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

[morrowcosom]

Homework Statement

Make four calculations in the group Z7
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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?

 

[Mark44]

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.

=2

5 \not \equiv 2 (mod 7), but 5 \equiv 2 (mod 7)

The program I am using for independent study says that the answer is 5, how is this so? Where did I screw up?

 

[morrowcosom]

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)

 

[Mark44]

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).

 

[Mentallic]

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.