By what rule or property this expression is equal to six?

Question

$$2^{-1} \equiv 6\mod{11}$$

Sorry for very strange question. I want to understand on which algorithm there is a computation of this expression. Similarly interested in why this expression is equal to two?

$$6^{-1} \equiv 2\mod11$$

Answer 1

It's because $2\cdot 6 \equiv 1 \mod 11$

Answer 2

First, we need to understand what $2^{-1}\mod 11$ stands for. Raising to the power $-1$ is usually used to denote the multiplicative inverse. That is, $b$ is a multiplicative inverse of $a$ if and only if $a\cdot b=1$. Thus, $2^{-1}\mod 11$ denotes the number such that $$2^{-1}\cdot2\equiv1\mod11$$ Since $2\cdot 6\equiv 12\equiv1\mod11$, we see $2^{-1}\equiv 6\mod11$. We can do the same to show $6^{-1}\equiv 2\mod 11$.

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For a reliable approach (that is, one that doesn't rely on simply testing $2\cdot 1,2\cdot 2,\cdots,2\cdot 10$ and see which one is $1$), we can use (where we're using that $\gcd(2,11)=1$) $$2^{\phi(11)}\equiv 1\mod 11$$ by Euler's Totient Function (or you could use Fermat's Little Theorem in this case, since $11$ is prime), and since $\phi(11)=10$, we have \begin{align} 2^{-1}&\equiv 2^{-1}\cdot 1\\ &\equiv 2^{-1}\cdot2^{\phi(11)}\\ &\equiv 2^{-1}\cdot2^{10}\\ &\equiv 2^{-1+10}\\ &\equiv 2^9\\ &\equiv 512\\ &\equiv 6\mod 11 \end{align}

Answer 3

You have $2\times 6=12$ which is $1$ mod $11$. Hence, we see that that $2\times 6=1$ mod $11$.

Now the set of congruence numbers modulo $11$ (often noted $\mathbb{Z}/11$) is a ring with addition and multiplication coming from $\mathbb{Z}$.

Saying that $6^{-1}\text{ mod }11=2$ is saying that $6$ mod $11$ is invertible as an element of the ring $(\mathbb{Z}/11,+,\times)$ and that its inverse is $2$ mod $11$.

Answer 4

The notation $2^{-1} \pmod{11}$ means the multiplicative inverse of $2$ modulo $11$, that is, a number $x$ satisfying the equation $2x \equiv 1 \pmod{11}$.

Since $2 \cdot 6 \equiv 12 \equiv 1 \pmod{11}$, $2^{-1} \equiv 6 \pmod{11}$. The same calculation shows that $6^{-1} \equiv 2 \pmod{11}$.

The multiplicative inverse of $6$ exists modulo $11$ since $\gcd(6, 11) = 1$.

We can find the multiplicative inverse of $6$ modulo $11$ using the extended Euclidean algorithm. \begin{align*} 11 & = 1 \cdot 6 + 5\\ 6 & = 1 \cdot 5 + 1\\ 5 & = 5 \cdot 1 \end{align*} We solve for $1$ as a linear combination of $6$ and $11$. \begin{align*} 1 & = 6 - 5\\ & = 6 - (11 - 6)\\ & = 2 \cdot 6 - 11 \end{align*} Hence, $2 \cdot 6 \equiv 1 \pmod{11} \iff 6^{-1} \equiv 2 \pmod{11}$.

Let's see what happens when we use the extended Euclidean algorithm to solve for $2^{-1}$ modulo $11$.

The multiplicative inverse of $2$ exists modulo $11$ since $\gcd(2, 11) = 1$. \begin{align*} 11 & = 5 \cdot 2 + 1\\ 2 & = 2 \cdot 1 \end{align*} Solving for $1$ as a linear combination of $2$ and $11$ yields $$1 = 11 - 5 \cdot 2$$ Thus, $$-5 \cdot 2 \equiv 1 \pmod{11} \iff 2^{-1} \equiv -5 \pmod{11}$$ What does this mean? It means that for each $t \in \mathbb{Z}$, $$2(-5 + 11t) \equiv 1 \pmod{11}$$ In particular, if $t = 1$, we obtain $$2 \cdot 6 \equiv 1 \pmod{11} \iff 2^{-1} \equiv 6 \pmod{11}$$ Why is $2(-5 + 11t) \equiv 1 \pmod{11}$ for $t \in \mathbb{Z}$? Observe that $$2(-5 + 11t) \equiv -10 + 22t \equiv 1 - 11 + 22t \equiv 1 + 11(2t - 1) \equiv 1 \pmod{11}$$

Answer 5

It is part of the collection of integer pairs such that:

$$xy\equiv1\mod p$$

in this case with $p=11$.

For $p=11$, they are $1\cdot1,2\cdot6,3\cdot4,5\cdot9,7\cdot8$ and $10\cdot10$

Except for $1$ and $p-1$, each $x$ is paired to a unique $y$.