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First, I need to evaluate left and right hand sides of '$\Rightarrow$' to use definition of implication (its truth table).

And, I'm simply lost in a question: "how should I evaluate truth or falsity of both $(\phi^2 > 2)$ and $(\phi > 1.4)$ ?

Both of left hand side and right hand side of implication depend on $\phi$

And any reasoning I came up with looks like: there will be intervals of $\phi$ where this statement is true and intervals where this statement is false...

We seek when LHS is True: $(\phi < - \sqrt 2) \cup (\phi > \sqrt 2)$. Then $\phi > 1.4$ is true for $\phi > \sqrt 2$ but false for $\phi < - \sqrt 2$.

So entire expression is True $\phi > \sqrt 2$ and false for $\phi < - \sqrt 2$

Now consider $ 1.4 < \phi <= \sqrt 2$. RHS is True and LHS is true. So entire expression is true.

For $ -\sqrt 2 \le \phi \le 1.4$, RHS is false and LHS is false as well, so entire expression is true.

So the final answer is: $(\phi^2 > 2) \Rightarrow (\phi > 1.4)$ is false for $\phi < - \sqrt 2$ and true for $\phi >= - \sqrt 2$

Did I get it right?

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    $\begingroup$ It depends on the value of $\phi$ $\endgroup$ Aug 16 at 11:13

3 Answers 3

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As you said, you will have four open sub-intervals given by:

$$ I_1 = \left(-\infty, -\sqrt{2}\right) $$ $$ I_2 = \left(-\sqrt{2}, \ 1.4\right) $$ $$ I_3 = \left(1.4, \ \sqrt{2}\right) $$ $$ I_4 = \left(\sqrt{2}, \ \infty\right) $$ $$ I = I_1 \cup I_2 \cup I_3 \cup I_4 = \mathbb{R} $$

Let's enumerate the statements $(1)$, $(2)$ and $(3)$

$$ \phi^2 > 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$

$$ \phi > 1.4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) $$ $$ \left(1\right) \Rightarrow \left(2\right) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) $$

The statement $(1)$ is true for $T_1$ and false for $F_1$

$$ \phi^2 > 2 \Rightarrow \begin{cases} \phi < -\sqrt{2} \\ \phi > \sqrt{2} \end{cases} \Rightarrow \begin{cases} T_1 = I_1 \cup I_4 \\ F_1 = I_2 \cup I_3 \end{cases} $$

The statement $(2)$ is true for $T_2$ and false for $F_2$

$$ \phi > 1.4 \Rightarrow \begin{cases} T_2 = I_3 \cup I_4 \\ F_2 = I_1 \cup I_2 \end{cases} $$

Then if we want to analyze what $(3)$ is, we want to know our objective.

Case 1: Know if $(3)$ is true for every $\phi$

$$ (1) \Rightarrow (2) \ \ \ \ \ \ \ \ \ \text{if} \ \ \ T_1 \subseteq T_2 $$

As we have

$$ \left(I_1 \cup I_4\right) \not\subseteq \left(I_3 \cup I_4\right) $$

Then the $(1) \Rightarrow (2)$ is False.

Case 2: Search for which values of $\phi$ the statement $(3)$ is valid.

The statement $(3)$ is true for $T_3$ and false for $F_3$. Then

$$ T_3 = \left[T_1 \cap T_2\right] \cup \left[F_1 \right] $$ $$ T_3 = \underbrace{\left[\left(I_1 \cup I_4\right) \cap \left(I_3 \cup I_4\right)\right]}_{I_4} \cup \left[I_2 \cup I_3\right] $$ $$ \boxed{T_3 = I_2 \cup I_3 \cup I_4} $$ $$ F_3 = I \setminus T_3 = I_1 $$

That means

$$ T_3 = \left(-\sqrt{2}, \ \infty\right) $$ $$ F_3 = \left(-\infty, \ -\sqrt{2}\right) $$

PS: I took out the exact values of $-\sqrt{2}$, $1.4$ and $\sqrt{2}$ just to make it easier to write. But for a more formal analysis you should include them.

EDIT: For the case

$$ (1) \Leftrightarrow (2) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4) $$

Then

$$ T_4 = \left[T_1 \cap T_2\right] \cup \left[F_1 \cap F_2\right] $$

$$ T_4 = \left(-\sqrt{2}, \ 1.4\right) \cup \left(\sqrt{2}, \ \infty\right) $$ $$ F_4 = \left(-\infty, \ -\sqrt{2}\right) \cup \left(1.4, \ \sqrt{2}\right) $$

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  • $\begingroup$ For case 2: (3) can't be false on $(1.4,\sqrt 2)$, as (1) will be false, and no matter what we derive from false - both false and true, is true. So (3) is true on $(1.4,\sqrt 2)$. So it seems there's some flaw in your reasoning somewhere $\endgroup$ Aug 20 at 13:13
  • $\begingroup$ I see. $T_3$ should be $ [T_1 \cap T_2] \cup F_1 = I_4 \cup I_2 \cup I_3$. Then $F_3$ = $\mathbb R \setminus T_3 = I_1$ $\endgroup$ Aug 20 at 13:36
  • $\begingroup$ as you provided clear reasoning, never the less actual answer is right, but reasoning is clear and what I was looking for! So I accept your answer:) $\endgroup$ Aug 20 at 13:37
  • $\begingroup$ Nice! I corrected what you said. I put the case where $T_4 = \left[T_1 \cap T_2 \right] \cup \left[F_1 \cap F_2\right]$ cause it happens only when $(1) \Leftrightarrow (2)$ $\endgroup$ Aug 26 at 13:02
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Assuming that $\phi$ is an arbitrary real number, then the assertion is false, since, for instance, $(-2)^2>1.4$, but it is not true that $-2>1.4$.

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    $\begingroup$ The subtle thing is: there is no such assumption in original problem statement. If there were an assumption like $\forall \phi$: <this statement>, then it would be such as you answered. That's why I asked this question. $\endgroup$ Aug 16 at 11:05
  • $\begingroup$ All statements formally need quantifiers and some mention of a universe where variables come from. Most often the qualifers are not explciitely given, and the universe is assumed to be something that is obvious from context. In this case, assuming $\varphi$ is a real number seems the "natural" interpretation, without further context. Also to me using the $\forall$ quantifier also seems natural, but you are of course righ that this is not stated. $\endgroup$
    – Ingix
    Aug 16 at 12:34
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In the standard interpretation (where the symbols have their usual arithmetical meanings) and with domain of discourse $\mathbb R,$ $$\forall \phi\;\Big((\phi^2 > 2 \to \phi > 1.4)\iff \phi\in[-\sqrt2,\infty)\Big).\tag1$$ So, for $$\phi^2 > 2 \implies \phi > 1.4$$ to be universally true, we can change the interpretation by changing the domain of discourse, by restricting it from $\mathbb R.$ There are infinitely many ways of doing this, and from $(1),$ the most conservative restriction is $$\mathbb R{\setminus}(-\infty,-\sqrt2).$$


Addendum

Here's one way of obtaining $(1):$ \begin{align}&\phi^2 > 2 \to \phi > 1.4 \\\iff{}&\phi^2 \le 2 \quad\lor\quad \phi > 1.4 \\\iff{}& -\sqrt2\le\phi \le \sqrt2 \quad\lor\quad \phi > 1.4 \\\iff{}& -\sqrt2\le\phi \end{align}

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  • $\begingroup$ while you've got the same answer as I have, the reasoning itself is essential for me to figure out if I've got correct answer. And I don't see any reasoning part in your answer, just statements. Could you elaborate a bit more on your answer, please $\endgroup$ Aug 20 at 13:18
  • $\begingroup$ @OleksandrKhryplyvenko At your request. I've revised the Answer. $\endgroup$
    – ryang
    Aug 20 at 13:48
  • $\begingroup$ This Answer is highly related. $\endgroup$
    – ryang
    Aug 21 at 4:51
  • $\begingroup$ elegant, indeed, thanks! $\endgroup$ Aug 21 at 20:34

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