$\newcommand{\nfrac}[2]{\frac{\displaystyle{#1}}{\displaystyle{#2}}}$

Problems

1

(b). No. 3/2 is not an integer.

(d). Yes. $cA$ is a totally valid matrix for any scalar $c$ or matrix $A$.

2

(a). Yes.

(b). No. $\frac{\displaystyle{1/2}}{\displaystyle{3}} = \frac{\displaystyle{1}}{\displaystyle{2}} \frac{\displaystyle{1}}{\displaystyle{3}} \neq \frac{\displaystyle{3}}{\displaystyle{2}} = \frac{\displaystyle{1}}{\displaystyle{2/3}}$.

(e). No. $(2^2)^3 = 2^6 \neq 2^8 = 2^{(2^3)}$

3

(c). No. $3(x^2) \neq 3^2 x^2 = (3x)^2$.

(d). No. Known from linear algebra.

5

(a). $20 - 13 = 7$.

(b). The problem is reduced to finding $x$ and $y$ such that $13x = 14y + 1$. In other familiar notation from chapter 1, $13x - 14y = 1$. Clearly $13(-1) + (-14)(-1) = 1$ so $13(-1 + 14) + (-14)(-1 + 13) = 1$. Thus the inverse of $13$ is $13$.

7

Not closed. $1 + 3 = 4$.

No inverse. $3 + x \neq 1$ for any odd integer $x$.

14

$(ab)^3 = ababab$.

$(ab^{-2}c)^{-2} = (ab^{-2}c)^{-1} (ab^{-2}c)^{-1} = c^{-1}b^{-3}a^{-1}c^{-1}b^{-3}a^{-1}$

16

Fact. $x^n$ is an odd integer for any odd $x$.

Fact. The summation of two even integers is even.

We take a different perspective of the problem by the set $\{ (5 \cdot 1), (5 \cdot 3), (5 \cdot 5), (5 \cdot 7) \}$ modulo $5 \cdot 8$. Upon multiplying any two elements we get the form $5 \cdot 5 \cdot x \cdot y$ where $x,y \in \{ 1, 3, 5, 7 \}$. Think of the output of multiplication as the factor of $5$ deciding the element.

Observe the element is decided by $5 \cdot x \cdot y \mod 8$. For example if we knew $5 \cdot 5 \cdot 5 \cdot 1 = (5)(8 + 8 + 8 + 1)$ then we can easily deduce the output of $\mod 5 \cdot 8$ operation is $(5)(1)$.

The numbers $1$, $3$, $5$, and $7$ are all odds. So whatever $x$ or $y$ chosen, $5 \cdot x \cdot y$ will be odd. It follows $\textit{odd} \mod 8 = \textit{odd} \in \{1, 3, 5, 7\}$. To see why note $8k + \textit{odd} = \textit{odd}$.

Lemma. The given set is closed under the given operation.

Lemma. The identity is $5 \cdot 5 = 25$.

Observe $5 \cdot 5 \cdot x \mod 8 = 24x + x \mod 8 = x \mod 8$ since $24x \mod 8 = 0$.

Lemma. THe inverse of $5x$ is $5x$ by computation on the given elements.

Lemma. Associativity is known from integers and modulus properties.

18

$(R_0)^2 = (R_{180})^2 = H^2 = V^2 = D^2 = (D')^2 = R_0$.

$(R_{90})^2 = (R_{270})^2 = R_{180}$.

So $K = \{ R_0, R_{180} \}$, and $L = \{ R_0, R_{180}, H, V, D, D' \}$.

33

Observe the group follows the same pattern as $\mathcal{Z}_4$.

eabcd
eeabcd
aabcde
bbcdea
ccdeab
ddeabc

inverses. Since $ad = e$, $d = a^{-1}$. Since $bc = e$, $c = b^{-1}$.

ab = c. $ab = (cc)b = c(cb) = ce = c$.

db = a. $db = d(aa) = (da)a = ea = a$.

cd = b. $cd = c(bb) = (cb)b = eb = b$.

dc = b. $dc = (bb)c = b(bc) = be = b$.

ac = d. $d = bb = (aa)(dc) = a(ad)c = ac$.

bd = a. $bd = (dc)(bb) = d(cb)b = db = a$.

dd = c. $dd = (ac)(bb) = a(cb)b = ab = c$

34

$(\leftarrow)$. Given $ab = ba$

$$ \begin{aligned} (ab)^2 &= (ab)(ab) \\ &= a(ba)b, \text{Associativity} \\ &= a(ab)b \\ &= (aa)(bb), \text{Associativity} \\ &= a^2b^2 \end{aligned} $$

$(\rightarrow)$. Given $(ab)^2 = a^2b^2$

$$ \begin{aligned} (ab)^2 &= (ab)(ab) \\ &= a(ba)b \\ &= aabb \\ ba &= ab, \text{Cancellation} \end{aligned} $$

$(\leftarrow)$. Given $ab = ba$

$$ \begin{aligned} (ab)^2 &= (ab)^{-1} (ab)^{-1} \\ &= b^{-1} a^{-1} b^{-1} a^{-1} \\ &= b^{-1} (ba)^{-1} a^{-1} \\ &= b^{-1} (ab)^{-1} a^{-1} \\ &= b^{-1} b^{-1} a^{-1} a^{-1} \\ &= (b)^{-2} (a)^{-2} \end{aligned} $$

$(\rightarrow)$. Given $(ab)^{-2} = b^{-2} a^{-2}$

$$ \begin{aligned} (ab)^{-1} (ab)^{-1} &= b^{-1} a^{-1} b^{-1} a^{-1} \\ &= b^{-1} b^{-1} a^{-1} a^{-1} \\ a^{-1} b^{-1} &= b^{-1} a^{-1}, \text{Cancellation} \\ (ba)^{-1} &= (ab)^{-1} \end{aligned} $$

Now observe by the definition of inverse, if $x = y^{-1}$ then $y = x^{-1}$. Therefore $ab = [ (ab)^{-1} ]^{-1}$ and $ba = [ (ba)^{-1} ]^{-1}$, and $ab = ba$.

47

Clearly $aabb = a^2b^2 = ee = e$, and $abab = (ab)^2 = e$. It follows $aabb = abab$, and by cancellation $ab = ba$.