find supremum and infimum of sets (and min/max if they exist)

Question

I need help with finding the supremum I don't really understand even how to start

$B=\lbrace{\frac{m}{m+n} : m,n \in N\rbrace}$

$C=\lbrace{\frac{mn}{4m^2+n^2}:m \in Z, n \in N\rbrace}$

Try substituting in different numbers, such as the smallest $m$ and $n$ and see what happens. — David, Nov 17 '19 at 20:09

score 0 · Answer 1 · answered Nov 17 '19 at 20:19

0

As $m,n>0$, we have $m<m+n$, and then $\frac{m}{m+n}<1$. Also $0$ is a lower bound (since everything is positive). And it is the infimum, as $1/(1+n)\to 0$. Try similarly to get $sup(C)$ and $inf(C)$

answered Nov 17 '19 at 20:19

emonHR

2,650

score 0 · Answer 2 · answered Nov 17 '19 at 20:39

0

I dont know why but my phone dont allow me to write a comment, so I decided to write here a hint for $C$. Note that $0\leq (2m+n)^2=4m^2+4mn+n^2$, therefore dividing by $ 4m^2+n^2 $ which is not zero, $$ -1\leq \frac{4mn}{ 4m^2+n^2} $$

answered Nov 17 '19 at 20:39

Senna

1,233

how you note that – ocram ali Nov 18 '19 at 09:20
Because $x^2\geq 0$ for every $x\in \mathbb{R}$. Now take $x=2m+n$. – Senna Nov 18 '19 at 09:25

find supremum and infimum of sets (and min/max if they exist)

2 Answers2