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I read definitions in Calculus books that often confuse me from a logical perspective. For example, the definition of a monotonic function, e.g. a strictly increasing function, is defined as follows.

$$ \forall x_1, x_2 \in A ~~~ x_1 < x_2 \Rightarrow f(x_1) < f(x_2) \tag{1}\label{1} $$

Now let's try to forget the intuitive knowledge of a strictly increasing function. This definition should pinpoint a class of functions that are OK and a class of functions that are KO. The functions that are OK are the ones for which the property \eqref{1} is TRUE for every $(x_1, x_2)$ couple in $A$. I've circled the TRUE cases in the attached truth table (only the $\Rightarrow$ part for now; 1) 2) 3) 4) and a) b) c) d) are labels).

TruthTable

I've been schematic and tried to represent each possible case in the following graphs (the KO cases are marked with an X).

enter image description here

The cases that puzzle me are the ones marked with question marks. For example, the $\delta)$ case is an allowed case since the property $\eqref{1}$ is TRUE. In fact, $x_1 < x_2$ is false and $f(x_1) < f(x_2)$ is true, giving us TRUE for $\Rightarrow$. This is counterintuitive and either is $\eqref{1}$ or my logic wrong. Another allowed case is $\theta)$, for which $x_1 < x_2$ is false and $f(x_1) < f(x_2)$ is false (in fact $f(x_1)=f(x_2)$) so $\Rightarrow$ is TRUE.

Admitting for a minute that $\eqref{1}$ may be wrong, I adjusted it as in $\eqref{2}$.

$$ \forall x_1, x_2 \in A ~~~ x_1 < x_2 \Leftrightarrow f(x_1) < f(x_2) \tag{2}\label{2} $$

This adjustment gets rid of the $\delta$ case but not the $\theta$ one, so I'm logically confused. What is the correct definition of a strictly increasing function if not $\eqref{1}$? How should I interpret definitions like $\eqref{1}$ from a logical point of view?

I've never studied logic before (but sooner or later I will)... for now please help me sleep peacefully tonight :D.

Thanks, Luca

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    $\begingroup$ You cannot forget $\forall x, y \ldots$. Consider case $\delta$: you have $x_2 < x_1$ and $f(x_2) > f(x_1)$ (assuming I've read well the sketch). This means that, with $x=x_2$ and $y=x_1$ we have a pair $x,y$ such that $x < y$ and not $f(x) < f(y)$. Thus the conditional is false and the definition implies that $f$ is not increasing. $\endgroup$ – Mauro ALLEGRANZA May 15 '16 at 13:20
  • $\begingroup$ You are right, I was half-forgetting $\forall x_1, x_2$. $\forall x_1, x_2$ means fixing a $x_1$ and a $x_2$, do the implication involved, and then connect with a $\wedge$ all the implications. For a function like $\delta$ there is a "moment" in which $x_1$ and $x_2$ swap places, and this is enough for the overall implication being wrong. The same for $\theta$. $\endgroup$ – the_eraser May 15 '16 at 13:50
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To be a strictly increasing function, every pair $(a,~b)$ must pass the test. The test is that if $a < b$ then $f(a) < f(b)$.

In your plot (d), you've shown a pair that passes the test: the pair $(x_1,~x_2)$. But the plot is clearly not strictly increasing. This suggests that there must be a pair that doesn't pass the test. What about $(x_2,~x_1)$?

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A statement of the form $\forall (x_1,x_2){\in}A^2~\big[P(x_1,x_2)\Rightarrow Q(x_1,x_2)\big]$ is satisfied if for every possible pair of values from the interval either $Q(x_1,x_2)$ is true or $P(x_1,x_2)$ is false.   That is to say when ever $P(x_1,x_2)$ is true, then $Q(x_1,x_2)$ must be true too — and when ever $P(x_1,x_2)$ is false then, the implication is satisfied.

Finding single examples of the implication holding (particularly those where the antecedent is false) will not suffice to demonstrate satisfaction of the whole statement; the implication has to hold for all cases.

(Although, a single counter example will suffice to show that the statement is unsatisfied.)

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