Integer Inequality Homework: Proving Existence of Integer m

[annoymage]

Homework Statement

let a,b be positive integer , c is real number, and [itex]-a<c<b[/itex]

i want to show there exist integer m, [itex]-a \leq m \leq b[/itex] such that [itex]m-1 \leq c<m[/itex]

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

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

So by contradiction,

suppose that for all m in S, [itex]m \leq c[/itex] or [itex]m-1>c[/itex]

If [itex]m \leq c[/itex] for all m in S, then i know b is in S, means [itex]b \leq c[/itex] which contradict [itex]c<b[/itex],

If [itex]m-1>c[/itex] 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

[tex] A=\{n\in \mathbb{Z}~\vert~n\geq c\} [/tex]

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

[itex]
m-1 \leq c \leq m[/itex] it's not the same as [itex]
m-1 \leq c<m[/itex] right?

 

[╔(σ_σ)╝]

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

 

[annoymage]

you mean this right? [tex]
A=\{n\in \mathbb{Z}~\vert~n> c\}
[/tex]

thanks you soo much

 

[╔(σ_σ)╝]

Yes. :-)