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Let mi EZ for i=1,2,...,k and gcd (mi, mj) = 1 for all i ‡ j Prove that for x,y e Zif x = y (mod m₂) then, where m = m1 m2 mk. x=y (mod m)

Let mi EZ for i=1,2,...,k and gcd (mi, mj) = 1 for all i ‡ j Prove that for x,y e Zif x = y (mod m₂) then, where m = m1 m2 mk. x=y (mod m)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let \( m_i \in \mathbb{Z} \) for \( i = 1, 2, \ldots, k \) and \(\gcd(m_i, m_j) = 1\) for all \( i \neq j \). Prove that for \( x, y \in \mathbb{Z} \) if \[ x \equiv y \pmod{m_i} \] then, \[ x \equiv y \pmod{m} \] where \( m = m_1 \cdot m_2 \cdot \ldots \cdot m_k \).
Transcribed Image Text:Let \( m_i \in \mathbb{Z} \) for \( i = 1, 2, \ldots, k \) and \(\gcd(m_i, m_j) = 1\) for all \( i \neq j \). Prove that for \( x, y \in \mathbb{Z} \) if \[ x \equiv y \pmod{m_i} \] then, \[ x \equiv y \pmod{m} \] where \( m = m_1 \cdot m_2 \cdot \ldots \cdot m_k \).
Expert Solution
Step 1: Define congruent

Given that m subscript i element of straight integer numbers for i=1, 2,..., k and g c d left parenthesis m subscript i comma m subscript j right parenthesis equals 1 for all i not equal to j

Also given that x identical to y left parenthesis m o d space m subscript i right parenthesis


Now, x identical to y left parenthesis m o d space m subscript i right parenthesis

This implies that m subscript i vertical line x minus y

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