# Set such that 1 = 0

• Jan 20th 2010, 12:15 PM
tunaaa
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
Drexel28
Quote:

Originally Posted by tunaaa
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

What other field axioms enable you to say that all elements of the field are zero?
• Jan 20th 2010, 12:30 PM
tunaaa
Well, above 0 = 1, the set simply needs to satisfy:

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
Drexel28
Quote:

Originally Posted by tunaaa
Well, above 0 = 1, the set simply needs to satisfy:

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

Ok...so use them. Which in particular look useful?
• Jan 20th 2010, 12:40 PM
luoginator
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
HallsofIvy
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
tunaaa
Thanks, but how does distributivity help me? Thanks
• Jan 20th 2010, 03:05 PM
Drexel28
Quote:

Originally Posted by tunaaa
Thanks, but how does distributivity help me? Thanks

You got to be kidding me.

$\displaystyle z=z\cdot 1=z\cdot0=0$