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I was wondering if this answer would be correct the multiplicative of $11$ modulo $59$ would be $5$ hence $5\cdot11 \equiv 4 \pmod{59}$.
Is this correct?

Bart Michels
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stacey
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2 Answers2

1

No, by definition, $ $ mod $\,59\!:\,\ n\equiv 11^{-1}\iff 11n\equiv 1.\ $ There are various way to compute modular inverses, e.g. one may use the extended Euclidean algorithm (see here for an especially easy version). Alternatively we can use Gauss's algorithm as below

$\quad {\rm mod}\ 59\!:\ \ \dfrac{1}{11} \,\equiv\, \dfrac{5}{55}\,\equiv\, \dfrac{64}{-4}\,\equiv\, {-16}.\,\ $ Equivalently, eliminating the fractions

$\quad 11x\equiv 1\!\color{#c00}\iff\! 55x\equiv 5\!\iff\! -4x\equiv 64\!\color{#0a0}\iff\! x\equiv -16$

Above we $\color{#c00}{\rm multiplied}$ a congruence by $\,5\,$ and $\color{#0a0}{\rm cancelled}$ $\,-4\,$ from another. Doing so yields an equivalent congruence because such operations are invertible, since $\,5\,$ and $\,-4\,$ are invertible, being coprime to the modulus. By Bezout, $\,a\,$ is invertible mod $\,m\iff \gcd(a,m)= 1.\,$ If so then $\,x\equiv y\iff ax\equiv ay\,$ since we can scale the RHS by $\,a^{-1}\,$ to get the LHS (recall that congruences are preserved under scalings by the Congruence Product Rule)

Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.

Community
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Bill Dubuque
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No. A multiplicative inverse $a$ has $5 \cdot a \equiv 1 \pmod{59}$ and $11$ has $5 \cdot 11 \equiv -4 \not\equiv 1\pmod{59}$. If you want to find the inverse, use the Euclidean algorithm, as you said: \begin{align*} 59 &= 5 \cdot 11 + 4\\ 11 &= 2 \cdot 4 + 3\\ 4 &= 1 \cdot 3 + 1\\ \end{align*} Hence \begin{align*} 1 &= -1 \cdot 3 + 4\\ &= -1 \cdot (11 - 2 \cdot 4) + 4\\ &= -1 \cdot 11 + 3 \cdot 4\\ &= -1 \cdot 11 + 3 \cdot (59 - 5 \cdot 11)\\ &= -16 \cdot 11 + 3 \cdot 59 \end{align*} Hence, $\pmod{59}$, we have $$ 1 \equiv -16 \cdot 11 \equiv 43 \cdot 11 \pmod{59} $$ So, the multiplicative inverse is $43$.

martini
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