Question

Let x be a real number. Show that$\lfloor 3x\rfloor =\lfloor x\rfloor +⌊x+\frac{1}{3}⌋+⌊x+\frac{2}{3}⌋$

Short answer

For any real number$\lfloor 3x\rfloor =\lfloor x\rfloor +⌊x+\frac{1}{3}⌋+⌊x+\frac{2}{3}⌋$

Step-by-step solution

Step: 1

If $\lfloor 3\mathrm{x}\rfloor =\lfloor \mathrm{x}\rfloor +\lfloor \mathrm{x}\rfloor +\lfloor \mathrm{x}\rfloor \mathrm{then}\mathrm{let}\lfloor \mathrm{x}\rfloor =\mathrm{a}$

It gives

$\begin{array}{r}3\mathrm{a}\le 3\mathrm{x}\le 3\mathrm{a}+1\\ \mathrm{a}\le \mathrm{x}\le \mathrm{a}+1/3\\ \mathrm{a}+1/3\le \mathrm{x}+1/3\le \mathrm{a}+2/3\\ \Rightarrow \mathrm{a}\le \mathrm{x}+1/3\le \mathrm{a}+1\\ \Rightarrow \mathrm{a}\le \mathrm{x}+2/3\le \mathrm{a}\\ \lfloor 3\mathrm{x}\rfloor =\lfloor \mathrm{x}\rfloor +\lfloor \mathrm{x}+1/3\rfloor +\lfloor \mathrm{x}+2/3\rfloor \end{array}$

Step: 2

If $\begin{array}{r}\lfloor 3\mathrm{x}\rfloor =\lfloor \mathrm{x}\rfloor +\lfloor \mathrm{x}\rfloor +\lfloor \mathrm{x}\rfloor +1\mathrm{let}\mathrm{x}\lfloor \mathrm{x}\rfloor =\mathrm{a}\\ \Rightarrow \mathrm{a}+1\le \mathrm{x}+2/3\le \mathrm{a}+2\end{array}$

It gives

$\lfloor 3x\rfloor =\lfloor x\rfloor +\lfloor x+1/3\rfloor +\lfloor x+2/3\rfloor =3a+1$

Step: 3

If $\begin{array}{r}\lfloor 3\mathrm{x}\rfloor =\lfloor \mathrm{x}\rfloor +\lfloor \mathrm{x}\rfloor +\lfloor \mathrm{x}\rfloor +2\mathrm{then}\mathrm{let}\lfloor \mathrm{x}\rfloor =\mathrm{a}\\ \Rightarrow \mathrm{a}+1\le \mathrm{x}+2/3\le \mathrm{a}+2\end{array}$

It gives $\lfloor 3x\rfloor =\lfloor x\rfloor +\lfloor x+1/3\rfloor +\lfloor x+2/3\rfloor =3a+2$

For any real number ,$\lfloor 3x\rfloor =\lfloor x\rfloor +⌊x+\frac{1}{3}⌋+⌊x+\frac{2}{3}⌋$