Troubleshooting Euclid's Lemma Proof in Modular Arithmetic

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SUMMARY

This discussion focuses on troubleshooting the proof of Euclid's Lemma in modular arithmetic, specifically within the context of prime numbers. The proof asserts that if \( p \) is a prime number, then \( 0 \mod p \) can only be achieved through multiplication by \( 0 \mod p \) in the set \( \mathbf{Z}_p \). A counterexample involving \( 2 \mod 4 \) is presented to illustrate potential pitfalls in understanding the lemma. The conclusion emphasizes the necessity of recognizing the properties of prime numbers in modular systems.

PREREQUISITES
  • Understanding of modular arithmetic
  • Familiarity with prime numbers and their properties
  • Knowledge of the notation and operations in \( \mathbf{Z}_p \)
  • Basic proof techniques in number theory
NEXT STEPS
  • Study the properties of prime numbers in modular arithmetic
  • Learn about the structure of the integers modulo \( p \) in \( \mathbf{Z}_p \)
  • Explore detailed proofs of Euclid's Lemma and its applications
  • Investigate counterexamples in number theory to strengthen understanding
USEFUL FOR

Mathematicians, students of number theory, and anyone interested in understanding modular arithmetic and the implications of Euclid's Lemma.

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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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