion. Something's in $C$ if and only if it has no 1's in its base 3 expansion.

Here's an explicit $t$ that I think does the job: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}\mathbf 1_{j\text{ odd}}. $$ In base 3 it's $$ t=0.1|00|0101|00000000|0101010101010101|00000000000000000000000000000000|01\ldots $$ (the $|$ are not part of the number, but just make it easier to read). Since $t$ does not have an ultimately periodic base 3 expansion, it's irrational.

Let's call the coordinate ranges where $t$ has 0's "0-blocks" (i.e. from $2^k$ to $2^{k+1}-1$ for $k$ odd) and the coordinate ranges where $t$ has 01's "01-blocks". Now if you form $x+t$ for any $x\in C$, $x+t$ has arbitrarily long blocks with no 1's in the base 3 expansion (corresponding to the blocks of 0's in $t$) - if you're unlucky, the last digit of $x+t$ in a 0-block might be a 1, but there are no others.

Hence if assume for a contradiction that $x+t$ is rational (and so has ultimately periodic base 3 expansion), the repeating block must contain only 0's and 2's.

Now consider the expansion of $x+t$ on the 01-blocks. A calculation shows that the only way to avoid having 1's in the sum in the interior of the 01-blocks is for $x$ to consist of a concatenation of 0022's and 2022's in those blocks and $x+t$ to consist of 0200's and 2200's there. Since the blocks occurring in $x+t$ have one less two than the corresponding blocks in $x$, we see that $x+t$ "looks different" on the 0-blocks than it does on the 01-blocks. Hence $x+t$ cannot have a periodic base 3 expansion, and must be irrational.

**EDIT: embarrassingly this is not correct as it stands (see comments below). I’m thinking about how to fix it. **I don't think this should be hard. Think base 3. Something is irrational if and only if it does not have a ultimately periodic base 3 expansion. Something's in $C$ if and only if it has no 1's in its base 3 expansion.

Here's an explicit $t$ that I think does the job: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}. $$ In base 3 it's $$ t=0.10011110000000011111111111111110000000000000000000000000000000011\ldots $$ Since $t$ does not have an ultimately periodic base 3 expansion, it's irrational.

Let's call the coordinate ranges where $t$ has 0's "0-blocks" (i.e. from $2^k$ to $2^{k+1}-1$ for $k$ odd) and the coordinate ranges where $t$ has 1's "1-blocks". Now if you form $x+t$ for any $x\in C$, $x+t$ has arbitrarily long blocks with no 1's in the base 3 expansion (corresponding to the blocks of 0's in $t$) - if you're unlucky, the last digit of $x+t$ in a 0-block might be a 1, but there are no others.

Hence if assume for a contradiction that $x+t$ is rational (and so has ultimately periodic base 3 expansion), the repeating block must contain only 0's and 2's.

Now consider the expansion of $x+t$ on the 1-blocks. A calculation shows that the only way to avoid having 1's in the interior of the blocks is for $x$ to have 2's in those blocks and $x+t$ to consist of 0's in the 1-blocks.

Now since $x+t$ has arbitrarily long blocks of 0's, we deduce that the periodic block must consist of all 0's -- that is $x+t$ is a triadic rational.

Since $1-t$ has ternary expansion $$ 1-t=0.122111122222222111111111111111122222222\ldots $$ it's easy to check that there's no $x\in C$ with the property that $x+t$ is a triadic rational.

EDIT: embarrassingly this is not correct as it stands (see comments below). I’m thinking about how to fix it.

I don't think this should be hard. Think base 3. Something is irrational if and only if it does not have a ultimately periodic base 3 expansion. Something's in $C$ if and only if it has no 1's in its base 3 expansion.

Here's an explicit $t$ that I think does the job: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}. $$ In base 3 it's $$ t=0.10011110000000011111111111111110000000000000000000000000000000011\ldots $$ Since $t$ does not have an ultimately periodic base 3 expansion, it's irrational.

Let's call the coordinate ranges where $t$ has 0's "0-blocks" (i.e. from $2^k$ to $2^{k+1}-1$ for $k$ odd) and the coordinate ranges where $t$ has 1's "1-blocks". Now if you form $x+t$ for any $x\in C$, $x+t$ has arbitrarily long blocks with no 1's in the base 3 expansion (corresponding to the blocks of 0's in $t$) - if you're unlucky, the last digit of $x+t$ in a 0-block might be a 1, but there are no others.

Hence if assume for a contradiction that $x+t$ is rational (and so has ultimately periodic base 3 expansion), the repeating block must contain only 0's and 2's.

Now consider the expansion of $x+t$ on the 1-blocks. A calculation shows that the only way to avoid having 1's in the interior of the blocks is for $x$ to have 2's in those blocks and $x+t$ to consist of 0's in the 1-blocks.

Now since $x+t$ has arbitrarily long blocks of 0's, we deduce that the periodic block must consist of all 0's -- that is $x+t$ is a triadic rational.

Since $1-t$ has ternary expansion $$ 1-t=0.122111122222222111111111111111122222222\ldots $$ it's easy to check that there's no $x\in C$ with the property that $x+t$ is a triadic rational.

I don't think this should be hard. Think base 3. Something is irrational if and only if it does not have a ultimately periodic base 3 expansion. Something's in $C$ if and only if it has no 1's in its base 3 expansion.

Here's an explicit $t$ that I think does the job: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}. $$ In base 3 it's $$ t=0.10011110000000011111111111111110000000000000000000000000000000011\ldots $$ Since $t$ does not have an ultimately periodic base 3 expansion, it's irrational.

Let's call the coordinate ranges where $t$ has 0's "0-blocks" (i.e. from $2^k$ to $2^{k+1}-1$ for $k$ odd) and the coordinate ranges where $t$ has 1's "1-blocks". Now if you form $x+t$ for any $x\in C$, $x+t$ has arbitrarily long blocks with no 1's in the base 3 expansion (corresponding to the blocks of 0's in $t$) - if you're unlucky, the last digit of $x+t$ in a 0-block might be a 1, but there are no others.

Hence if assume for a contradiction that $x+t$ is rational (and so has ultimately periodic base 3 expansion), the repeating block must contain only 0's and 2's.

Now consider the expansion of $x+t$ on the 1-blocks. A calculation shows that the only way to avoid having 1's in the interior of the blocks is for $x$ to have 2's in those blocks and $x+t$ to consist of 0's in the 1-blocks.

Now since $x+t$ has arbitrarily long blocks of 0's, we deduce that the periodic block must consist of all 0's -- that is $x+t$ is a triadic rational.

Since $1-t$ has ternary expansion $$ 1-t=0.122111122222222111111111111111122222222\ldots $$ it's easy to check that there's no $x\in C$ with the property that $x+t$ is a triadic rational.

**EDIT: embarrassingly this is not correct as it stands (see comments below). I’m thinking about how to fix it. **I don't think this should be hard. Think base 3. Something is irrational if and only if it does not have a ultimately periodic base 3 expansion. Something's in $C$ if and only if it has no 1's in its base 3 expansion.

Here's an explicit $t$ that I think does the job: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}. $$ In base 3 it's $$ t=0.10011110000000011111111111111110000000000000000000000000000000011\ldots $$ Since $t$ does not have an ultimately periodic base 3 expansion, it's irrational.

Let's call the coordinate ranges where $t$ has 0's "0-blocks" (i.e. from $2^k$ to $2^{k+1}-1$ for $k$ odd) and the coordinate ranges where $t$ has 1's "1-blocks". Now if you form $x+t$ for any $x\in C$, $x+t$ has arbitrarily long blocks with no 1's in the base 3 expansion (corresponding to the blocks of 0's in $t$) - if you're unlucky, the last digit of $x+t$ in a 0-block might be a 1, but there are no others.

Hence if assume for a contradiction that $x+t$ is rational (and so has ultimately periodic base 3 expansion), the repeating block must contain only 0's and 2's.

Now consider the expansion of $x+t$ on the 1-blocks. A calculation shows that the only way to avoid having 1's in the interior of the blocks is for $x$ to have 2's in those blocks and $x+t$ to consist of 0's in the 1-blocks.

Now since $x+t$ has arbitrarily long blocks of 0's, we deduce that the periodic block must consist of all 0's -- that is $x+t$ is a triadic rational.

Since $1-t$ has ternary expansion $$ 1-t=0.122111122222222111111111111111122222222\ldots $$ it's easy to check that there's no $x\in C$ with the property that $x+t$ is a triadic rational.