I have to check if the equivalence class has an inverse(without calculations). If yes, I have to find it. $$[7] \in \mathbb{Z}_{36}$$ We know that $[a] \in \mathbb{Z}_m$ has an inverse $\Leftrightarrow (a,m)=1$. In this case, knowing that $(7,36)=1$ we conclude that $[7]$ has an inverse. $$\text{So there is a } x \text{ such that } [7][x]=[1] \text{ in } \mathbb{Z}_{36}.$$ But how can I find this $x$? Do I have to check all integers that are in $\mathbb{Z}_{36}$, that means all integers in $\{0,1,...,35\}$?
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Depends on whether you are supposed to use "general" methods or not. If not, we have $5\cdot 7=35\equiv -1\pmod{36}$, so the equivalence class of $-5$ is the inverse. We could also call it the equivalence class of $31$.
André Nicolas
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You can use the Extended Euclidean algorithm to compute integers $u, v$ such that $$ a \cdot u+ m \cdot v = \gcd(a, m) = 1 $$ If follows that $$ a \cdot u = 1 \pmod{m} $$ which means that $[u]$ is the inverse to $[a]$ in $\mathbb Z_m$.
Martin R
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A way to find this inverse is to compute $[7^{\phi(36)-1}] = [7^{11}] = [31]$, using Euler's theorem, which states that $[a^{\phi(m)}] = [1]$ in $\mathbb{Z}_m$.
user133281
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Why do you calculate $[7^{\phi(36)-1}] = [7^{11}] = [31]$?? Since Euler's theorem states that $[a^{\phi(m)}] = [1]$ in $\mathbb{Z}_m$ shouldn't it be $[7^{\phi(36)}]$? – Mary Star Mar 07 '14 at 19:09
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4Since $[a^{\phi(m)}] = [1]$ we have $[a][a^{\phi(m)-1}]=[1]$, hence $[a^{\phi(m)-1}]$ is the multiplicative inverse of $[a]$. – user133281 Mar 07 '14 at 19:10
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Hint $\ {\rm mod}\ 36\!:\,\ 7x\equiv 1\,\Rightarrow\, x\,\equiv\, \dfrac{1}7 \,\equiv\, \dfrac{-35}7 \,\equiv\, -5.\,$
Or $\bmod 6^2\!:\,\ \dfrac{1}{1+6}= 1-6\ $ by simpler multiples or Hensel lifting or Newton's method.
More generally we can employ the Extended Euclidean Algorithm and related techniques to compute modular inverses. A few worked examples using a handful of methods are here and here and here.
Bill Dubuque
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