Question Number 159171 by LEKOUMA last updated on 13/Nov/21
$${Prove}\:{by}\:{absurd}\:{that}\:\mathrm{log}\:\mathrm{2}\:{is}\:{the} \\ $$$${number}\:{irrational} \\ $$
Answered by mr W last updated on 14/Nov/21
$${say}\:\mathrm{log}\:\mathrm{2}=\frac{{p}}{{q}}\:{with}\:{p},{q}\in\mathbb{Z} \\ $$$$\mathrm{2}=\mathrm{10}^{\frac{{p}}{{q}}} \\ $$$$\mathrm{2}^{{q}} =\mathrm{10}^{{p}} \\ $$$${the}\:{last}\:{digit}\:{of}\:\mathrm{2}^{{q}} \:{can}\:{only}\:{be}\:\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}, \\ $$$${but}\:{the}\:{last}\:{digit}\:{of}\:\mathrm{10}^{{p}} \:{is}\:{always}\:\mathrm{0}. \\ $$$$\Rightarrow{contradiction}! \\ $$$${or} \\ $$$$\mathrm{2}^{{q}} =\mathrm{2}^{{p}} \mathrm{5}^{{p}} \\ $$$$\mathrm{2}^{{q}−{p}} =\mathrm{5}^{{p}} \\ $$$${even}={odd} \\ $$$$\Rightarrow{contradiction}! \\ $$
Commented by gsk2684 last updated on 14/Nov/21
$${nice} \\ $$