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I was reading the book Homomorphic Encryption and Applications when I saw a modular equation involvind non-integer numbers.

In short,

on page 59 the book define the set $y$ as $\{y_1, y_2, .., y_\Theta\}$, where each $y_i$ is a positive number smaller than 2, with $k$ bits of precision after the binary point. Then, it says that $\sum_{i \in S} y_i (\mod 2) = \frac{1}{p} - \Delta_p$, with $|\Delta_p| < 2^{-k}$.

In details, the book defines

(I) $p$ is an big odd integer

(II) each $y_i$ is of the form $\frac{u_i}{2^{k}}$

(III) where these $u_i$ have this property: $\sum_{i \in S} u_i = [ \frac{2^k}{p} ] (\mod 2^{k+1})$

Then, it says that

$\sum_{i \in S} y_i = \sum_{i \in S} \frac{u_i}{2^k} = \frac{[\frac{2^k}{p} ] + \alpha 2^{k+1}}{2^k} = \frac{[\frac{2^k}{p} ]}{2^k} + 2\alpha = \frac{1}{p} - \frac{\Delta}{2^k} + 2\alpha = \frac{1}{p} - \Delta_p (\mod 2)$

The first equality comes from (II), the third comes from (III)... It's Ok, but I don't understand the last one:

What do the authors mean with this $\mod 2$ if the equation involves non-integers values?

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It means the difference between the left and right hand sides is a multiple of 2 - that is, an even whole number.

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  • $\begingroup$ So, did they just use this notation to "hide" the $2\alpha$ factor on the last equality ? Thank you in advance. $\endgroup$ – Hilder Vitor Lima Pereira Dec 5 '14 at 10:57

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