0
$\begingroup$

Let $n,m\in\mathbb{N}$ and $A_{p,m}=\{ n\in[|1,m|]; p\mid n\}$

Could someone explain to me why we have the cardinal of $A_{p,m}$ equal:

$$\fbox{$|A_{p;m}|=\left\lfloor \dfrac{m}{p} \right\rfloor$}$$

i tired using Quantified expressions

\begin{aligned} n\in A_{p,m}&\iff \begin{cases}n\in[|1,m|] & \\ p\mid n &\end{cases} \\ &\iff \begin{cases} n\in[|1,m|] & \\\exists k\in\mathbb{Z}\, : n=kp &\end{cases} \\ &\iff 1\leq kp \leq m \mbox{ with } k\in\mathbb{Z} \\ &\iff \frac{1}{p}\leq k \leq \frac{m}{p} \mbox{ with } k\in\mathbb{Z}\\ &\iff 0<\frac{1}{p}\leq k \leq \frac{m}{p} \mbox{ with } k\in\mathbb{Z}\\ &\iff 0 < k \leq \left \lfloor \frac{m}{p} \right\rfloor \mbox{ with } k\in\mathbb{Z} \\ \end{aligned}

1 Answers 1

1

The multiples of $p$ in $[1,m]$ are precisely $p$, $2p$, $3p$, and so on, up to $kp$ where $k$ is the largest integer such that $kp\leq m$. There are, therefore, $k$ such multiples; and, of course, $$ k=\left\lfloor\frac{m}{p}\right\rfloor. $$

  • 0
    Could you please explain it to me using Quantified expressions as i tired – 2017-01-12
  • 0
    It seems you have already done it (although I would say $n\in A_{p,m}$, not $n\in \lvert A_{p,m}\rvert$). – 2017-01-12
  • 0
    Saying you're stuck isn't helpful -- what don't you understand? Help me help you. – 2017-01-12
  • 0
    Why are you trying so hard to do this symbolically? You're losing the forest for the trees. – 2017-01-12