Proof related to monotone functions and induction

Question

1)$f : N => R$ is monotone increasing $<=> f(n) $$\leq$$ f(n+1)$ for all $n$ in $N$

2) The function f: R=>R / f(x) = 2x $ - \lfloor\ x \rfloor$ checks that $f(x)$$\leq$$f(x+1)$ however it isn't monotone increasing

Answer 1

1) If $f$ is monotone increasing obviously $f(n) \leq f(n + 1)$ for all $n \in \mathbb{N}$. Let's prove the other implication. Taken $n, m \in \mathbb{N}$ such that $n < m$ we have that $m = n + k$ for a $k >0$.

We proceed by induction on $k$. The case $k = 1$ is guaranteed by our hypothesis. We assume the thesis holds for $k$ and we prove it for $k + 1$: $$f(n + k + 1) \geq f(n + k) \geq f(n)$$ We used the hypothesis for the first inequality and the induction for the second. We proved our proposition.

2) Knowing that $\lfloor x + 1 \rfloor =\lfloor x \rfloor + 1$ we can see that $$ f(x + 1) = 2x + 2 - \lfloor x + 1 \rfloor = 2x + 1 - \lfloor x \rfloor = f(x) + 1 > f(x) $$ So we proved that for every $x \in \mathbb{R}$ we have $f(x) \leq f(x+1)$.

But $f$ is not monotone increasing: $\frac{3}{4} < 1$ but $$f\left({\frac{3}{4}}\right) = \frac{3}{2} > 1 = f(1)$$.