MHB Troubleshooting Euclid's Lemma Proof in Modular Arithmetic

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The discussion focuses on proving Euclid's lemma in modular arithmetic, highlighting challenges faced in the proof. A counterexample is provided, illustrating that \(2 \mod 4 \times 2 \mod 4 = 0 \mod 4\). The proof emphasizes that since \(p\) is prime, any multiple of \(p\) must include \(p\) as a factor, resulting in \(0 \mod p\). It concludes that the only way to achieve \(0 \mod p\) through multiplication in \(\mathbf{Z}_p\) is by multiplying with \(0 \mod p\). This reinforces the validity of Euclid's lemma in the context of modular arithmetic.
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Encountered difficulties in proving the attached image. Greatly appreciate for the help!
 

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Alexis87 said:
Encountered difficulties in proving the attached image. Greatly appreciate for the help!

A counter example is $\displaystyle \begin{align*} 2\,\textrm{mod}\,4 \times 2\,\textrm{mod}\,4 = 0\,\textrm{mod}\,4 \end{align*}$.
 
Alexis87 said:
Encountered difficulties in proving the attached image. Greatly appreciate for the help!

As for the proof: Since $\displaystyle \begin{align*} p \end{align*}$ is prime, it is a positive integer.

$\displaystyle \begin{align*} 0\,\textrm{mod}\,p = k \, p \end{align*}$, where $\displaystyle \begin{align*} k \end{align*}$ is some integer.

As $\displaystyle \begin{align*} p \end{align*}$ is prime, the only way to get a multiple of $\displaystyle \begin{align*} p \end{align*}$ is if that number has $\displaystyle \begin{align*} p \end{align*}$ as a factor. But any multiple of $\displaystyle \begin{align*} p \end{align*}$ is itself $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$, and thus the only way to get $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$ through multiplication in $\displaystyle \begin{align*} \mathbf{Z}_p \end{align*}$ is to multiply by $\displaystyle \begin{align*} 0\,\textrm{mod}\,p \end{align*}$.
 

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