For all $x,y$ in the real numbers [$(x>y)$ implies (there exists an e in the positive real numbers $x \ge y+e$)]

I can't see which of the real number axioms of addition, multiplication and order will help me prove this theorem.

note: this statement was deduced using the logical equivalence of the contrapositive so if there are any problems with the statement itself let me know.


If $x>y$ then $x-y>y-y=0$, so setting $e=x-y$ we get $$ x=y+(x-y)=y+e$$ with $e>0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.