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So here I wanted for fun to infer that: $$(\lvert a\rvert<\varepsilon) \rightarrow (a > -\varepsilon) \land (a < \varepsilon)$$

In other words, if it is true that the absolute value of $a$ is less than $\varepsilon$, then $a$ is in between $-\varepsilon$ and $\varepsilon$.

So I started by the definition of $\lvert a \rvert$. Which says that: $$if\; a > 0 \; then \; \lvert a\rvert=a$$ $$if\; a < 0 \; then \; \lvert a\rvert=-a$$

Which I know them to be true (by definition). So then I do the following logical steps: $$suppose\;\lvert a\rvert<\varepsilon$$ $$a>0 \rightarrow \;a<\varepsilon$$ $$a<0 \rightarrow \;-a<\varepsilon$$

from here I am literally stuck. I do not know how to infer/conclude that $a > -\varepsilon \land a < \varepsilon $.

Where I get confused is in here: If I suppose that $a$ is greater than $0$, then I get that $a<\varepsilon$. If I also suppose that $a$ is less than $0$, then I get $-a<\varepsilon$ or $a>-\varepsilon$.

But how can $a$ be both positive and negative? In other words how can this statement: $a>0\,\land\;a<0$ be true?

I want them to be both true so that through modus ponens I infer $a > -\varepsilon \land a < \varepsilon $.

So long story short, I'd like to see this proof done by only involving logical steps.

Mozibur Ullah
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gmmk
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3 Answers3

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If $a < b$ and $r > 0$, then $ar < br.$

If $a < b$ and $r < 0$, then $ar > br$.

In other words, the rule is that when you multiply the inequality by a negative number, you must reverse the direction of the inequality.

So, $-a < \epsilon \implies (a) = [(-a) \times (-1)] > [\epsilon \times (-1)] = -\epsilon.$

Therefore, under the premises that

  • $a < 0$

  • $|a| < \epsilon$,

you can conclude that $a > -\epsilon.$ Combining this with the premise that $a < 0$, you have that then $-\epsilon < a < 0.$

user2661923
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    How can it be true that $If,a<b,and,r>0,,then,ar>br$?? Shouldn't it be $If,a<b,and,r>0,,then,ar<br$ ? – gmmk Sep 22 '22 at 09:21
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    @gmmk : +1 : good catch; was typo(s) on my part. Answer edited. – user2661923 Sep 22 '22 at 09:25
  • Also I don't want to show that $-\varepsilon<a<0$, but that: $-\varepsilon<a<\varepsilon$. – gmmk Sep 22 '22 at 09:34
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    @gmmk I intentionally truncated my answer to dovetail with the work that you had already posted. There are $(3)$ possibilities: either $~a < 0, ~a = 0,~$ or $~a > 0.$ $a = 0$ is clearly a satisfying value. Your posting already showed that if $a > 0,$ that $a < \epsilon.$ My work explains why, if $a < 0$, that $a > - \epsilon.$ Putting this all together, the three mutually exclusive subsets of satisfying values for $(a)$ are $$-\epsilon < a < 0 ~\text{or}~ ~a = 0 ~\text{or}~ 0 < a < \epsilon.$$ – user2661923 Sep 22 '22 at 09:38
  • Though I am having troubles to get exaclty to this this statement: $-\varepsilon<a<\varepsilon ,$ from the set of premises. Now I've worked it out by following your ideas, and I get what you mean and what's going on under the scenes. By I do not know how I can formally convert those 'or's' in 'and's' and conclude EXACLTY this statement $-\varepsilon<a,\land,a<\varepsilon ,$. – gmmk Sep 22 '22 at 10:32
  • In other words, instead of concluding $-\varepsilon<a<0$ or $a=0$ or $0<a<\varepsilon;$ I want to conclude exactly this statement: $-\varepsilon < a\land a<\varepsilon$ – gmmk Sep 22 '22 at 10:40
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    @gmmk Let $S_1$ denote the set ${a~: ~-\epsilon < a < 0}.$ Let $S_2$ denote the set ${a~: ~a=0}.$ Let $S_3$ denote the set ${a ~: 0 < a < \epsilon}.$ Let $T$ denote the set ${a~: -\epsilon < a < \epsilon}.$ You want to prove that $$S_1 \cup S_2 \cup S_3 = T.$$ This is routine. When you have two sets $A$ and $B$, and you want to prove that $A = B$, you simply have to prove two things: [1] $A \subseteq B.$ [2] $B \subseteq A.$ In general, you prove that $A \subseteq B$ by assuming that $x$ is an element in $A$ and deducing that this implies that $x$ must be an element of $B$. – user2661923 Sep 22 '22 at 12:51
  • Alright. So if set theory didn't exist, how would I have approached to the same conclusion? (By the way I know what you're talking about, I studied all of this and so on). Just wanted to use pure logic. But it helped. – gmmk Sep 22 '22 at 13:24
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    @gmmk From a formal perspective, you can not attack the problem without Set theory, because you have to consider the cases of $~a < 0, a = 0,~$ and $~a > 0~$ separately. This is because of how the absolute value function is defined. So, you can not avoid having $(3)$ mutually exclusive sets that you have to take the union of. – user2661923 Sep 22 '22 at 17:10
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HINT

As long as $\varepsilon \geq 0$, we can square both sides of the proposed inequality which holds an equivalent transformation. Having said that, we can reach the desired result: \begin{align*} |a| < \varepsilon & \Longleftrightarrow |a|^{2} \leq \varepsilon^{2}\\\\ & \Longleftrightarrow |a|^{2} - \varepsilon^{2} \leq 0\\\\ & \Longleftrightarrow a^{2} - \varepsilon^{2} \leq 0\\\\ & \Longleftrightarrow (a - \varepsilon)(a + \varepsilon)\leq 0 \end{align*} The last inequality is equivalent to the following collection of systems of inequalities: \begin{align*} \begin{cases} a - \varepsilon \geq 0\\\\ a + \varepsilon \leq 0 \end{cases}\quad\vee\quad \begin{cases} a - \varepsilon \leq 0\\\\ a + \varepsilon \geq 0 \end{cases} \end{align*}

Can you take it from here?

Átila Correia
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  • Yes this is totally useful. But I wanted to show the result only from the definition of the absolute value function. – gmmk Sep 22 '22 at 09:40
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As you've observed, if $a > 0$, $a < \epsilon$ and if $a < 0$, $a > -\epsilon$. Now $\epsilon > 0$ (you haven't said this explicitly, but it's implicitly stated in the inequality $|a| < \epsilon$), so the following statements are also true:

  1. If $a < 0$, $a < \epsilon$

  2. If $a > 0$, $a > -\epsilon$

  3. If $a = 0$, $-\epsilon < a < \epsilon$

So, regardless of whether $a$ is positive, negative, or $0$, $-\epsilon < a < \epsilon$.

user1090793
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  • I think there's a mistake in statement 1 and 2. If $a<0,,a>-\varepsilon$ and If $a>0,,a<\varepsilon$ – gmmk Sep 22 '22 at 09:41
  • @gmmk user's point is that both are true, which is what you want to prove – No Name Sep 22 '22 at 10:21
  • Ok but how can I conclude this result without using words like 'regardless of', and only through a logical sounding argument? I just want to arrive to this statement $-\varepsilon<a<\varepsilon ,$ from the set of premises in a real premise -> conclusion way and as much formal as possible. – gmmk Sep 22 '22 at 10:25