Day: September 27, 2020

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202009272336 Homework 1 (Q3)

Verify the following statements by definition of limit.

[Hint: Use the (\epsilon ,\,\delta )-language.]

i. \displaystyle{\lim_{n\to \infty}\frac{2n^2-1}{4n^2+2}=\frac{1}{2}};

ii. \displaystyle{\lim_{n\to \infty}\frac{\sqrt{n^2+n}}{n}=1};

iii. \displaystyle{\lim_{n\to \infty}\sqrt[n]{n+2}=1}.

Verification.

i. Given

\displaystyle{\lim_{n\to \infty}\frac{2n^2-1}{4n^2+2}=\frac{1}{2}},

or,

\displaystyle{\lim_{n\to \infty}f(n)=\frac{1}{2}},

in which

\displaystyle{f(n)=\frac{2n^2-1}{4n^2+2}}.

Informally speaking of it, the function f(n) shall go to the limit 1/2 if its variable n goes to the (+\textrm{ve}) infinity.

Remark.

In other words, if a function f tends to its limit L, by that it is to say, there exists some output value(s) f(x) sufficiently close to the number L.

Make-up arguments.

\begin{aligned} & \quad\enspace \bigg| \frac{2n^2-1}{4n^2+2} - \frac{1}{2} \bigg| \\ & = \bigg| \frac{(2n^2-1)-(2n^2+1)}{4n^2+2} \bigg| \\ & = \bigg| \frac{-2}{4n^2+2}\bigg| \\ & = \frac{1}{2n^2+1} \qquad\qquad \big( < \frac{1}{2n^2} \big) \\ \end{aligned}

As needs

\big| f(n)-L\big| = \displaystyle{\frac{1}{2n^2+1}} < \epsilon,

so let

\begin{aligned} \epsilon & = \frac{1}{2n^2} \\ n & = \frac{\sqrt{2}}{2\sqrt{\epsilon}} \\ \end{aligned}

Roughwork. (Instructive)

\forall\, \epsilon >0,

\exists\, n>N(\epsilon )=\displaystyle{\frac{\sqrt{2}}{2\sqrt{\epsilon}}} s.t.

\begin{aligned} & \quad\enspace \bigg| \frac{2n^2-1}{4n^2+2} - \frac{1}{2} \bigg| \\ & = \frac{1}{2n^2+1} \\ & < \frac{1}{2n^2} \\ & < \frac{1}{2(\frac{\sqrt{2}}{2\sqrt{\epsilon}})^2} \\ & = \epsilon \\ \end{aligned}

(to be continued)

Definition. (Limit) Let f be a real-valued function defined on a subset D of the real numbers \mathbb{R}. Let c be a limit point of D and let L be a real number. Symbolically:

\begin{aligned} & \qquad\enspace \lim_{x\to c} f(x) = L \\ & \Longleftrightarrow \bigg( \forall\, \epsilon >0,\, \exists\, \delta >0,\, \forall\, x\in D, 0<|x-c|<\delta \Rightarrow \big| f(x)-L \big| < \epsilon \bigg) \\ \end{aligned}

Wikipedia on (\epsilon ,\,\delta )-definition of limit

(continue)

The presentation below is based on the Suggested Solution:

First,

\bigg| \displaystyle{\frac{2n^2-1}{4n^2+2}-\frac{1}{2}}\bigg| = \bigg| \frac{1}{2n^2+1}\bigg| <\bigg| \frac{1}{n^2} \bigg|.

Then, just do let

N=N(\epsilon )=\bigg[ \displaystyle{\frac{1}{\sqrt{\epsilon}}} \bigg] +1.

Lastly, \forall\, \epsilon >0, it stands that

\bigg| \displaystyle{\frac{2n^2-1}{4n^2+2}-\frac{1}{2}}\bigg| <\epsilon

whenever n\ge N.

QED

Part ii. and part iii. are noteworthy exercises that have yet to be done.

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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.

\begin{aligned} (2r)^2 & = 6q^2 \qquad\qquad (\exists\, r\in\mathbb{Z}) \\ 4r^2 & = 6q ^2 \\ 3q^2 & = 2r^2 \\ 2\big| 3q^2 & \stackrel{\textrm{why?}}{\Longrightarrow} 2\big| q^2 \\ & \Rightarrow 2\big| q \\ \end{aligned}

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.