Home page for accesible maths E Solutions E.7 Online-assessed exercises from week 3 E.9 Tutor-assessed exercises from week 2

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E.8 Tutor-assessed exercises from week 1

Exercise 1.16. The two statements are logically equivalent because the $7^{\text{th}}$ and $9^{\text{th}}$ columns of the following truth table are the same.

$p$	$q$	$r$	$p\Rightarrow q$	$q\Rightarrow r$	$(p\Rightarrow q)\Rightarrow(q\Rightarrow r)$	$\neg\bigl((p\Rightarrow q)\Rightarrow(q\Rightarrow r)\bigr)$	$\neg r$	$q\ \&\ (\neg r)$
$T$	$T$	$T$	$T$	$T$	$T$	$F$	$F$	$F$
$T$	$T$	$F$	$T$	$F$	$F$	$T$	$T$	$T$
$T$	$F$	$T$	$F$	$T$	$T$	$F$	$F$	$F$
$T$	$F$	$F$	$F$	$T$	$T$	$F$	$T$	$F$
$F$	$T$	$T$	$T$	$T$	$T$	$F$	$F$	$F$
$F$	$T$	$F$	$T$	$F$	$F$	$T$	$T$	$T$
$F$	$F$	$T$	$T$	$T$	$T$	$F$	$F$	$F$
$F$	$F$	$F$	$T$	$T$	$T$	$F$	$T$	$F$

Exercise 1.17.

(a)

$\displaystyle{\bigl(\forall m\in\{-3,-2,-1,1,2,3\}\bigr)(\exists\,q\in\mathbb{% Q})\Biggl((q>0)\ \ \&\ \ \biggl(\frac{1}{m}=q-\frac{3}{2}\biggr)\Biggr)}$ .
(b)

Negation: $\displaystyle{\bigl(\exists\,m\in\{-3,-2,-1,1,2,3\}\bigr)(\forall q\in\mathbb{% Q})\Biggl((q\leqslant 0)\ \ \text{or}\ \ \biggl(\frac{1}{m}\neq q-\frac{3}{2}% \biggr)\Biggr)}$ .
(c)

The statement in part (a) is true. Indeed, suppose that $m\in\{-3,-2,-1,1,2,3\}$ , and let

$q=\frac{1}{m}+\frac{3}{2}$

(found by solving the equation $\frac{1}{m}=q-\frac{3}{2}$ for $q$ ). Then certainly the equation $\frac{1}{m}=q-\frac{3}{2}$ holds, $q$ is rational because $\frac{1}{m}$ and $\frac{3}{2}$ are, and $q$ is positive because $\frac{1}{m}\geqslant-1$ for each $m\in\{-3,-2,-1,1,2,3\}$ , so that

$q=\frac{1}{m}+\frac{3}{2}\geqslant-1+\frac{3}{2}=\frac{1}{2}>0.$

Bonus exercise 1.23. We claim that ‘‘ $p\Rightarrow q$ ’’ is logically equivalent to ‘‘ $\neg\bigl(p\ \&\ (\neg q)\bigr)$ ’’ for all statement variables $p$ and $q$ . This follows from the fact that the third and sixth column of the following truth table are the same.

$p$	$q$	$p\Rightarrow q$	$\neg q$	$p\ \&\ (\neg q)$	$\neg\bigl(p\ \&\ (\neg q)\bigr)$
$T$	$T$	$T$	$F$	$F$	$T$
$T$	$F$	$F$	$T$	$T$	$F$
$F$	$T$	$T$	$F$	$F$	$T$
$F$	$F$	$T$	$T$	$F$	$T$

Bonus exercise 1.24.

(\exists\,x\in\mathbb{R})(\exists\,y\in\mathbb{R})\bigl((x\neq y)\ \&\ (2x^{3}% +x^{2}-4x+4=0)\ \&\ (2y^{3}+y^{2}-4y+4=0)\bigr).

Alternatively, we may combine the first two quantifiers and write this statement as

(\exists\,x,y\in\mathbb{R})\bigl((x\neq y)\ \&\ (2x^{3}+x^{2}-4x+4=0)\ \&\ (2y% ^{3}+y^{2}-4y+4=0)\bigr).

This statement is false. Indeed, by experimentation we find that $-2$ is a root of the polynomial $2x^{3}+x^{2}-4x+4$ , and the division algorithm then shows that

2x^{3}+x^{2}-4x+4=(x+2)(2x^{2}-3x+2),

where the quadratic factor $2x^{2}-3x+2$ has no real roots because its discriminant $D=(-3)^{2}-4\cdot 2\cdot 2=-7$ is negative. Hence $2x^{3}+x^{2}-4x+4$ has exactly one real root (namely $-2$ ).