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

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- Dec 20th 2006, 07:17 PM #1

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- Dec 20th 2006, 07:30 PM #2

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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 #3

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- Dec 20th 2006, 07:45 PM #4

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- Dec 20th 2006, 07:49 PM #5

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- Dec 20th 2006, 07:53 PM #6

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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.

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