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Demonstrate-that-a-b-N-if-a-b-can-not-be-simplified-then-a-b-a-2-ab-b-2-can-not-also-be-simplified-




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

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