Suppose that R is a ring and that a2=a for all a∈RZ. Show that R...

For any ${\displaystyle a\in R}$, we have,
${\displaystyle (a+a)={(a+a)}^2}$
${\displaystyle \Rightarrow (a+a)={\left(2a\right)}^2}$
${\displaystyle \Rightarrow (a+a)=4a^2}$
${\displaystyle \Rightarrow (a+a)=a^2+a^2+a^2+a^2}$
${\displaystyle \Rightarrow (a+a)=a+a+a+a\dots \left(1\right)}$
Now, let ${\displaystyle a,b\in R}$
${\displaystyle a+b={(a+b)}^2}$
${\displaystyle =a^2+ab+ba+b^2}$
${\displaystyle =a+ab+ba+b}$
${\displaystyle \Rightarrow ab+ba=0}$
Adding ba on both sides of above equation, we get,
$ab+(ba+ba)=ba$
From equation 1, we can observe that same terms on both sides of equation get cancelled out.
So, applying same observation on equation 2, we get, $ab=ba$
Hence, Ring R is commutative.