# Thread: Subtraction in F3 Field 1-2-1

1. ## Subtraction in F3 Field 1-2-1

Hi,

consider the field $\mathbb{F}_{3}$

The composition table should look like:

0+0=0
0+1=1+0=1
0+2=2+0=2
1+1=2
1+2=2+1=0
2+2=1

0*0=0=0*1=0*2=1*0=2*0
1*1=1
1*2=2*1=2
2*2=1

No I have the case that I have to calculate:

1-2-1=?

I guess it would be zero but I am not sure

How to do this?

Thank you for helping
greetings

2. Hello,

1-2-1=(1-1)-2=-2=1

3. Hi,

first: Thanks for help

second: Could you explain me why this is one? I don't got it

greetings

4. Originally Posted by Herbststurm
Hi,

first: Thanks for help

second: Could you explain me why this is one? I don't got it

greetings
The fundamental problem with the subtraction operator is that it is NOT ASSOCIATIVE! For example, $1 - (2 - 1) \neq (1 - 2) - 1$
Thus parantheses cant be dropped.

So I will interpret 1 - 2 -1 as $1 + (-2) + (-1) = 1 + 1 + 2 = 1$

5. Originally Posted by Isomorphism
The fundamental problem with the subtraction operator is that it is NOT ASSOCIATIVE! For example, $1 - (2 - 1) \neq (1 - 2) - 1$
Thus parantheses cant be dropped.

So I will interpret 1 - 2 -1 as $1 + (-2) + (-1) = 1 + 1 + 2 = 1$
Where did anyone mention associativity ?

6. Originally Posted by Moo
Where did anyone mention associativity ?
Well the problem is how you interpret 1-2-1. Are you thinking of "-" as an operator OR are you thinking of "-x" as a representation for additive inverse?

How do you justify the sequence of operations in your asnwers 1-2-1=(1-1)-2=-2=1?

7. Originally Posted by Isomorphism
Well the problem is how you interpret 1-2-1. Are you thinking of "-" as an operator OR are you thinking of "-x" as a representation for additive inverse?

How do you justify the sequence of operations in your asnwers 1-2-1=(1-1)-2=-2=1?