202009270104 Homework 1 (Q1)

i. Prove that $\sqrt{6}$ is irrational; and

ii. Determine whether $\sqrt{2}+\sqrt{3}$ is rational or not.

Solution.

i. Assume on the contrary that $\sqrt{6}$ be rational, may I rewrite it in terms of a quotient (or a fraction) where the numerator and the nonzero denominator are a pair of coprime integers.

I.e., $\sqrt{6}=\displaystyle{\frac{p}{q}}$ where $p,q \in\mathbb{Z}$ and $(p,q)=1$ .

Roughwork.

$\begin{aligned} \sqrt{6} & = \frac{p}{q} \\ 6 & = \frac{p^2}{q^2} \\ p^2 & = 6q^2 \\ & = 2(3q^2) \\ \therefore \qquad 2\big| p^2 & \Rightarrow 2\big| p \textrm{\qquad(why?)} \\ \end{aligned}$

From $p^2=6q^2$ we have $2\big| p$ .

So let $p=2r$ for some integer $r$ .

Roughwork.

We have also seen $2\big| q$ .

If and when both statements $2\big| p$ and $2\big| q$ meet, it implies $p$ and $q$ are no more coprime, and thus a contradiction.

Definition. (Coprime) Two integers $a$ and $b$ are said to be coprime if the only positive integer that divides them both is one. Equivalently speaking, the greatest common divisor (gcd) of $a$ and $b$ is 1,i.e., $\textrm{gcd}(a,b)=1$ , or written simply, $(a,b)=1$ . Synonymous with coprime' are relatively prime’, `mutually prime’, and the like.

ii.

To prove or disprove from scratch, assume that $\sqrt{2}+\sqrt{3}$ be rational, and see what happens. Let

$\sqrt{2}+\sqrt{3}=\displaystyle{\frac{p}{q}}$

for some coprimes $p,\, q\in\mathbb{Z}$ s.t. $(p,q)=1$ .

$\begin{aligned} \sqrt{3} & = \frac{p}{q} - \sqrt{2} \\ 3 & = \frac{p^2}{q^2} - 2\bigg(\frac{p}{q}\bigg)(\sqrt{2}) +2 \\ 2(\sqrt{2})\bigg(\frac{p}{q}\bigg) & = \frac{p^2}{q^2} - 1 = \frac{p^2-q^2}{q^2} \\ \sqrt{2} & = \frac{p^2-q^2}{2pq} \\ \sqrt{2} & = \frac{p}{2q} - \frac{q}{2p} \in \mathbb{Q} \\ \textrm{As is proven,\enspace} & \sqrt{2}\in \mathbb{R}\backslash\mathbb{Q} \\ \end{aligned}$

Contradiction arises $( \Rightarrow \Leftarrow )$ .

$\therefore$ $\sqrt{2}+\sqrt{3}$ is irrational.

(to be continued)

(doing another way around)

To make use of part i., one might find the Lemma below useful:

Lemma. The product of any two rational numbers is again one rational number.

Proof. Let $\displaystyle{\frac{a}{b}}$ and $\displaystyle{\frac{c}{d}}$
be rational numbers in their simplest forms reducible.
$\begin{aligned} \bigg( \frac{a}{b}\bigg) \bigg( \frac{c}{d}\bigg) & = \bigg( \frac{ac}{bd}\bigg) \\ \because \quad a,b,c,d\in\mathbb{Z} & \Rightarrow ac,\, bd\in\mathbb{Z}\\ \therefore \bigg( \frac{a}{b}\bigg) , \bigg( \frac{c}{d}\bigg) \in \mathbb{Q} & \Rightarrow \bigg( \frac{ac}{bd}\bigg) \in \mathbb{Q} \\ \end{aligned}$

If $\sqrt{2}+\sqrt{3}$ were rational, again were its square rational in view of the aforementioned Lemma. Just make an experiment in so doing:

$\begin{aligned} (\sqrt{2}+\sqrt{3})^2 & = (\sqrt{2})^2 + 2(\sqrt{2})(\sqrt{3}) + (\sqrt{3})^2 \\ & = 5 + 2\sqrt{6} \\ \end{aligned}$

On logic,

$\begin{aligned} & \quad (\sqrt{2}+\sqrt{3})^2 \textrm{\quad is rational} \\ & \Rightarrow 5+2\sqrt{6} \textrm{\quad is rational} \\ & \stackrel{\textrm{why?}}{\Longrightarrow} \sqrt{6} \textrm{\quad is rational} \\ \end{aligned}$

However, as shown in part i., $\sqrt{6}$ is irrational. The assumption that $\sqrt{2}+\sqrt{3}$ be rational has been contradicted.

The contrary is true that $\sqrt{2}+\sqrt{3}$ is $\textrm{\scriptsize{NOT}}$ rational.

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