Prove that $\forall \epsilon>0, \epsilon>a \implies 0 \geq a$

Question

I am doing a course on basic real analysis in which firstly i am emphasising on real numbers. My book says that real number satisfies the following axioms.

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1)Field Axiom 2)Extend Axiom 3)Order Axiom 4)Completeness axiom.

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While doing the exercise of 3) I found following sets of questions.

$\forall \epsilon>0, \epsilon>a \implies 0 \geq a$

$\forall \epsilon<0, \epsilon<a \implies a \geq 0$

For the first part I think like this: however small is $\epsilon$ if positive and as $\epsilon>a$ so $a$ has to be negative. But I am unable to write the formal proof.

Answer 1

I can't see your axioms, nor can I parse the strange order of your highlighted line, but I'd prove the title theorem's contrapositive. If $a> 0$ (which by order, I guess, is equivalent to $0\not\ge a$) then the positive choice $\epsilon:=a/2$ contradicts $\epsilon>a$ (by whatever axiomatic proof you'd manage of $a>0\implies a>a/2>0$).

Answer 2

If $a>0$, then take $\epsilon=a/2,$ so $0<\epsilon<a,$ so it is not true that for all $\epsilon>0$, $\epsilon>a$. The contrapositive of that is what you wanted for the first part.

If $a<0$, then take $\epsilon=a/2$ so $0>\epsilon>a$ so it is not true that for all $\epsilon<0$, $\epsilon<a$. The contrapositive of that is that if for all $\epsilon<0$, $\epsilon<a$, then $a\ge0$.