Let $m$ and $n$ be positive integers with no common factor. Prove that if $\sqrt{m / n}$ is rational, then $m$ and $n$ are both perfect squares, that is to say there exist integers $p$ and $q$ such that $m=p^2$ and $n=q^2$. (This is proved in Proposition 9 of Book $\mathrm{X}$ of Euclids Elements).

Assume $\sqrt{m / n}$ is rational. Then there exist positive integers $M$ and $N$ with no common factor such that $\sqrt{m / n}=M / N$ and so $m N^2=n M^2$.
Claim: $M^2$ divides $m$ and $N^2$ divides $n$.
Assume the claim for now. Then
$$
m=M^2 m^{\prime} \text { and } n=N^2 n^{\prime} \text { for some } m^{\prime} \text { and } n^{\prime} \text {. }
$$
Substituting we obtain $M^2 m^{\prime} N^2=N^2 n^{\prime} M^2$ which gives $m^{\prime}=n^{\prime} . m^{\prime}=n^{\prime}$ divides $m$ and $n$ so $m^{\prime}=n^{\prime}=1$ and we have shown $m$ and $n$ are perfect squares.

Proof of claim: We show that $M^2$ divides $m$; the argument that $N^2$ divides $n$ is identical. Write $M$ as a product of primes $p_1 \cdots p_r$ and note that no $p_i$ divides $N$. Assume inductively that $p_1^2 \cdots p_t^2$ divides $m$. Then
$$
p_{t+1}^2\left|\frac{M^2}{p_1^2 \cdots p_t^2}\right| \frac{m}{p_1^2 \cdots p_t^2} N^2
$$
Since $p_{t+1}$ does not divide $N^2$ we see
$$
p_{t+1}^2 \mid \frac{m}{p_1^2 \cdots p_t^2}, \text { which gives } p_1^2 \cdots p_{t+1}^2 \mid m \text {. }
$$
The inductive hypothesis holds when $t=0$; the empty product is 1 . Thus, by induction $p_1^2 \cdots p_r^2=M^2$ divides $m$.

We define
$$
F=\mathbb{Q}={a+\sqrt{2} b \mid a, b \in \mathbb{Q}} \subset \mathbb{R}
$$
We wish to show that $F$ is a subfield of $\mathbb{R}$. In order to show this, we need to show that a) $0,1 \in F$; b) $F$ is closed under addition and multiplication; and c) if $x \in F$ and $x \neq 0$, then $-x \in F$ and $1 / x \in F$. The commutative, associate, and distributive properties all follow from the corresponding properties on $\mathbb{R}$.
a), b), and the first half of c) are straightforward; we have $0=0+0 \sqrt{2} \in F$ and $1=1+0 \sqrt{2} \in F$. For b), we have
$$
(a+b \sqrt{2})+(c+d \sqrt{2})=(a+c)+(b+d) \sqrt{2} \in F
$$
and
$$
(a+b \sqrt{2})(c+d \sqrt{2})=(a b+2 c d)+(a d+b c) \sqrt{2} \in F
$$
If $x=a+b \sqrt{2}$, then $-x=(-a)+(-b) \sqrt{2} \in F$. So the only fact remaining to show is that $F$ is closed under multiplicative inverses.
To prove this, we need the following
Fact: if $0=a+b \sqrt{2} \in F$, then $a=b=0$

Proof: Suppose $b \neq 0$. Then $\sqrt{2}=-a / b \in \mathbb{Q}$, a contradiction. So we must have $b=0$, and then $0=a+0=a$.

Now take $x=a+b \sqrt{2} \in F, x \neq 0$. By the above fact, $a-b \sqrt{2}$ is also nonzero, and hence
$$
a^2-2 b^2=(a+b \sqrt{2})(a-b \sqrt{2}) \neq 0
$$
Since the product of non-zero real numbers is non-zero.
So we can define $c=a /\left(a^2-2 b^2\right) \in \mathbb{Q}, d=-b /\left(a^2-2 b^2\right) \in \mathbb{Q}$, and $y=c+d \sqrt{2} \in F$. I claim that $x y=1$, so $y=1 / x$ and $F$ contains multiplicative inverses. Indeed,
$$
(a+b \sqrt{2})(c+d \sqrt{2})=\frac{1}{a^2-2 b^2}(a+b \sqrt{2})(a-b \sqrt{2})=\frac{a^2-2 b^2}{a^2-2 b^2}=1
$$
and we are done.

We have a metric space $(X, d)$, and define the function $d^{\prime}(x, y)=\sqrt{d(x, y)}$. We wish to show that $\left(X, d^{\prime}\right)$ is also a metric space with the same open sets as $(X, d)$. We first check that $d^{\prime}$ is a metric.
(a) If $x \neq y$, then $d^{\prime}(x, y)=\sqrt{d(x, y)}>0$ since $d(x, y)>0$, and similarly $d^{\prime}(x, x)=0$
(b) $d^{\prime}(x, y)=\sqrt{d(x, y)}=\sqrt{d(y, x)}=d^{\prime}(y, x)$
(c) For the triangle inequality, we first need the following elementary
Fact: If $a, b \geq 0$, then $\sqrt{a+b} \leq \sqrt{a}+\sqrt{b}$
Indeed, squaring the right hand side gives $a+b+2 \sqrt{a b} \geq a+b$, and the square root function is order preserving. Using this fact, for $x, y, z \in X$ we have
$$
d^{\prime}(x, z)=\sqrt{d(x, z)} \leq \sqrt{d(x, y)+d(y, z)} \leq \sqrt{d(x, y)}+\sqrt{d(y, z)}=d^{\prime}(x, y)+d^{\prime}(y, z)
$$
Now, let $E$ be an open set for $E$. We need to show that it is open for $d^{\prime}$. Let $x \in E$. Then there is some $r>0$ such that the ball of radius $r$ around $x$ is contained in $E$, where the ball is taken with respect to $d$, i.e. $N_r(x) \subset E$. But the ball of radius $r$ with respect to $d$ is the ball of radius $\sqrt{r}$ with respect to $d^{\prime}$, so there is a neighbourhood of $x$ with respect to $d^{\prime}$ contained in $E$. In other words, $E$ is open with respect to $d^{\prime}$. Similarly, a set that is open with respect to $d^{\prime}$ is also open with respect to $d$.

Let $c, t, m, b$ denote the number of cars, trucks, motorcycles, and bicycles. Then the statements from the problem yield the equations:
$$
\begin{aligned}
c+t+m+b & =66 \
c-4 t & =0 \
4 c+4 t+2 m+2 b & =252
\end{aligned}
$$
We form the augmented matrix for this system and row-reduce
$$
\left[\begin{array}{ccccc}
1 & 1 & 1 & 1 & 66 \
1 & -4 & 0 & 0 & 0 \
4 & 4 & 2 & 2 & 252
\end{array}\right] \stackrel{\text { RREF }}{\longrightarrow}\left[\begin{array}{ccccc}
1 & 0 & 0 & 0 & 48 \
0 & 1 & 0 & 0 & 12 \
0 & 0 & 1 & 1 & 6
\end{array}\right]
$$
The first row of the matrix represents the equation $c=48$, so there are 48 cars. The second row of the matrix represents the equation $t=12$, so there are 12 trucks. The third row of the matrix represents the equation $m+b=6$ so there are anywhere from 0 to 6 bicycles. We can also say that $b$ is a free variable, but the context of the problem limits it to 7 integer values since you cannot have a negative number of motorcycles.

Using the independence of orders the chance that only the first two stores place orders is $0.35^2 \cdot 0.65^8$. As there are $10 \times 9 / 2=45$ distinct pairs of stores that could order we have
$$
P(X=2)=450.35^2 0.65^8=0.1757
$$
A similar argument works for any number of orders. We say that the number of orders placed has the $\operatorname{Bin}(10,0.35)$ distribution. The formula for $x$ orders is
$$
P(X=x)=\left(\begin{array}{c}
10 \
x
\end{array}\right) 0.35^x 0.65^{10-x}
$$
The most probable number of orders is 3 (either calculate $P(X=x)$ for a few different $x$ values or look at binomial tables in a textbook) and $P(X=3)=120(0.35)^3(0.65)^7 \approx 0.2522$. The expected number of orders is
$$
\sum_0^{10} x \cdot P(X=x)=1 \cdot 0.0725+2 \cdot 0.1757+3 \cdot 0.2522+4 \cdot 0.2377+\cdots
$$
which (barring numericals errors) will give the same answer as the formula $E(X)=n p=10 \times 0.35=$ $3.5$.
(The problem of which number is most likely for general $n$ and $p$ was not set but is not all that hard – show that $P(X=x+1)(n+1) p$ and think about what that means $)$

The condition about the multiple of the column of constants will allow you to show that the following values form a solution of the system $\mathcal{C S}(A, \mathbf{b})$,
$$
x_1=0 \quad x_2=0 \quad \ldots \quad x_{j-1}=0 \quad x_j=\alpha \quad x_{j+1}=0 \quad \ldots \quad x_{n-1}=0 \quad x_n=0
$$
With one solution of the system known, we can say the system is consistent (Definition CS).
A more involved proof can be built using Theorem RCLS. Begin by proving that each of the three row operations (Definition RO) will convert the augmented matrix of the system into another matrix where column $j$ is $\alpha$ times the entry of the same row in the last column. In other words, the “column multiple property” is preserved under row operations. These proofs will get successively more involved as you work through the three operations.