Question Number 31596 by Nayon.Sm last updated on 11/Mar/18
Commented by Nayon.Sm last updated on 11/Mar/18
Proof that it is irrational
Commented by rahul 19 last updated on 11/Mar/18
$${such}\:{type}\:{of}\:{questions}\:{are}\:{easily}\:{done}\: \\ $$$${by}\:{contradiction}\:! \\ $$
Answered by Joel578 last updated on 11/Mar/18
$$\mathrm{If}\:\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:\mathrm{is}\:\mathrm{rational},\:\mathrm{then}\:\mathrm{there}\:\mathrm{is}\:\mathrm{number}\:{a}\:\mathrm{and}\:{b} \\ $$$$\mathrm{satisfy}\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:=\:\frac{{a}}{{b}},\:\mathrm{where}\:{a},\:{b}\:\in\:\mathbb{Z},\:{b}\:\neq\:\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:=\:\frac{\mathrm{a}}{\mathrm{b}}\:\:\rightarrow\:\:\mathrm{2}^{\frac{{a}}{{b}}} \:=\:\mathrm{5}\:\:\rightarrow\:\:\mathrm{2}^{{a}} \:=\:\mathrm{5}^{{b}} \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{number}\:\mathrm{that}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{because}\:\mathrm{LHS}\:\mathrm{always}\:\mathrm{even}\:\mathrm{and}\:\mathrm{RHS}\:\mathrm{always}\:\mathrm{odd} \\ $$$$\mathrm{Hence},\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:\mathrm{is}\:\mathrm{irrational} \\ $$
Answered by mrW2 last updated on 11/Mar/18
$${assume}\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:{is}\:{rational},\:{i}.{e}.\:{it}\:{can} \\ $$$${be}\:{expressed}\:{as}\: \\ $$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{5}=\frac{{m}}{{n}} \\ $$$${where}\:{m},\:{n}\:{are}\:{integers}\:{and}\:{gcd}\left({m},{n}\right)=\mathrm{1}. \\ $$$$\Rightarrow\mathrm{2}^{\frac{{m}}{{n}}} =\mathrm{5} \\ $$$$\Rightarrow\mathrm{2}^{{m}} =\mathrm{5}^{{n}} \\ $$$${since}\:\mathrm{2}\:{and}\:\mathrm{5}\:{are}\:{co}−{prime}, \\ $$$$\Rightarrow{m}={n}=\mathrm{0}. \\ $$$${i}.{e}.\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:{can}\:{not}\:{be}\:{expressed}\:{as}\:\frac{{m}}{{n}}, \\ $$$${it}'{s}\:{irrational}. \\ $$