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

<br /> m-1 \leq c \leq m it's not the same as <br /> m-1 \leq c&lt;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 &lt; n_0. Since n_0 is the smallest integer with this property we know that n_0 -1 \leq c &lt; n_0.

 

[annoymage]

you mean this right? <br /> A=\{n\in \mathbb{Z}~\vert~n&gt; c\} <br />

thanks you soo much

 

[╔(σ_σ)╝]

Yes. :-)