#### Exercise 1.2.1:

Verify that the set of complex numbers described in Example 4 is a subfield of $\mathbb C$.

Solution: Let $F=\{x+y\sqrt{2}\mid x,y\in\mb Q\}$. Then we must show six things:

- $0$ is in $F$
- $1$ is in $F$
- If $x$ and $y$ are in $F$ then so is $x+y$
- If $x$ is in $F$ then so is $-x$
- If $x$ and $y$ are in $F$ then so is $xy$
- If $x\not=0$ is in $F$ then so is $x^{-1}$

For 1, take $x=y=0$.

For 2, take $x=1$, $y=0$.

For 3, suppose $x=a+b\sqrt{2}$ and $y=c+d\sqrt{2}$. Then $x+y=(a+c)+(b+d)\sqrt{2}\in F$.

For 4, suppose $x=a+b\sqrt{2}$. Then $-x=(-a)+(-b)\sqrt{2}\in F$.

For 5, suppose $x=a+b\sqrt{2}$ and $y=c+d\sqrt{2}$. Then $$xy=(a+b\sqrt{2})(c+d\sqrt{2})=(ac+ 2bd) + (ad+bc)\sqrt{2}\in F.$$For 6, suppose $x=a+b\sqrt{2}$ where at least one of $a$ or $b$ is not zero. Let $n=a^2-2b^2$. Since $\sqrt{2}$ is not rational, we have $n\ne 0$. Let $y=a/n + (-b/n)\sqrt{2}\in F$. Then $$xy=\frac1n(a+b\sqrt{2})(a-b\sqrt{2})=\frac1n(a^2-2b^2)=1.$$ Thus $y=x^{-1}$ and $y\in F$.

#### Exercise 1.2.2:

Let $F$ be the field of complex numbers. Are the following two systems of linear equations equivalent? If so, express each equation in each system as a linear combination of the equations in the other system.

Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the second, and conversely.

\begin{alignat*}{2}

3x_1+x_2&=\frac13(x_1-x_2)+\frac43(2x_1+x_2)\\

x_1+x_2 &= \frac{-1}3(x_1-x_2)+\frac23(2x_1+x_2)\\

x_1-x_2 &= (3x_1+x_2)-2(x_1+x_2)\\

2x_1+x_2&=\frac12(3x_1+x_2)+\frac12(x_1+x_2)

\end{alignat*}

#### Exercise 1.2.3:

Test the following systems of equations as in Exercise 2.

\begin{alignat*}{6}

-x_1 &+ x_2 &+ 4x_3 &=0\quad\quad x_1 & &-x_3 &=0\\

x_1 &+ 3x_2 &+ 8x_3 &=0 & x_2 &+ x_3 &=0\\

{\scriptstyle \frac12}x_1 &+ x_2 &+{\scriptstyle\frac52} x_3 &=0 & & &

\end{alignat*}Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the second, and conversely.

\begin{alignat*}{1}

x_1-x_3 &= {\frac{-3}4}(-x_1+x_2+4x_3) + {\frac14}(x_1+3x_3+8x_3)\\

x_2+3x_3&={\frac14}(-x_1+x_2+4x_3) + {\frac14}(x_1+3x_3+8x_3)

\end{alignat*}

and

\begin{alignat*}{1}

-x_1+x_2+4x_3 &= -(x_1-x_3) + (x_2+3x_3)\\

x_1+3x_2+8x_3&= (x_1-x_3) + 3(x_2+3x_3)\\

{\frac12}x_1+x_2+{\frac52}x_3&= {\frac12}(x_1-x_3) + (x_2+3x_3)

\end{alignat*}

#### Exercise 1.2.4:

Test the following systems as in Exercie 2.

\begin{alignat*}{8}

2x_1 &+ (-1+i) x_2 & &+ x_4 =0 \quad\quad (1+{\scriptstyle\frac{i}2})x_1&+8x_2 &-ix_3 &-x_4 &=0\\

& & 3x_2 – 2ix_3 &+5x_4 =0\quad\quad {\scriptstyle\frac23}x_1 &-{\scriptstyle\frac12}x_2 &+ x_3 &+ 7x_4 &=0

\end{alignat*}Solution: These systems are not equivalent.

Call the two equations in the first system $E_1$ and $E_2$ and the equations in the second system $E’_1$ and $E’_2$. Then if $E’_2=aE_1+bE_2$ since $E_2$ does not have $x_1$ we must have $a=1/3$. But then to get the coefficient of $x_4$ we’d need $$7x_4=\dfrac13x_4+5bx_4.$$ That forces $b=\dfrac43$. But if $a=\dfrac13$ and $b=\dfrac43$ then the coefficient of $x_3$ would have to be $-2i\dfrac43$ which does not equal $1$. Therefore the systems cannot be equivalent.

#### Exercise 1.2.5:

Let $F$ be a set which contains exactly two elements, $0$ and $1$. Define an addition and multiplication by the tables:

$$\begin{array}{c|cc}

+ & 0 & 1\\ \hline

0 & 0 & 1\\

1 & 1 & 0

\end{array}

\quad\quad

\begin{array}{c|cc}

\cdot & 0 & 1\\ \hline

0 & 0 & 0\\

0 & 0 & 1

\end{array}

$$Solution: We must check the nine conditions on pages 1-2:

- An operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right. This is true for the addition table so addition is commutative.
- There are eight cases. But if $x=y=z=0$ or $x=y=z=1$ then it is obvious. So there are six non-trivial cases. If there’s exactly one $1$ and two $0$’s then both sides equal $1$. If there are exactly two $1$’s and one $0$ then both sides equal $0$. So addition is associative.
- By inspection of the addition table, the element called $0$ indeed acts like a zero, it has no effect when added to another element.
- $1+1=0$ so the additive inverse of $1$ is $1$. And $0+0=0$ so the additive inverse of $0$ is $0$. In other words $-1=1$ and $-0=0$. So every element has an additive inverse.
- As stated in 1, an operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right. This is true for the multiplication table so multiplication is commutative.
- As with addition, there are eight cases. If $x=y=z=1$ then it is obvious. Otherwise at least one of $x$, $y$ or $z$ must equal $0$. In this case both $x(yz)$ and $(xy)z$ equal zero. Thus multiplication is associative.
- By inspection of the multiplication table, the element called $1$ indeed acts like a one, it has no effect when multiplied to another element.
- There is only one non-zero element, $1$. And $1\cdot1=1$. So $1$ has a multiplicative inverse. In other words $1^{-1}=1$.
- There are eight cases. If $x=0$ then clearly both sides equal zero. That takes care of four cases. If all three $x=y=z=1$ then it is obvious. So we are down to three cases. If $x=1$ and $y=z=0$ then both sides are zero. So we’re down to the two cases where $x=1$ and one of $y$ or $z$ equals $1$ and the other equals $0$. In this case both sides equal $1$. So $x(y+z)=(x+y)z$ in all eight cases.

#### Exercise 1.2.6:

Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.

Solution: Write the two systems as follows:

$$

\begin{array}{c}

a_{11}x+a_{12}y=0\\

a_{21}x+a_{22}y=0\\

\vdots\\

a_{m1}x+a_{m2}y=0

\end{array}

\quad\quad

\begin{array}{c}

b_{11}x+b_{12}y=0\\

b_{21}x+b_{22}y=0\\

\vdots\\

b_{m1}x+b_{m2}y=0

\end{array}

$$ Each system consists of a set of lines through the origin $(0,0)$ in the $x$-$y$ plane. Thus the two systems have the same solutions if and only if they either both have $(0,0)$ as their only solution or if both have a single line $ux+vy-0$ as their common solution. In the latter case all equations are simply multiples of the same line, so clearly the two systems are equivalent. So assume that both systems have $(0,0)$ as their only solution. Assume without loss of generality that the first two equations in the first system give different lines. Then

\begin{equation}

\frac{a_{11}}{a_{12}}\not=\frac{a_{21}}{a_{22}}

\label{eq1}

\end{equation} We need to show that there’s a $(u,v)$ which solves the following system:

$$\begin{array}{c}

a_{11}u+a_{12}v=b_{i1}\\

a_{21}u+a_{22}v=b_{i2}

\end{array}$$ Solving for $u$ and $v$ we get

$$u=\frac{a_{22}b_{i1}-a_{12}b_{i2}}{a_{11}a_{22}-a_{12}a_{21}}$$ $$v=\frac{a_{11}b_{i2}-a_{21}b_{i1}}{a_{11}a_{22}-a_{12}a_{12}}$$ By (\ref{eq1}) $a_{11}a_{22}-a_{12}a_{21}\not=0$. Thus both $u$ and $v$ are well defined. So we can write any equation in the second system as a combination of equations in the first. Analogously we can write any equation in the first system in terms of the second.

#### Exercise 1.2.7:

Prove that each subfield of the field of complex numbers contains every rational number.

Solution: Every subfield of $\C$ has characterisitc zero since if $\mb F$ is a subfield then $1\in \mb F$ and $n\cdot 1=0$ in $\mb F$ implies $n\cdot1=0$ in $\C$. But we know $n\cdot1=0$ in $\C$ implies $n=0$. So $1,2,3,\dots$ are all distinct elements of $\mb F$. And since $\mb F$ has additive inverses $-1,-2,-3,\dots$ are also in $\mb F$. And since $\mb F$ is a field also $0\in \mb F$. Thus $\Z\subseteq \mb F$.

Now $\mb F$ has multiplicative inverses so $\pm\frac1n\in \mb F$ for all natural numbers $n$. Now let $\frac mn$ be any element of $\mb Q$. Then we have shown that $m$ and $\frac1n$ are in $\mb F$. Thus their product $m\cdot\frac1n$ is in $\mb F$. Thus $\frac mn\in \mb F$. Thus we have shown all elements of $\mb Q$ are in $\mb F$.

#### Exercise 1.2.8:

Prove that each field of characteristic zero contains a copy of the rational number field.

Solution: Call the additive and multiplicative identities of $\mb F$ $0_{\mb F}$ and $1_{\mb F}$ respectively. Define $n_{\mb F}$ to be the sum of $n$ $1_{\mb F}$’s. So $$n_{\mb F}=1_{\mb F}+1_{\mb F}+\cdots+1_{\mb F}$$ ($n$ copies of $1_{\mb F}$).

Define $-n_{\mb F}$ to be the additive inverse of $n_{\mb F}$. Since ${\mb F}$ has characteristic zero, if $n\not=m$ then $n_{\mb F}\not=m_{\mb F}$.

For $m,n\in\Z$, $n\not=0$, let $\left(\dfrac{m}{n}\right)_{\mb F}=m_{\mb F}\cdot n_{\mb F}^{-1}$. Since ${\mb F}$ has characteristic zero, if $\dfrac{m}{n}\not=\dfrac{m’}{n’}$ then $$\left(\dfrac{m}{n}\right)_{\mb F}\not=\left(\dfrac{m’}{n’}\right)_{\mb F}.$$ Therefore the map $\dfrac mn\mapsto\left(\dfrac{m}{n}\right)_{\mb F}$ gives a one-to-one map from $\mb Q$ to $\mb F$.

Call this map $h$. Then $h(0)=0_{\mb F}$, $h(1)=1_{\mb F}$ and in general $h(x+y)=h(x)+h(y)$ and $h(xy)=h(x)h(y)$. Thus we have found a subset of $\mb F$ that is in one-to-one correspondence to $\mb Q$ and which has the same field structure as $\mb Q$.

From http://greggrant.org