Results 1 to 8 of 8

- Jan 20th 2010, 12:15 PM #1

- Joined
- Nov 2009
- Posts
- 27

## Set such that 1 = 0

Hi, I've been asked to find a set such that all other field axioms hold except for that 1 = 0 (multiplicative neutral = additive neutral). Currently, I reckon that the set {0} satisfies this condition, but I've also been asked to show that set is the only such example - how can I do this?

Any help is much appreciated! Thanks

- Jan 20th 2010, 12:21 PM #2

- Jan 20th 2010, 12:30 PM #3

- Joined
- Nov 2009
- Posts
- 27

Well, above 0 = 1, the set simply needs to satisfy:

A1 Additive commutativity

A2 Additive associativity

A3 Existence of additive neutral in set

A4 Existence of additive inverse for each element in set

M1 Multiplicative commutativity

M2 Multiplicative associativity

M3 Existence of multiplicative neutral in set

M4 Existence of multiplicative inverse for each element in set

D Distributivity of multiplication over addition.

Thanks

- Jan 20th 2010, 12:31 PM #4

- Jan 20th 2010, 12:40 PM #5

- Joined
- Mar 2009
- Posts
- 14

suppose for some integer k such that Z(k)={0,1,.......,(k-1)}

if k-1=0, then k=1 as we required Z(1)={0}.

suppose there are some other integer set such that 0 is the only element, then k-1 doesn't not equal to 0, which means k doesn't equal to 1. But if k doesn't equal to one, then k-1>0 so k>1. so k-1 must be at least 1, then there are two or more elements in this set. Therefore it is a contradiction.

so Z(1)={0} is the only case.

- Jan 20th 2010, 01:11 PM #6

- Joined
- Apr 2005
- Posts
- 18,288
- Thanks
- 2458

Why are you assuming that this is "Z modulo" some integer? 0 and 1 make sense in any ring with identity, not necessarily just integers modulo a specific integer.

For any ring with identity, z*1= z for any z. Also, you should be able to prove, by looking at z*(y+ 0)= z*y+ z*0, that z*0= 0 for all z. Now, put those together.

- Jan 20th 2010, 02:26 PM #7

- Joined
- Nov 2009
- Posts
- 27

- Jan 20th 2010, 03:05 PM #8