How to find the modulo multiplicative inverse?

Question

"Find the multiplicative inverse $(\overline {47})^{-1}$ in $\mathbb{Z}_{53}$"

My attempt so far is to use the Euclidean algorithm to establish that (-1/6)(47) + (1/6)(53) = 1

However I'm not exactly sure how this relates to the multiplicative inverse.

Would the multiplicative inverse simply be (-1/6)?

Answer 1

If you apply the Euclidean correctly, you'll have two integers $a$ and $b$ so that $$a\cdot 47 + b\cdot 53 =1.$$ What does this equation tell you if you interpret it mod $53$?

Answer 2

The identity element is $1$ because $(1\times a)\mod b=a$ A multiplication inverse is such that $(a\times a^{-1})\mod b=1$ For $47^{-1} $ in $ \mathbb{Z}_{53}$ i.e. We have to find the element a such that $47a=53b+1\implies a=\frac{53b+1}{47}$. Now the integer value of $b \mod{53}$ for which a is an integer is the answer. That is what the euclidean algorithm suggests.

Answer 3

the Euclidean gives $8\cdot53 - 9\cdot47 =1$, and thus considering the equality in $\mod 53$, and assuming the fact that $-9$ in $\mathbb{Z}_{53}$ equals to $44$. We're done.