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Question Number 5738 by FilupSmith last updated on 26/May/16

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{having}\mathrm{trouble}\mathrm{understanding}\mathrm{why}:$$$$2^{-1}\equiv 3\left(\mathrm{mod}5\right)$$$$$$$$\mathrm{Can}\mathrm{someone}\mathrm{explain},\mathrm{please}?\mathrm{Thank}\mathrm{you}!$$

Commented by Rasheed Soomro last updated on 26/May/16

$$6\equiv 1\left(mod5\right)$$$$\frac{6}{2}\equiv \frac{1}{2}\left(mod5\right)$$$$3\equiv 2^{-1}\left(mod5\right)$$$$2^{-1}\equiv 3\left(mod5\right)$$$$Thisisanincorrectresult.Dividing$$$$bothsidesbyanumberissomewhat$$$$conditional.$$$$Divisibility{isn}^{\prime }tdefinedforrational$$$$numbers.$$

Commented by FilupSmith last updated on 26/May/16

$$\mathrm{Hmm}...\mathrm{I}\mathrm{am}\mathrm{going}\mathrm{to}\mathrm{have}\mathrm{to}\mathrm{study}\mathrm{further}$$$$\mathrm{because}\mathrm{I}\mathrm{still}\mathrm{cannot}\mathrm{understand}$$

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