if we are doing modular arith. why will (a)mod p always have an inverse if p is prime??? Please help Thanks
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Hello, A number a has an inverse modulo p if a is coprime with p. so if $\displaystyle a \neq p^n$, where p is prime, then a has an inverse modulo p ! especially if $\displaystyle a<p$, then it has an inverse.
Originally Posted by Moo Hello, A number a has an inverse modulo p if a is coprime with p. so if $\displaystyle a \neq \color{red}np$, where p is prime, then a has an inverse modulo p ! especially if $\displaystyle a<p$, then it has an inverse. Just a minor correction.
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