How to show either $\left \lfloor {\frac{m-1}{2}} \right \rfloor$ or $\left \lfloor {\frac{m+1}{2}} \right \rfloor$ odd and other is even?

Question

I was working in my Calculus assignment and came across this and proving this would mean that I will come to a conclusion in greater part of the task but I couldn't achieve any solution. Intuitively seems correct of course there is a chance that my intuition is wrong. Question is as stated in the title, I will rephrase anyway

How to show either $\left \lfloor {\frac {m-1}{2}} \right \rfloor$ or $\left \lfloor {\frac{m+1}{2}} \right \rfloor$ odd and other is even?

Thanks

Answer 1

Note that $\left \lfloor {\frac{m-1}{2}} \right \rfloor + 1 = \left \lfloor {\frac{m-1}{2}} + 1 \right \rfloor = \left \lfloor {\frac{m+1}{2}} \right \rfloor$

So if $\left \lfloor {\frac{m-1}{2}} \right \rfloor$ is an even/odd integer, then $\left \lfloor {\frac{m+1}{2}} \right \rfloor$ will be an odd/even integer.

Answer 2

Hint $:$ Break the entire problem in two cases. Where $m$ is odd and $m$ is even.

If $m$ is odd. Then $m$ is of the form $2k+1$ for some $k \in \Bbb Z.$ Then observe that $\left \lfloor {\frac {m-1} {2}} \right \rfloor = k$ and $\left \lfloor {\frac {m+1} {2}} \right \rfloor = k+1.$ Now note that whatever $k \in \Bbb Z$ is we always have one of $k$ and $k+1$ is odd and the other is even. Similarly do it when $m$ is even.

Answer 3

Floor function is the same as( GIF [x]), there could be two cases:

(1) Let $x=2m+q, m \in I, 0\le q<1$, then $[\frac{x-1}{2}]=[m+(q-1)/2]=m-1,$ $[\frac{x+1}{2}]=[m+(q+1)/2]=m.$ If one of this is even, the other one is odd.

(2) Let $x=2m+1+q, q<1$, then $[(x-1)/2]=[(2m+1+q-1)/2]=[m+q/2]=m,$ $[(x+1)/2]=[(2m+1+q+1)/2]=[m+1+q/2]=m+1.$ If one of them is even, then other one is odd