Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

E.5 Online-assessed exercises from week 1

Exercise 1.18. We construct the truth table; as there are $3$ statement variables involved, it has $2^{3}=8$ rows.

$p$	$q$	$r$	$\neg r$	$q$ or $(\neg r)$	$p\Rightarrow\bigl(q\ \text{or}\ (\neg r)\bigr)$	$\neg q$	$\Bigl(p\Rightarrow\bigl(q\ \text{or}\ (\neg r)\bigr)\Bigr)\ \&\ (\neg q)$
$T$	$T$	$T$	$F$	$T$	$T$	$F$	$F$
$T$	$T$	$F$	$T$	$T$	$T$	$F$	$F$
$T$	$F$	$T$	$F$	$F$	$F$	$T$	$F$
$T$	$F$	$F$	$T$	$T$	$T$	$T$	$T$
$F$	$T$	$T$	$F$	$T$	$T$	$F$	$F$
$F$	$T$	$F$	$T$	$T$	$T$	$F$	$F$
$F$	$F$	$T$	$F$	$F$	$T$	$T$	$T$
$F$	$F$	$F$	$T$	$T$	$T$	$T$	$T$

Thus (E) is the correct answer.

Exercise 1.19. As there are $3$ variables involved, the parity table must have $2^{3}=8$ rows and the following columns.

Thus (C) is the correct answer.

Exercise 1.20. The correct answer is (D). The reason is as follows.

First, since $3n-1\in{\mathbb{Z}}$ for each $n\in{\mathbb{Z}}$ , we have

(3n-1\in L)\ \Leftrightarrow\ (-1\leqslant 3n-1\leqslant 2)\ \Leftrightarrow\ % (0\leqslant 3n\leqslant 3)\ \Leftrightarrow\ (0\leqslant n\leqslant 1),

so that $M=\{0,1\}$ .

Second, by definition we have

N=\{2(-1)+1,2\cdot 0+1,2\cdot 1+1,2\cdot 2+1\}=\{-1,1,3,5\}.

Exercise 1.21. (D) is the correct answer.

Exercise 1.22. The negation of

(\forall x\in\mathbb{R})(\exists\,y\in\mathbb{Q})(\forall z\in\mathbb{R})\bigl% ((z>x)\Rightarrow(z>y)\bigr)

(\exists\,x\in\mathbb{R})(\forall y\in\mathbb{Q})(\exists\,z\in\mathbb{R})% \Bigl(\neg\bigl((z>x)\Rightarrow(z>y)\bigr)\Bigr),

and $\neg\bigl((z>x)\Rightarrow(z>y)\bigr)$ is logically equivalent to $(z>x)\ \&\ (z\leqslant y)$ , so that (A) is the correct answer.