Is $N=\{0;1;2;...\}$ ? (because each country has a different definition of $N$

I do not know, I think this comes from Taiwan, I just asked for details

wardtnt1234 wrote: Is $N=\{0;1;2;...\}$ ? (because each country has a different definition of $N$ Positive integers.

parmenides51 wrote: Find all injective function $f: N \to N$ satisfying that for all positive integers $m,n$, we have: $f(n(f(m)) \le nm$ Not sure about this one. I will consider $\mathbb N\in \{1,2,...\}$ Let $n=1$ then $f(f(m))\leq m$ We will induct that $f(f(n))=n, \forall n\in \mathbb N$ For $n=1$, it's true as $f(f(1))\geq 1$ and $f(f(1))\leq 1 \Rightarrow f(f(1))=1$ Assume that $f(f(k))=k, \forall k\in\{1,2,...,n\}$, now we prove that it's true to $n+1$, which is $f(f(k+1))=k+1$ We can see that $f(f(k+1))\leq k+1$ so $f(f(k+1))\in \{1,2,...,k+1\}$ If there exists $l<k$ that $f(f(k+1))=l$ then $f(f(k+1))=f(f(l))$ and so by injectivity $k+1=l$, contradicts. Therefore $f(f(n))=n, \forall n\in \mathbb N$. Let $m=f(1)$ we have $f(nf(f(1)))\leq f(1)n \Rightarrow f(n)\leq f(1)n$ (1) Plugging $n$ by $f(n)$ in (1) we have $n\leq f(1)f(n)$ (2) Hence $\frac{f(n)}{f(1)}\leq f(n)f(1) \Rightarrow f(1)=1$ (because $f(n)>0$) Plugging $f(1)=1$ into (1) and (2) we have $f(n)\geq n$ and $f(n)\leq n$ so $f(n)=n,\forall n\in \mathbb N$, which is satisfied function.

IMOStarter wrote: Hence $\frac{f(n)}{f(1)}\leq f(n)f(1) \Rightarrow f(1)=1$ (because $f(n)>0$). Uhhh ? your inequality just implies $f(1)\ge 1$ (which is quite obvious), and not at all $f(1)=1$

parmenides51 wrote: Find all injective function $f: N \to N$ satisfying that for all positive integers $m,n$, we have: $f(n(f(m)) \le nm$ Let $P(x,y)$ be the assertion $f(xf(y))\le xy$ Let $c=f(1)$ $P(1,x)$ $\implies$ $f(f(x)\le x$ and a simple induction using the fact that $f(f(x))$ is injective gives $f(f(x))=x$ $P(x,c)$ $\implies$ $\frac{f(x)}x\le c$ $\forall x$ $\frac x{f(x)}=\frac{f(f(x))}{f(x)}\le c$ $\implies$ $\frac{f(x)}x\ge\frac 1c$ $P(x,1)$ $\implies$ $\frac{f(cx)}{cx}\le \frac 1c$ And so $\frac{f(cx)}{cx}=\frac 1c$ and $f(cx)=x=f(f(x))$ and injectivity implies $f(x)=cx$ Plugging this in original inequation, we get $c=1$ and so $\boxed{f(x)=x\quad\forall x\in\mathbb Z_{>0}}$, which indeed fits

parmenides51 wrote: Find all injective function $f: N \to N$ satisfying that for all positive integers $m,n$, we have: $f(n(f(m)) \le nm$ It's easy to see/prove $f(f(n))=n$ now Claim -: $f(n)=nf(1)$ Proof-: let $P(n,m)$ be the assertion to given FE $P(2,1)\implies f(2f(1))\leq 2$ Now by Injectivity $2f(1)=f(1)$ or $2f(1)=f(2)$ as $f(1)\neq 0, 2\neq 1$ so $f(2)=2f(1)$ now assume the statement is true $\forall 1\leq n\leq k$ then $P(k+1, 1) \implies f((k+1)f(1))\leq k+1$ now if $(k+1)f(1)=f(p)$ fir some $p<k+1$ then by Injectivity $(k+1)f(1)=pf(1)=f(p)\implies k+1=p$ which is contradiction so $f(k+1)=(k+1)f(1)$ hence by Induction our claim is proved. Hence $f(n)=nf(1)$ so $f(f(n))=n=f(n)f(1)=nf(1)^2$ so $f(1)^2=1\implies f(1)=1$ so $\boxed{f(n)=n}$