Determine the least upper and the greatest lower bounds of the following sets Determine the least upper and the greatest lower bounds of the following sets:

$a)$ A={${\frac{m} {n} : m,n \in N ,m&lt;2n}$}

$b) $B={${n^{1/2} -[n^{1/2}] : n \in N}$}

my attempts :  for option a) if m= n=1  then greatest lower bound will be  1 and if m = 2n ,the the least upper bound will be 2

for option b) here greatest lower bound will be 0... that is $n^{1/2} -n^{1/2}$=0 here i don't know the what is least upper bounds

Pliz verified its  and tell me the solution

thanks in advance
Consider (a). All elements of $A$ are between $0$ and $2$. Now think about limits.
a) Supremum will be 2. As choose any $\epsilon &gt;0$ there exist a rational number such that $m-\epsilon &lt;m/n&lt;2$ 
And Infimum will be 0. As by Archimedean property for any $\epsilon&gt;0$ there exist $m\in N$ such that $0&lt;1/m&lt;\epsilon$

b)note that $0\leq b&lt;1$ for all $b\in B$ so Infimum is 0 as 0 is the element of B for any square number.
Now Supremum is 1(we show that to you)
Now choose $\epsilon&gt;0$ and let $y=1-\epsilon$ and $1&gt;y&gt;0$ consider any integer $m$ such that $m&gt;\frac{y^2}{2(1-y)}$ so then $y&lt;\sqrt{m^2+2m}-m$ now note that, $\sqrt{m^2+2m}-m =\sqrt{m^2+2m}-[\sqrt{m^2+2m}]&lt;1$ 

so given any $y&gt;0$ we find a number $m'$ in $B$ such that $1-\epsilon&lt;m'&lt;1$ so Supremum is 1.

To understand the solution, you must know the definition of Supremum and Infimum.