Understanding left and right adjoint

Question

I'm doing some self-study and found this question.

Let N be the set of natural numbers with its usual notion of ≤. There's a function f:N→N with f(x)=2x. This function doesn't have an inverse. But:

1. Find a right adjoint for f: that is, a function g:N→N with f(m)≤n if and only if m≤g(n) for all m,n∈N. How many right adjoints can you find?

2. Find a left adjoint for f: that is, a function g:N→N with g(m)≤n if and only if m≤f(n) for all m,n∈N. How many left adjoints can you find?

It seems like the answer will be $\lfloor n/2 \rfloor$ and $\lceil n/2 \rceil$ or something like that. But trying to hold all the objects and relations in my head is somehow more than I can handle. There are four quantities $(m,n,f(m),g(n))$, plus the logical relations $f(m)\le n$,....

I tried making a table to make it easier to see, and I think that the last column being all-true means that floor of n/2 works for right adjoint.

| m | n | 2m | n/2 | floor(n/2) | f(m)<=n | m<=floor | f_le_impl_le_floor | floor_le_impl_f_le | iff  |
|---+---+----+-----+------------+---------+----------+--------------------+--------------------+------|
| 0 | 0 |  0 |   0 |          0 | TRUE    | TRUE     | TRUE               | TRUE               | TRUE |
| 1 | 0 |  2 |   0 |          0 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 2 | 0 |  4 |   0 |          0 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 3 | 0 |  6 |   0 |          0 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 0 | 1 |  0 | 0.5 |          0 | TRUE    | TRUE     | TRUE               | TRUE               | TRUE |
| 1 | 1 |  2 | 0.5 |          0 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 2 | 1 |  4 | 0.5 |          0 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 3 | 1 |  6 | 0.5 |          0 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 0 | 2 |  0 |   1 |          1 | TRUE    | TRUE     | TRUE               | TRUE               | TRUE |
| 1 | 2 |  2 |   1 |          1 | TRUE    | TRUE     | TRUE               | TRUE               | TRUE |
| 2 | 2 |  4 |   1 |          1 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 3 | 2 |  6 |   1 |          1 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 0 | 3 |  0 | 1.5 |          1 | TRUE    | TRUE     | TRUE               | TRUE               | TRUE |
| 1 | 3 |  2 | 1.5 |          1 | TRUE    | TRUE     | TRUE               | TRUE               | TRUE |
| 2 | 3 |  4 | 1.5 |          1 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |
| 3 | 3 |  6 | 1.5 |          1 | FALSE   | FALSE    | TRUE               | TRUE               | TRUE |

But I can't tell really, I'm just getting confused. Is there another technique or way to visualize some small examples that will make it simpler? I also tried graphing $y=2x$ with $y=\lfloor 2x \rfloor$ but it was still too much. Or maybe it's just a matter of breaking down the problem into smaller pieces?

Answer 1

All you have to do is verify that for all natural numbers $m$ and $n$, $2m\leq n$ if and only if $m\leq \lfloor n/2\rfloor$.

You cannot prove this by looking at a table, since a table only checks finitely many values of $m$ and $n$.

Hint: If $m$ is a natural number and $r$ is a real number, then $m \leq r$ if and only if $m \leq \lfloor r\rfloor$.