-1
$\begingroup$

There was a lot of confusion in the past few posts on this question and so I hope to simplify things and try to make it more clear.

Question - Let $X$ be the set with only one element, zero. Given the binary operations $0+0=0$ and $0\cdot 0=0$. Let $X^+$ be the empty set.

So using my class's Integer Axiom definitions there were a few aspects of this question that have me confused:

  1. Using the binary operation $\cdot$, is it possible to have the identity element equal to $1$?

    I think not as $e \cdot 0 = 0 \cdot e = 0$.

  2. Given the subset $X^+$ does it follow the following property?

There is a subset $Z^+$ called the positive integers and it satisfies the following: for every $a \in Z$, exactly one of the following holds:

$a$ is in $Z^+$, $-a$ is in $Z^+$, or $a = 0$.

If I am looking at this correctly, this property does work as the empty set contains no elements and only $a = 0$ is true?

All help is greatly appreciated.

Thanks!

1 Answers 1

0

Yes to both questions. Specifically,

  1. The identity element must belong to $X$. In fact, $0$ is both the additive and the multiplicative identity. That is, if you write $1$ for the multiplicative identity, then $1=0$ in this case.

  2. The definition is satisfied by simply noting that the only element of $X$, $0$, satisfies the third condition.