# 线性代数代写Linear Algebra|MATH 19620 University of Chicago Assignment

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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 X of Euclid’s Elements).

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 $m$.

Consider the subset of $\mathbb{R}$ defined by
$$\mathbb{Q}(\sqrt{2})={a+\sqrt{2} b: a, b \in \mathbb{Q}}$$
with the usual addition and multiplication. Show that this is a field (you may use all properties of the real numbers).

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.

Let $(X, d)$ be a metric space. Show that $d^{\prime}(x, y)=\sqrt{d(x, y)}$ is also a metric on $X$, and that the open sets for $d^{\prime}$ are the same as the open sets for $d$.

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 $d$.