Question

Define a new multiplication in $\mathit{Z}$by the rule:$\mathit{a}\mathit{b}\mathbf{=}\mathbf{1}$for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{Z}$. With ordinary addition and new multiplication, is $\mathit{Z}$ is a ring?

Short answer

No, $Z$is a ring under ordinary addition and new multiplication

Step-by-step solution

Prove under addition

As we know, $Z$under addition satisfies closure for addition, associative addition, commutative addition, and additive identity. The property which states that for each$a\in Z$ , the equation$a+x=0_Z$ has a solution.

Now, prove under multiplication.

Prove under multiplication

Prove that is $Z$closed on multiplication and it follows the associative and distributive law:

1. Closed under multiplication:

Consider $a,b\in Z$this implies that $ab=0\in Z$. Therefore, it is closed under multiplication.

2. Associative property:

Consider localid="1659328971869" $a,b,c\in Z$this implies that $a\left(bc\right)=1=\left(ab\right)c$. Therefore, it satisfies associative property under multiplication.

3. Distributive Law:

Consider $a,b,c\in Z$this implies that:

$a\left(b+c\right)=1$

$$\begin{gathered} ab+ac=1+1 \\ =2 \end{gathered}$$

Therefore, it does not satisfy the distributive property under multiplication.

Hence, it is proved that $Z$is not a ring.