Question Number 124693 by mathocean1 last updated on 05/Dec/20
$${Demonstrate}\:{that}\:\forall\:{a},{b}\:\in\mathbb{N}^{\ast} \:{if}\:\frac{{a}}{{b}} \\ $$$${can}\:{not}\:{be}\:{simplified},\:{then}\:\frac{{a}+{b}}{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} } \\ $$$${can}\:{not}\:{also}\:{be}\:{simplified}. \\ $$$$ \\ $$
Answered by MJS_new last updated on 05/Dec/20
$$\frac{{a}}{{b}}\:\mathrm{can}\:\mathrm{not}\:\mathrm{be}\:\mathrm{simplified}\:\Leftrightarrow\:{a}\neq{bp}\wedge{b}\neq{aq} \\ $$$$\frac{{a}+{b}}{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }\:\mathrm{can}\:\mathrm{not}\:\mathrm{be}\:\mathrm{simplified}\:\Leftrightarrow \\ $$$$\Leftrightarrow\:\frac{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }{{a}+{b}}\:\mathrm{can}\:\mathrm{not}\:\mathrm{be}\:\mathrm{simplified} \\ $$$$\frac{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }{{a}+{b}}={a}+\frac{{b}^{\mathrm{2}} }{{a}+{b}}={b}+\frac{{a}^{\mathrm{2}} }{{a}+{b}} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}+{b}}\:\mathrm{can}\:\mathrm{not}\:\mathrm{be}\:\mathrm{simplified}\:\mathrm{if}\:{a}\neq{bp} \\ $$$$\frac{{a}^{\mathrm{2}} }{{a}+{b}}\:\mathrm{can}\:\mathrm{not}\:\mathrm{be}\:\mathrm{simplified}\:\mathrm{if}\:{b}\neq{aq} \\ $$