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What do you think about my first simple direct proof? What mark/grade would you give me? Besides, I am curious about whether you like the style.


Theorem

Let $I = [a,b]$ be a non-empty closed interval. A difference between elements of $I$ is less than or equal to $b - a$.

Proof

Let $I = [a,b]$ be a non-empty closed interval. Let $x_1, x_2 \in I$. We consider the difference $d = x_1 - x_2$. It remains to prove that $d \le b - a$. We do so by deducing that there is a non-negative real number - which will be denoted by $\varepsilon\;$ - such that $d + \varepsilon = b - a$.

By the definitions of $I$ and $x_1$, we have $x_1 \le b$. Thus there is a non-negative real number $\varepsilon_1$ such that $x_1 = b - \varepsilon_1$.

Similarly, by the definitions of $I$ and $x_2$, we have $x_2 \ge a$. Thus there is a non-negative real number $\varepsilon_2$ such that $x_2 = a + \varepsilon_2$.

By the definition of $d$ and the results of the previous two paragraphs, the following holds: $d = b - \varepsilon_1 - (a + \varepsilon_2)$. Applying an appropriate equivalent transformation, we see that $$d = b - a - (\varepsilon_1 + \varepsilon_2)\text{.}$$ The parenthesized sum is a real number, since i) the addends are real numbers and ii) the set of real numbers is closed under addition. Moreover, the sum is a non-negative real number, since i) the addends are non-negative real numbers and ii) a sum of non-negative real numbers is non-negative. Introducing yet another symbol, we call this sum $\varepsilon$, and we have $d = b - a - \varepsilon$.

By another appropriate transformation, we deduce that $d + \varepsilon = b - a$. As $\varepsilon$ is a non-negative real number, we may state that $d \le b - a$. This is the statement that remained to prove. QED

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    $\begingroup$ Looks fine to me. Almost too precise. XD Unless the intention is to make it as precise as humanly possible, a lot of steps you included would just be assumed (i.e. that the sum of real numbers is a real number or the ability to subtract on both sides of an equation). $\endgroup$ – Hayden Mar 31 '14 at 17:33
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    $\begingroup$ Why not just add the inequalities $x_1 \leq b$ and $-x_2 \leq -a$? $\endgroup$ – user133281 Mar 31 '14 at 17:58
  • $\begingroup$ I agree with user133281 - your proof doesn't really achieve much, in that you translate the language of inequalities into equalities with $\epsilon$-slack, and then back again into the language of inequalities, without actually benefiting from the reformulation (there's nothing you can do there that you can't do be manipulating inequalities directly). It's somewhat like proving that $3+5=8$ by saying that $3+5=3+4+(-4)+5=7+1=8$ - completely valid, but unnecessarily obscure. As such, I dislike this proof, although it is clearly written and (afaict) correct. $\endgroup$ – Joshua Pepper Mar 31 '14 at 18:06
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    $\begingroup$ @user133281: Thanks for commenting. I did not add the inequalities, since the initial knowledge about inequalities (which I did not state in the question) was artificially limited to the following two definitions: Definition 1: $r_1 < r_2 \Leftrightarrow \exists \epsilon \in \mathbb{R}, \epsilon > 0, \; r_1 + \epsilon = r_2$. Definition 2: $r_1 \le r_2 \Leftrightarrow \exists \epsilon \in \mathbb{R}, \epsilon \ge 0, \; r_1 + \epsilon = r_2$. Hence, before adding, I possibly would have had to prove that adding inequalities is valid. Such a sub-proof seemed to lengthen the proof significantly. $\endgroup$ – DracoMalfoy Apr 1 '14 at 7:41

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