# Section 2.1: Problem 1 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Assume that we have a language with the following parameters: $\forall$ , intended to mean “for all things”; $N$ , intended to mean “is a number”; $I$ , intended to mean “is interesting”; $<$ , intended to mean “is less than”; and $0$ , a constant symbol intended to denote zero. Translate into this language the English sentences listed below. If the English sentence is ambiguous, you will need more than one translation.
(a) Zero is less than any number.
(b) If any number is interesting, then zero is interesting.
(c) No number is less than zero.
(d) Any uninteresting number with the property that all smaller numbers are interesting certainly is interesting.
(e) There is no number such that all numbers are less than it.
(f) There is no number such that no number is less than it.
I saw some ambiguity in (b) only.
(a) $\forall x(Nx\rightarrow0 .
(b) Here “any number” may be understood either as “some number” or as “every number”. Correspondingly, we have $\exists x(Nx\wedge Ix)\rightarrow I0$ (with $\exists$ expressed via $\forall$ : $\neg\forall x(\neg Nx\vee\neg Ix)\rightarrow I0$ ) or $\forall x(Nx\rightarrow Ix)\rightarrow I0$ .
(c) $\forall x(Nx\to\neg x<0)$ .
(d) $\forall x(Nx\wedge\neg Ix\wedge\forall y(Ny\wedge y .
(e) $\neg\exists x(Nx\wedge\forall y(Ny\rightarrow y (with $\exists$ expressed via $\forall$ : $\forall x(Nx\to\neg\forall y(Ny\rightarrow y ).
(f) $\neg\exists x(Nx\wedge\neg\exists y(Ny\wedge y (with $\exists$ expressed via $\forall$ : $\forall x(Nx\to\neg\forall y(Ny\to\neg y ).