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Rewrite each of the following sentences using logical connectives. Assume that each symbol $f, x_0, n, x, S, B$ represents some fixed object.

(a) If $f$ has a relative minimum at $x_0$ and if $f$ is differentiable at $x_0$, then $f’(x_0)=0$

a) (($f$ has a relative minimum at $x_0$) $\land$ ( $f$ is differentiable at $x_0$))$ \implies$ $f’(x_0)=0$

(b) If $n$ is prime, then $n = 2$ or $n $ is odd.

b) $n$ is prime $\implies$ (($n = 2$) $\lor$ ($n$ is odd))

(c) A number $x$ is real and not rational whenever $x$ is irrational.

c) ((A number $x$ is real) $\land$ $\lnot$ (rational)) $\iff$ $x$ is irrational.

(d) If $x=1$ or $x=−1$,then $|x| =1$.

d) (($x=1$) $\lor$ ($x=−1$)) $\implies$ $|x| =1$

(e) $f$ has a critical point at $x_0$ iff $f’(x_0) = 0$ or $f ′(x_0)$ does not exist.

e) ($f$ has a critical point at $x_0$) $\iff$ (($f’(x_0) = 0$) $\lor$ ($f ′(x_0)$ does not exist))

(f) $S$ is compact iff $S$ is closed and bounded.

f) $S$ is compact $\iff$ (($S$ is closed) $\land$ (bounded)).

(g) $B$ is invertible is a necessary and sufficient condition for $\det B$ not equal to $0$.

g) ($B$ is invertible) $\iff$ $\lnot (\det B= 0)$.

(h) $6\geq n−3$ only if $n>4$ or $n>10$.

h) (($n>4$) $\lor$ ($n>10$)) $\implies$ (($6> n−3$) $\lor$ ($6=n-3$))

(i) $x$ is Cauchy implies $x$ is convergent.

i) $x$ is Cauchy $\implies$ $x$ is convergent.

(j) $f$ is continuous at $x_0$ whenever $\lim_{x\to x_0} f(x)=f(x_0)$

j) $f$ is continuous at $x_0$ $\iff$ \$lim_{x\to x_0} f(x)=f(x_0)$

(k) If $f$ is differentiable at $x_0$ and $f$ is strictly increasing at $x_0$, then $f′(x_0) > 0$.

k) (( $f$ is differentiable at $x_0$) $\land$ ($f$ is strictly increasing at $x_0$)) $\implies$ $f′(x_0) > 0$.

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    $\begingroup$ There is no indication in the sentence (c) that x is irrational being a necessary condition for the x is real and not rational, otherwise the sentence would look like x is real and not rational when and only when x is irrational. So implication should be used instead of biconditional connective. $\endgroup$ – anfauglit Sep 23 '18 at 21:53
  • $\begingroup$ @anfauglit you are right, thanks $\endgroup$ – Dima Sep 24 '18 at 11:14
  • $\begingroup$ @Dima - In my answer I assumed that you are working in propositional logic, and then quantifiers are not allowed. This is consistent with the fact that "each symbol $f, x_0, n, x, S, B$ represents some fixed object". $\endgroup$ – Taroccoesbrocco Sep 24 '18 at 16:26
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They are correct, except (c), (h) and (j).

In (c) and (j), "whenever" stands for "if", i.e. it expresses a sufficient condition, which in general is not necessary. "$A$ whenever $B$" has the same meaning as "if $B$ then $A$". See here for more details.

In (h), "only if" expresses a necessary condition, which in general is not sufficient. It is the dual of "if": "$A$ only if $B$" has the same meaning as "if $A$ then $B$". A necessary and sufficient condition is expressed by "if and only if". See here for more details.

The correct translations of the sentences (c), (h) and (j) are the following (I assume you are not allowed to use quantifiers):

(c) A number $x$ is real and not rational whenever $x$ is irrational.

c) A number $x$ is irrational $\implies$ (($x$ is real) $\land$ $\lnot$ ($x$ is rational))

or more precisely,

c') $x$ is a number $\implies$ ($x$ is irrational $\implies$ (($x$ is real) $\land$ $\lnot$ ($x$ is rational))).

(h) $6\geq n−3$ only if $n>4$ or $n>10$.

h) (($6> n−3$) $\lor$ ($6=n-3$)) $\implies$ (($n>4$) $\lor$ ($n>10$)).

(j) $f$ is continuous at $x_0$ whenever $\lim_{x\to x_0} f(x)=f(x_0)$

j) $\lim_{x\to x_0} f(x)=f(x_0)$ $\implies$ $f$ is continuous at $x_0$.

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