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  1. Every integer is a rational number -> false -- correct?

  2. Let r = true; s = true. Is this statement true or false? $$\lnot [r \lor (\lnot s \lor r)];\quad$$ -> true -- correct?

  3. Let p = true; r = false; q = false. Is the following statement true or false? $$\lnot [\lnot q \lor(p \lor \lnot r)$$ -> false -- correct?

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    $\begingroup$ For #1, you are permitted to express an integer $ \ n \ $ as $ \ \frac{n}{1} \ . $ $\endgroup$ – colormegone Jul 18 '13 at 22:04
  • $\begingroup$ Every integer is a rational number. $\endgroup$ – Ittay Weiss Jul 18 '13 at 22:04
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    $\begingroup$ @Ittay Weiss: that issue is one of the well known intractable problems. on this site. In elementary math people learn that $\mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R} \subseteq \mathbb{C}$ but in the usual presentations that construct each of these from the previous one (e.g. in abstract algebra), those subset relations are all literally false, e.g. an integer is an equivalence class of pairs of natural numbers, and so a natural number is not an integer according to those definitions. It's completely a matter of viewpoint. $\endgroup$ – Carl Mummert Jul 19 '13 at 12:03
  • $\begingroup$ @CarlMummert I assumed OP was talking about number systems here. Which set theoretic constructions is used to produce the next number system from the previous one in the tower you mention is irrelevant. All that matters is the axioms that the system satisfy and of course that the rationals contain an isomorphic copy of the integers, and then immediately the integers are identified with certain rationals. I don't anybody actually thinks of rationals as equivalence classes of pairs of integers, nor of the reals as Dedekind cuts or Cauchy sequences.... $\endgroup$ – Ittay Weiss Jul 19 '13 at 22:51
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    $\begingroup$ @IttayWeiss There have existed some logicians who at least seem to have thought that an isomorphism (in the sense of category theory) isn't enough... at least from their language. At least from what I can tell, certain logicians of the Warsaw-Lvov school wouldn't even call this "c" here and this "c" here the same, I guess since they occupied differently places (perhaps also because if you look at them with a powerful enough microscope, you might start to see differences). They would just call the first "c" and the second "c" equiform. That may seem hairsplitting, but... $\endgroup$ – Doug Spoonwood Jul 20 '13 at 0:46
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You didn't strike out, but only one of your answers is correct:

  1. Note that any integer $n$ can be expressed as the fraction: $\dfrac n1$. (In all fairness, the answer depends on how the rational numbers are defined. If we define the integers as a proper subset of the rationals, as suggested by my note, then your answer should be true.)

  2. This is in fact false. $$\lnot[T \lor (F \lor T)] \equiv \lnot [T \lor T] \equiv \lnot [T] = F$$

  3. Correct: "false" is the correct answer. $$\lnot [\lnot q \lor(p \lor \lnot r)] \equiv \lnot[\lnot F \lor T \lor\lnot F] \equiv \lnot[T] \equiv F$$

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  • $\begingroup$ Because the question was edited I was able to correct my vote. Thanks, $\endgroup$ – Carl Mummert Jul 19 '13 at 23:04
  • $\begingroup$ @CarlMummert Yes, I edited in part due to a previous comment you left me (different post), about the integers in relation to the rationals. I wanted to be sure here not to be too emphatic about $\mathbb Z \subset \mathbb Q$ ;-) $\endgroup$ – Namaste Jul 19 '13 at 23:08

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