Integer Inequality Homework: Proving Existence of Integer m

[annoymage]

Homework Statement

let a,b be positive integer , c is real number, and $-a<c<b$

i want to show there exist integer m, $-a \leq m \leq b$ such that $m-1 \leq c<m$

i don't know any easy method, but this is where i got now,

Let set $S=[m|-a \leq m \leq b]$

So by contradiction,

suppose that for all m in S, $m \leq c$ or $m-1>c$

If $m \leq c$ for all m in S, then i know b is in S, means $b \leq c$ which contradict $c<b$,

If $m-1>c$ for all m in S and i stuck somewhere. Any hint T_T, or easier any easier method, I'm thinking of well ordering principle, but it i can't see it for now

 

[micromass]

Take the set

$$A=\{n\in \mathbb{Z}~\vert~n\geq c\}$$

Show that A is nonempty and bounded by below. This implies that A has a minimal element m. Show that this m satisfies all your conditions.

 

[annoymage]

hmm then i got

$m-1 \leq c \leq m$ it's not the same as $m-1 \leq c<m$ right?

 

[╔(σ_σ)╝]

Remove the $\geq$ and replace it with a > sign. Well when you apply the modified suggestion given you should get $c < n_0$. Since $n_0$ is the smallest integer with this property we know that $n_0 -1 \leq c < n_0$.

 

[annoymage]

you mean this right? $$A=\{n\in \mathbb{Z}~\vert~n> c\}$$

thanks you soo much

 

[╔(σ_σ)╝]

Yes. :-)