(a) If $a=bc$ with b and c non-units, then d(b) is less than d(a)
(b) If d(a)=0 then a is a unit
(c) If d(a)=1 then a is either a unit or irreducible
So I know a Euclidean Domain is when there exists a function $d$: $D$-{0} to $Z$, s.t. $d(a)$ is less than or equal to $d(ab)$ for all $a,b∈ D$-{$0$} and if $a,b∈ D$ and $b≠0$ then there exists $q,r∈ D$ s.t. $a=bq+r$ where $r=0$ or $d(r)<d(b)$. I am not sure really how to approach these problems. Looking at the first question since $d(ab)≥d(a)$ for the domain I don't see how this is possible. The second question since $d(a)=0$ then if $d=0$ then $a$ could be a unit however if $a=0$ then a couldn't be a unit so that is confusing me. I have a similar logically understanding for the third question. I don't really know how to approach these problems though
(a) If $a=bc$, then $d(a) = d(bc)\ge d(b)$ by definition. We need to prove $d(a)\ne d(b)$. If $d(a)=d(b)$, then we can perform division. We get $b=aq+r$, with $r=0$ or $d(r)<d(b)$. However $a=bc$, so we have $b=bcq+r$. Thus $b\mid r$, so $r=0$ or $d(r)\ge d(b)$. Thus we must have $r=0$. Hence $b=bcq$, so $1=cq$, contradicting $c$ being a nonunit.
(b) If $d(a)=0$, then $d(a)\le d(1)$. So by division, we have for some $q,r$, $1=aq+r$ with $r=0$ or $d(r)<d(a)=0$. Since $d(r)<0$ is impossible, $r=0$. Thus $1=aq$ for some $q$, so $a$ is a unit.
(c) If $d(a)=1$, and $a=bc$, then if $b$ and $c$ were both nonunits, we'd have $d(b)=d(c)=0$ by (a), but then they are units by (b), so this is impossible. Thus $a$ is not reducible. Hence it is either a unit or irreducible.