Does remainder of 2^-1 divide by 7 exists? (mod 7) Decimal?

Question

I am confused as to how decimal plays a part compared to the multiplicative inverse.

For example, I know that $2^2\equiv 2^5\equiv 4\bmod 7$ (the pattern is 1,2,4, for every power of 3)

This then implies that $2^{-1}\equiv 4 \bmod 7 .$

However $2^{-1}$ is a decimal, and the definition of the divides I know is specific for integers. When searched online, it says the remainders for decimals do not exist.

So does $2^{-1}$ divide by 7 exists? Is it different from $2^{-1}$ mod 7?

Answer 1

$2^{-1}\equiv 4 \bmod 7$ makes perfect sense because $2 \cdot 4 \equiv 1 \bmod 7$.