Is it possible to derive $N= \left\lceil R \right\rceil$ from $R\leq N < R+1$ when $R$ positive real number and N is positive integer?

Question

I have the following expression

\begin{eqnarray*} &&R\leq N < R+1 \end{eqnarray*}

where $N$ is a positive integer and $R$ positive real number. Is it correct to derive $N= \left\lceil R \right\rceil$. As we know that $E\leq N< E+1$, we can derived $N=E$ if E is a positive integer.

similarly

\begin{eqnarray*} &&R-1< N \leq R \end{eqnarray*}

where $N$ is a positive integer and $R$ positive real number. Is it correct to derive $N=\left\lfloor R \right\rfloor$. As we know that $E-1< N \leq E$, we can derived $N=E$ if E is a positive integer.

Example :

\begin{eqnarray*} && x+(N-1+1)\times K \leq y,\\ && x+(N+1)\times K> y \end{eqnarray*}

Is it correct to derive $N=\left\lfloor \frac{y-x}{K} \right\rfloor$ if $x,y,K,N$ are positive integers.

Answer 1

The question is not very clearly formulated, so forgive me if I have misinterpreted it.

There appear to be three similar but separate sub-questions. As the third does not appear to be an "example" of either of the first two, I shall treat the three questions separately.

1. If $N$ is a positive integer, and $R$ is a positive real number, and $$ R \leq N < R + 1, $$ then $$ N - 1 < R \leq N, $$ whence by definition $N = \lceil R \rceil,$ just as you suspected.

2. If $N$ is a positive integer, and $R$ is a positive real number, and $$ R - 1 < N \leq R, $$ then $$ N \leq R < N + 1, $$ whence by definition $N = \lfloor R \rfloor,$ again just as you suspected.

3. If $x, y, K, N$ are positive integers, and $$ x + NK \leq y < x + (N + 1)K, $$ then $$ N \leq \frac{y - x}K < N + 1, $$ and once again you are proved correct, because the definition of the floor function now implies $$ N = \left\lfloor\frac{y - x}K\right\rfloor. $$