(a) The inequalities follow from $0\le h^{2}\le h$ .

(b) One needs to take $h>0$ sufficiently small to ensure $h(2x)<h(2x+1)<|a-x^{2}|$ , and use (a) here.

(c) If $a<1$ then $a\in B$ , as $a<1$ implies $a^{2}<a$ , and $B$ is bounded by $1$ , as $x^{2}<a<1$ implies $x<1$ . If $a\ge1$ then $(0,1)\subset B$ , as $0<x<1$ implies $x^{2}<1\le a$ , and $B$ is bounded by $a$ , as $x\ge a\ge1$ implies $x^{2}\ge xa\ge a$ . Now, suppose that $b^{2}<a$ , then, using (b), for some $h>0$ , $(b+h)^{2}<a$ , and $b+h>b$ is in $B$ , hence, $b$ is not an upper bound of $B$ . Now, suppose that $b^{2}>a$ , then, using (b), for some $h>0$ , $(b-h)^{2}>a$ , and for every $x\in B$ , $x<b-h$ (otherwise, $x\ge b-h>0$ , and $x^{2}\ge(b-h)x\ge(b-h)^{2}>a$ ), hence, $b$ is not the least upper bound of $B$ . Therefore, $b^{2}=a$ .

(d) If both are positive and one is less than the other, say $0<b<c$ , then $b^{2}<bc<c^{2}$ . We can only add that even if $b$ and $c$ are known to be nonnegative, $b^{2}=c^{2}$ still implies $b=c$ , as $x^{2}=0$ iff $x=0$ .