Solving Modulus Math Problems with Negative Numbers | Step-by-Step Guide

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Homework Help Overview

The discussion revolves around understanding modulus operations, particularly with negative numbers, and how congruences are established in modular arithmetic. Participants are exploring specific examples involving congruences with the modulus of 26.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to clarify how to derive congruences involving negative numbers and the modulus operation. Questions are raised about the definition of modulus and how to handle negative inputs in this context.

Discussion Status

Some participants are providing insights into the properties of modulus, particularly focusing on the requirement that remainders must be positive. There is a mix of interpretations regarding the congruences, and while some definitions are being clarified, no consensus has been reached on the specific examples presented.

Contextual Notes

There appears to be some confusion regarding the definitions and properties of modulus, particularly in relation to negative numbers. Participants are working within the constraints of their current understanding and are seeking further clarification.

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Homework Statement



Im confused as to how you obtain that,

77 is congruent to -1 mod 26
-77 is congruent to 1 mod 26
-11 is congruent to 15 mod 26

Homework Equations





The Attempt at a Solution



Some help would be great thanks
 
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haven't worked a heap with mod, but i think the trick is that the remainder is positive, so for the first
77 = 2x26 + 25, so 77mod26 = 25

simlarly
-1 = (-1)*26 + 25, then (-1)mod26 = 25
 
Last edited:


a is congruent to b mod 26 if a-b is divisible by 26. 77-(-1)=78, -77-1=(-78), -11-15=(-26). All of the differences are evenly divisible by 26.
 


to clrify, i think mod is defined by the fact that if
a mob n = b
then (b-a) is a multiple of n

so take a poitive number c, such that c<n, clearly
c mod n = c
now consider the negative case, say
(-c) mod n = b
then as c<n, and b>0, we must have
(-c)-b = (-1)n
so
b = n -c
 

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