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

In summary, when making calculations in the group Z7, it is important to remember that the possible equivalence classes are 0, 1, 2, 3, 4, 5, and 6. When finding the remainder, it must fall within these equivalence classes. In the example given, the answer is 5 because -9 is not a valid remainder in Z7 and must be adjusted by adding 7.
  • #1
morrowcosom
54
0

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?
 
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  • #2
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?
 
  • #3
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)
 
  • #4
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).
 
  • #5
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.
 

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

What is Group Z7 in arithmetic?

Group Z7 refers to the set of integers modulo 7, also known as the integers modulo 7 group. This means that all calculations are done using the numbers 0, 1, 2, 3, 4, 5, and 6, and any result greater than 6 is reduced by subtracting 7 until it falls within this range.

How do you calculate 6-3*5 in Group Z7?

To calculate 6-3*5 in Group Z7, we first evaluate the multiplication 3*5, which gives us 15. Since this number is greater than 6, we then subtract 7 from it until it falls within the range of 0-6. This gives us an intermediate result of 1. We then subtract 1 from 6, giving us a final result of 5.

Can 6-3*5 have more than one answer in Group Z7?

No, in Group Z7, every calculation will have a unique answer within the range of 0-6. This is because any result greater than 6 is continually reduced by subtracting 7 until it falls within this range, resulting in only one possible answer.

Why is the answer to 6-3*5 either 5 or 2 in Group Z7?

Since we are working within the set of integers modulo 7, any calculation will result in a number between 0 and 6, inclusive. Therefore, when we subtract 3*5 (15) from 6, we get either 5 or 2. This is because 15 is equivalent to 1 in Group Z7, so subtracting 1 from 6 gives us 5, and subtracting 2 from 6 gives us 4. Both 5 and 2 are within the range of 0-6, making them the only possible answers.

What is the purpose of using Group Z7 in arithmetic?

Group Z7 is used in arithmetic for its properties of closure, associativity, identity, and inverses, which make it a useful tool for solving equations and performing calculations. It is also used in cryptography and coding theory due to its modular arithmetic principles.

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