Question Number 5738 by FilupSmith last updated on 26/May/16
$$\mathrm{I}'\mathrm{m}\:\mathrm{having}\:\mathrm{trouble}\:\mathrm{understanding}\:\mathrm{why}: \\ $$$$\mathrm{2}^{−\mathrm{1}} \equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{someone}\:\mathrm{explain},\:\mathrm{please}?\:\mathrm{Thank}\:\mathrm{you}! \\ $$
Commented by Rasheed Soomro last updated on 26/May/16
$$\mathrm{6}\equiv\mathrm{1}\left({mod}\mathrm{5}\right) \\ $$$$\frac{\mathrm{6}}{\mathrm{2}}\equiv\frac{\mathrm{1}}{\mathrm{2}}\left({mod}\:\mathrm{5}\right) \\ $$$$\mathrm{3}\equiv\mathrm{2}^{−\mathrm{1}} \left({mod}\:\mathrm{5}\right) \\ $$$$\mathrm{2}^{−\mathrm{1}} \equiv\mathrm{3}\left({mod}\:\mathrm{5}\right) \\ $$$${This}\:{is}\:{an}\:{incorrect}\:{result}.{Dividing} \\ $$$${both}\:{sides}\:{by}\:{a}\:{number}\:{is}\:{somewhat} \\ $$$${conditional}. \\ $$$${Divisibility}\:{isn}'{t}\:{defined}\:{for}\:{rational} \\ $$$${numbers}. \\ $$
Commented by FilupSmith last updated on 26/May/16
$$\mathrm{Hmm}…\:\mathrm{I}\:\mathrm{am}\:\mathrm{going}\:\mathrm{to}\:\mathrm{have}\:\mathrm{to}\:\mathrm{study}\:\mathrm{further} \\ $$$$\mathrm{because}\:\mathrm{I}\:\mathrm{still}\:\mathrm{cannot}\:\mathrm{understand} \\ $$