# z mod 6

• Dec 20th 2006, 07:17 PM
tummbler
z mod 6
why isnt Z mod 6 a field?

this is an extra credit problem in my pre-calc class...

I know that it must abide to the field axioms(am i right?)

but mostly i dont understand the form Z mod 6
• Dec 20th 2006, 07:30 PM
ThePerfectHacker
Quote:

Originally Posted by tummbler
why isnt Z mod 6 a field?

this is an extra credit problem in my pre-calc class...

I know that it must abide to the field axioms(am i right?)

but mostly i dont understand the form Z mod 6

Okay, the basic answer is that it is not a power of a prime. But that is a more advanced answer.

By you can do it by checking whether it is a field or not.
The set is,
\$\displaystyle \{0,1,2,3,4,5\}\$
It froms a group under addition.
Furthermore, it is a ring (just check the definitions for a ring).
Furthermore, it is a commutative ring.
Furthermore, it has unity (1).
Thus, it is a commutative ring with unity.
Now, if every non-zero element has an inverse the proof is complete.
Thus, we need to show every element has a multiplicative inverse.
The element 2, has no inverse!
Just check each one and convince thyself.
• Dec 20th 2006, 07:42 PM
tummbler
what parts of the set are a and b..

im sorry if i sound stupid. ive just neever seen this type of problem before
• Dec 20th 2006, 07:45 PM
ThePerfectHacker
Quote:

Originally Posted by tummbler
what parts of the set are a and b..

im sorry if i sound stupid. ive just neever seen this type of problem before

I do not understand what you are saying.

Did you understand anything I said?

You mentioned it was for a Pre-Calculus class? That make no sense.
• Dec 20th 2006, 07:49 PM
tummbler
this is a extra credit prob for precalc.

in the definition of the addition field axiom is has the variables a and b. same in alot of the definitions of the terms u used in your original explanation. I used mathworld.com btw..
• Dec 20th 2006, 07:53 PM
ThePerfectHacker
Quote:

Originally Posted by tummbler
this is a extra credit prob for precalc.

in the definition of the addition field axiom is has the variables a and b. same in alot of the definitions of the terms u used in your original explanation. I used mathworld.com btw..

It is not a fair question for a pre-calculus class.
Just write on your paper, that not all elements (numbers in Z6) has an inverse, for example 2 is in Z6 and it got no inverse.
Inverse means we need to find a number 'x' (in Z6) such that,
2x=1
Now, just check all the possibilities,
2(0)=0
2(1)=2
2(2)=4
2(3)=6=0 (because we take the mod of 6)
2(4)=8=2 (same reason)
2(5)=10=4
Note, we went through all possibilities because,
Z6= { 0,1,2,3,4,5 }
And none of them produce 1.