Question Number 164623 by mathocean1 last updated on 19/Jan/22 | ||
$${Given}\:{a},\:{b}\:\in\:\mathbb{R}. \\ $$$${Show}\:{that}\:: \\ $$$$\left[{a}\right]+\left[{b}\right]\leqslant\left[{a}+{b}\right]\leqslant\left[{a}\right]+\left[{b}\right]+\mathrm{1} \\ $$ | ||
Answered by mahdipoor last updated on 19/Jan/22 | ||
$$\left[{b}\right]\leqslant{b}<\left[{b}\right]+\mathrm{1}\Rightarrow{a}+\left[{b}\right]\leqslant{a}+{b}<{a}+\left[{b}\right]+\mathrm{1}\Rightarrow \\ $$$$\left[{a}+\left[{b}\right]\right]\leqslant\left[{a}+{b}\right]\leqslant\left[{a}+\left[{b}\right]+\mathrm{1}\right]\Rightarrow \\ $$$$\left[{a}\right]+\left[{b}\right]\leqslant\left[{a}+{b}\right]\leqslant\left[{a}\right]+\left[{b}\right]+\mathrm{1} \\ $$ | ||
Commented by mathocean1 last updated on 19/Jan/22 | ||
$${thanks} \\ $$ | ||