Solve Arithmetic in Group Z7: Calculating 6-3*5 = 5 or 2? Find Out Here!

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Homework Help Overview

The discussion revolves around performing arithmetic operations in the group Z7, specifically calculating the expression 6 - 3*5. Participants are exploring the implications of modular arithmetic and equivalence classes within this group.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to calculate the expression and question the correctness of their results, particularly the equivalence of different values under modulo 7. There are discussions about the interpretation of congruences and the nature of remainders in modular arithmetic.

Discussion Status

Some participants have provided clarifications regarding the nature of congruences in Z7, noting that one cannot have two different equivalences for the same integer. There is ongoing exploration of the arithmetic process and its implications, with no clear consensus on the interpretation of results yet.

Contextual Notes

Participants are working under the constraints of modular arithmetic, specifically within the group Z7, and are questioning the validity of their calculations and interpretations based on the rules of this mathematical structure.

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Homework Statement


Make four calculations in the group Z7
--------------------------------------------------------------------------------

First, calculate in Z7
6 - 3*5 =



Homework Equations





The Attempt at a Solution


6-15=6-1=5
=2
The program I am using for independent study says that the answer is 5, how is this so? Where did I screw up?
 
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morrowcosom said:

Homework Statement


Make four calculations in the group Z7
--------------------------------------------------------------------------------

First, calculate in Z7
6 - 3*5 =

6-15=6-1=5
Stop. You're done.
morrowcosom said:
=2
5 [itex]\not \equiv[/itex] 2 (mod 7), but 5 [itex]\equiv[/itex] 2 (mod 7)
morrowcosom said:
The program I am using for independent study says that the answer is 5, how is this so? Where did I screw up?
 
Originally Posted by morrowcosom
1. Homework Statement
Make four calculations in the group Z7
--------------------------------------------------------------------------------

First, calculate in Z7
6 - 3*5 =

6-15=6-1=5

Stop. You're done.

Why am I done? Could you explain?

My original answer was =2, below you put:
5 not congruent 2 (mod 7), but 5 is congruent 2 (mod 7)
I am pretty sure one of these is a typo, but could explain why for example, in my answer 2 is not congruent to 5 (mod 7)
 
You got an answer of 5, which is correct, and agrees with the answer in your book. You continued working, and got 2, which is incorrect.

For a given integer in Z, its equivalence class in Z7 is the remainder when you divide by 7. When you divide by 7, there are seven possible remainders or equivalence classes: 0, 1, 2, 3, 4, 5, and 6. Each integer falls into one of these equivalence classes. If a number is congruent to 5 (mod 7), it can't also be congruent to 2 (mod 7).
 
I understand modulo as merely being the remainder, and the remainder needs to lie in the ring.

6-3*5=-9

-9/7=1 remainder -2

But remainder -2 isn't in the ring, so we simply add 7 to make it so. The answer is 5.
 

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