Show that if a < b + ε for every ε>0 then a ≤ b

[kmikias]

Homework Statement

Show that if a < b + ε for every ε>0 then a ≤ b

Homework Equations

I am not sure if this is a right way to do it? I just want to know if it does make sense

The Attempt at a Solution

proof.
a < b + ε → if a is bounded above by b+ε then b is the least upper bound for a.
which means a ≤ b.
ε is the upper bound of a since b≤ε.

 

[Dick]

No, that doesn't work. Try formulating it as a proof by contradiction.

 

[kmikias]

No, that doesn't work. Try formulating it as a proof by contradiction.

ok! so i tried it this way...

Suppose a > b+ε for every ε>0,then a≤b
let ε= 1/2a-1/2b since a-b>ε→→→a-b >1/2a-1/2b (true)

so a> b+ 1/2a-1/2b
1/2a>1/2b
a>b which contradict from the first agreement ∅

 

[gb7nash]

ok! so i tried it this way...

Suppose a > b+ε for every ε>0,then a≤b

Unfortunately, no. To do a proof by contradiction, you need to assume the hypothesis and the negation of the conclusion.

 

[kmikias]

Unfortunately, no. To do a proof by contradiction, you need to assume the hypothesis and the negation of the conclusion.

ok..another try...i won't stop until i get this ...

suppose a>b and let ε = 1/2 (a-b) ...(i think you know where i get that)

so a < b+ε → a< b+(1/2a-1/2b)
then a-1/2 < b-1/2 → 1/2a<1/2b→ a<b CONTRADICTION ∅

 

[kmikias]

I think that's is the answer...what do you think

 

[gb7nash]

ok..another try...i won't stop until i get this ...

suppose a>b and let ε = 1/2 (a-b) ...(i think you know where i get that)

so a < b+ε → a< b+(1/2a-1/2b)
then a-(1/2)a < b-(1/2)b → 1/2a<1/2b→ a<b CONTRADICTION ∅

It looks fine, just fixed a minor typo on your part.