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In Example 1.80 we found a left adjoint for the monotone map \( ( 3 \times - ) : \mathbb{N} \rightarrow \mathbb{R} \). Now find a right adjoint for the same map.

**Example 1.80**
Consider the map
\( ( 3 \times - ) : \mathbb{N} \rightarrow \mathbb{R} \)
which sends \( x \in \mathbb{N} \) to \(3x\),
which we can consider as a real number
\( 3x \in \mathbb{N} \subseteq \mathbb{R} \) .

Write \( \lceil z \rceil \) for the smallest natural number greater than \( z \in \mathbb{R} \), and write \( \lfloor z \rfloor \) for the largest natural number smaller than \( z \in \mathbb{R} \), e.g. \( \lceil 3.14 \rceil = 4 \) and \( \lfloor 3.14 \rfloor = 3 \). As the left adjoint \( \mathbb{R} \rightarrow \mathbb{N} \) , let’s see if \( \lceil −/3 \rceil \) works. It is easily checked that $$ x \le 3y \text{ if and only if } \lceil x/3 \rceil \le y . $$ Success! Thus we have a Galois connection between \( \lceil −/3 \rceil \) and \( 3 \times - \).

## Comments

Since a right adjoint is a "best approximation from below", I'd guess \( \lfloor x/3\rfloor \). Can someone check this?

`Since a right adjoint is a "best approximation from below", I'd guess \\( \lfloor x/3\rfloor \\). Can someone check this?`

I printed out a different version of the book (from the arxiv) and this is Exercise 1.75 there.

Notation: \( f\dashv g\) means that \( f\) is the left adjoint and \( g\) is the right adjoint.So here is what I got for this problem: \(\lceil -/3\rceil\dashv 3\times-\) (this was the example given).

\(3\times-\dashv\lfloor -/3\rfloor \) (Checked!)

Then I tried to find a right adjoint for \(\lfloor -/3\rfloor \) and came up with the function \(3\times-+2 \).

Finally, \(3\times-+2 \) does not seem to have a right adjoint. Here's my argument. If \(g\) were the right adjoint, then \(x\le g(0)\) if and only if \(3x+2\le 0\), and this cannot happen for any natural number \(x\).

`I printed out a different version of the book (from the arxiv) and this is Exercise 1.75 there. _Notation_: \\( f\dashv g\\) means that \\( f\\) is the left adjoint and \\( g\\) is the right adjoint. So here is what I got for this problem: \\(\lceil -/3\rceil\dashv 3\times-\\) (this was the example given). \\(3\times-\dashv\lfloor -/3\rfloor \\) (Checked!) Then I tried to find a right adjoint for \\(\lfloor -/3\rfloor \\) and came up with the function \\(3\times-+2 \\). Finally, \\(3\times-+2 \\) does not seem to have a right adjoint. Here's my argument. If \\(g\\) were the right adjoint, then \\(x\le g(0)\\) if and only if \\(3x+2\le 0\\), and this cannot happen for any natural number \\(x\\).`

I think I saw elsewhere on this site that \(3\times-:\mathbb{N}\to\mathbb{R}\) does not have a right adjoint. In any case, I wanted to make sure that that fact is mentioned here, and to say why.

By definition of right adjoint, if such a function \(R:\mathbb{R}\to\mathbb{N}\) were to exist, it would have to send \(z\in\mathbb{R}\) to the largest natural number smaller than \(\frac{z}{3}\). But if \(z\leq-1\) no such natural number exists, so there is no right adjoint.

If we restrict the codomain to \(3\times-:\mathbb{N}\to(-1,\infty)\), then \(\lfloor\frac{-}{3}\rfloor\) is the right adjoint.

`I think I saw elsewhere on this site that \\(3\times-:\mathbb{N}\to\mathbb{R}\\) does not have a right adjoint. In any case, I wanted to make sure that that fact is mentioned here, and to say why. By definition of right adjoint, if such a function \\(R:\mathbb{R}\to\mathbb{N}\\) were to exist, it would have to send \\(z\in\mathbb{R}\\) to the largest natural number smaller than \\(\frac{z}{3}\\). But if \\(z\leq-1\\) no such natural number exists, so there is no right adjoint. If we restrict the codomain to \\(3\times-:\mathbb{N}\to(-1,\infty)\\), then \\(\lfloor\frac{-}{3}\rfloor\\) is the right adjoint.`