Question

Starting from the definition of $\mathbf{x}\mathbf{\equiv }\mathbf{ymodN}$(namely, that $\mathbf{N}$divides $\mathbf{x}\mathbf{-}\mathbf{y}$), justify the substitution rule $\mathbf{x}\mathbf{\equiv }\mathbf{x}\mathbf{\text{'}}\mathbf{modN}\mathbf{,}\mathbf{y}\mathbf{\equiv }\mathbf{y}\mathbf{\text{'}}\mathbf{modNx}\mathbf{+}\mathbf{y}\mathbf{\equiv }\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{modN}$,and also the corresponding rule for multiplication.

Short answer

The substitution rule for multiplication can be defined as it is the process of finding out the value of one variable in terms of the second variable and putting the same value to calculate the second variable.

Where, the rule for multiplication is, $\mathrm{x}\times \mathrm{y}=\mathrm{x}\text{'}\times \mathrm{y}\text{'}\left(\mathrm{modN}\right)$.

Step-by-step solution

Introduction

The primary distinction between addition reactions and substitution reactions is that the former entail the joining of two or more atoms or functional groups, whilst the latter involve the displacement of an atom or functional group by another functional group.

Proof of the substitution rule of addition  

Assume that, $\mathrm{x}\mathrm{mod}\mathrm{N}=\mathrm{p}\mathrm{and}\mathrm{y}\mathrm{modN}=\mathrm{q}$.

Then,

$\mathrm{x}+\mathrm{y}\mathrm{modN}=\mathrm{p}+\mathrm{q}$

Now,

$p$ can be written as $\mathrm{p}=\mathrm{x}\text{'}\mathrm{modN}$, since we have $\mathrm{x}=\mathrm{x}\text{'}\left(\mathrm{modN}\right)$.

In the same way,

$q$can be written as$\mathrm{q}=\mathrm{y}\text{'}\mathrm{modN}$ , since we have $\mathrm{y}=\mathrm{y}\text{'}\left(\mathrm{modN}\right)$.

By using the above equations,

$(\mathrm{x}+\mathrm{y})\mathrm{modN}=(\mathrm{x}\text{'}+\mathrm{y}\text{'})\mathrm{modN}$

Therefore,

$\mathrm{x}+\mathrm{y}=\mathrm{x}\text{'}+\mathrm{y}\text{'}\left(\mathrm{modN}\right)$

Substitution Rule Multiplication Proof

For Multiplication rule we have to prove $\mathrm{x}*\mathrm{y}\equiv \mathrm{x}\text{'}*\mathrm{y}\text{'}\left(\mathrm{mod}\mathrm{N}\right)$

Assume that,

$\begin{array}{l}\mathrm{x}\mathrm{mod}\mathrm{N}=\mathrm{p}\\ \mathrm{y}\mathrm{mod}\mathrm{N}=\mathrm{q}\end{array}$

By multiplying $\mathrm{p}\mathrm{and}\mathrm{q}$.

$\mathrm{x}\times \mathrm{y}\left(\mathrm{modN}\right)=\mathrm{a}\times \mathrm{b}$

Now,

$p$can be written as$\mathrm{p}=\mathrm{x}\text{'}\mathrm{modN}$ since we have$\mathrm{x}=\mathrm{x}\text{'}\left(\mathrm{modN}\right)$ .

In the same way,

$q$can be written as$\mathrm{q}=\mathrm{y}\text{'}\mathrm{modN}$ since we have $\mathrm{y}=\mathrm{y}\text{'}\left(\mathrm{modN}\right)$

By using the above equation,

$(\mathrm{x}\times \mathrm{y})\mathrm{modN}=(\mathrm{x}\text{'}\times \mathrm{y}\text{'})\mathrm{modN}$

The above expression is,

$\mathrm{x}\times \mathrm{y}=\mathrm{x}\text{'}\times \mathrm{y}\text{'}\left(\mathrm{modN}\right)$

Therefore, the solution is$\mathrm{x}\times \mathrm{y}=\mathrm{x}\text{'}\times \mathrm{y}\text{'}\left(\mathrm{modN}\right)$ .