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July 12th, 2016, 12:53 PM   #1
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Lightbulb Direct proof for this inequality theorem

Theorem:$\displaystyle a\leqslant b+\epsilon, \forall \epsilon > 0 \Rightarrow a\leqslant b $ where $\displaystyle a,b\in \mathbb{R}$

I have seen proof by contradiction for this. But I tried to give a direct proof as follows;

Let $\displaystyle a,b\in \mathbb{R} \wedge \epsilon > 0$
$\displaystyle a\leqslant b+\epsilon \Rightarrow a-b\leqslant \epsilon$
That is $\displaystyle \epsilon_{min}=a-b$
But since $\displaystyle \epsilon > 0$,
$\displaystyle \epsilon_{min} > 0 \geqslant 0$
$\displaystyle \therefore a-b\geqslant 0 \Rightarrow a\geqslant b$

Here I obtained a complete opposite of the theorem.
Could someone please point me where I have gone wrong?
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July 12th, 2016, 01:05 PM   #2
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You cannot write "$\displaystyle \epsilon_{min}= b- a$" unless you first show that there is a minimum value for $\displaystyle \epsilon$ (and there isn't).
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Last edited by Country Boy; July 12th, 2016 at 01:10 PM.
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July 12th, 2016, 10:53 PM   #3
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Thanks!
Is that the condition "$\displaystyle \epsilon > 0$" which keeps us away from concluding that $\displaystyle \epsilon_{min}=a-b$ ?
And what if we didn't have that condition and the statement was as follows?
$\displaystyle a\leqslant b+\epsilon, \forall \epsilon > a-b \Rightarrow a\leqslant b $ where $\displaystyle a,b\in \mathbb{R}$
Is the above statement legit?
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