MATH2340 Week 3

## Cauchy sequences
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is *Cauchy* if

for all $\varepsilon>0$,   $\exists k\in\mathbb N$ such that $m,n>k \implies$ $\lvert a_n-a_m \rvert < \varepsilon$.

Informally, we think of a sequence being Cauchy meaning "the terms of the sequence get closer and closer to one another".
            But we need a more precise understanding for this course.

Note: there is no concept of limit in this definition (yet!).
 
 ---
 ## Cauchy sequences
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is *Cauchy* if

for all $\varepsilon>0$,   $\exists k\in\mathbb N$ such that $m,n>k \implies$ $\lvert a_n-a_m \rvert < \varepsilon$.

#### Or, in words,
 No matter how small we pick $\varepsilon>0$,

we can always go far enough (ie. $k$ far) down the sequence to ensure that

every term after that point (ie. after $k$),

is within $\varepsilon$ of every other term.
 #### How to prove Cauchyness
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 
 ---
 ## Proof of Cauchyness: example 1
 #### Theorem
 Let $a_1=4$ and $a_2=14$, with $a_n = (a\_{n-1}+a\_{n-2})/2$.
 Then $\\{a_n\\}\_{n=1}^\infty$ is Cauchy.
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is *Cauchy* if

#### How to prove Cauchyness
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Proof
 For all $n\geq2$, $\lvert a\_{n+1}-a_n\rvert=\frac12\lvert a_n-a\_{n-1}\rvert$, so $\lvert a\_{n+1}-a_n\rvert=2\^{-(n-1)}\lvert a_2-a_1\rvert$.
 But also $a_1 < a_3 \< a_5 < \ldots < a_6 < a_4 < a_2$.
 Fix any $\varepsilon>0$.
 Let $k=\lceil-\log_2(\frac\varepsilon{10})\rceil$.
 If $m,n>k$, then, we can assume $m\geq n$, so
 $\lvert a_m - a_n \rvert$ $\leq \lvert a\_{n+1}-a_n \rvert = 2\^{-(n-1)}\lvert a_2-a_1\rvert$ $<$ $2\^{-k}\times10 = 2\^{-\lceil-\log_2(\frac\varepsilon{10})\rceil}\times10 \leqslant 2\^{\log_2(\frac\varepsilon{10})}\times10 =$ $\varepsilon$.   □

Note:
            We can't use monotone convergence here, because the sequence is neither increasing nor decreasing.

It has increasing and decreasing subsequences (even and odd indexed terms) but the whole sequence is neither.
 
 ---
 ## Proof of Cauchyness: example 2
 #### Theorem
 Let $a_n=\sum\_{k=1}^n \frac1k$.
 Then $\\{a_n\\}\_{n=1}^\infty$ is not Cauchy.
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is *Cauchy* if

Therefore:

A sequence $\\{a_n\\}\_{n=1}^\infty$ is **not** Cauchy if

there exists $\varepsilon>0$ such that,   $\forall k\in\mathbb N$, $m,n>k \\;\not\nobreak\\!\\!\\!\\!\implies$ $\lvert a_n-a_m \rvert < \varepsilon$.

#### How to prove not Cauchyness
 Pick an $\varepsilon$ as small as you need.
 Then construct a machine that tells you $m,n$ for any given $k$, with $m,n>k$, yet $\lvert a_n-a_m \rvert>\varepsilon$.
 
 ---
 ## Proof of Cauchyness: example 2
 #### Theorem
 Let $ a_n=\sum\_{k=1}^n \frac1k$.
 Then $\\{a_n\\}\_{n=1}^\infty$ is not Cauchy.
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is **not** Cauchy if

#### How to prove not Cauchyness
 Pick an $\varepsilon$ as small as you need.
 Then construct a machine that tells you $m,n$ for any given $k$, with $m,n>k$, yet $\lvert a_n-a_m \rvert>\varepsilon$.
 #### Proof
 Let $\varepsilon=\frac12$.
 For any $k\in\mathbb N$,
 let $n>k$ and $m=2n>k$.

Then
 $\lvert a_m - a_n \rvert$ $= \frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}$ $>$ $\frac n{2n} = $ $\varepsilon$.   □
 
 ---
 ## Why do we care about Cauchyness?
 #### Theorem
 Every Cauchy sequence is convergent, and every convergent sequence is Cauchy.
 #### Proof
 See lecture notes.
 #### Application
 We can prove a sequence is convergent by (often easier) showing it is Cauchy.

We can prove a sequence is divergent by (often easier) showing it is not Cauchy.
            #### Examples
            The sequence $\\{a_n\\}\_{n=1}^\infty$ given by $a_1=4$ and $a_2=14$, with $a_n = (a\_{n-1}+a\_{n-2})/2$ for $n>2$ is convergent.

The sequence $\\{s_n\\}\_{n=1}^\infty$ given by $s_n=\sum\_{k=1}^n \frac1k$ is divergent.
 
 ---
 ## Infinite series
 We want to generalise the concept of $\displaystyle \sum\_{k=1}^n a_k$ to give meaning to $\displaystyle \sum\_{k=1}^\infty a_k$.

Define by $\displaystyle \sum\_{k=1}^\infty a_k = \lim_{n\to\infty} \sum\_{k=1}^n a_k$.

So the *partial sums* $s_n = \displaystyle \sum\_{k=1}^n a_k$ are important, because $\displaystyle \sum\_{k=1}^\infty a_k = \lim_{n\to\infty} s_n$.

#### Definition
 An infinite series is *convergent*/*divergent* if $\\{s_n\\}\_{n=1}^\infty$ is *convergent*/*divergent*.
 #### Example
 $\displaystyle \sum\_{k=1}^\infty \frac1k$ is divergent, by earlier example.
 
 ---
 ## Example of convergent infinite series
 In
 $\displaystyle \sum\_{k=1}^\infty \frac1{k(k+1)}$, the terms in the sum are $\displaystyle \frac1{k(k+1)} = \frac1k-\frac1{k+1}$. The partial sums are

$s_n = \left[ \frac11-\frac12 \right] + \left[ \frac12-\frac13 \right] + \ldots + \left[ \frac1n-\frac1{n+1} \right] = 1 - \frac1{n+1} \to 1$ as $n\to\infty$.

So $\displaystyle \sum\_{k=1}^\infty \frac1{k(k+1)}=1$.
 
 ---
 ## Telescoping sum calculation
 $s_n = \left[ \frac11-\frac12 \right] + \left[ \frac12-\frac13 \right] + \ldots + \left[ \frac1n-\frac1{n+1} \right] = 1 - \frac1{n+1} \to 1$ as $n\to\infty$.

Here we used the cancellation
 \\[\begin{alignedat}{8}
 &\tfrac11 & &- \tfrac12 \\\\
 & & &+ \tfrac12 & & - \tfrac13 \\\\
 & & & & &+ \tfrac13 & & - \tfrac14 \\\\
 & & & & & & & & \hspace{1em}\vdots\hspace{1em} \\\\
 & & & & & & & & & &+ \tfrac1{n-1} & & - \tfrac1n \\\\
 & & & & & & & & & & & &+ \tfrac1{n} & & - \tfrac1{n+1} = \tfrac11 - \tfrac1{n+1}.
 \end{alignedat}\\]
 
 ---
 ## Telescoping sum calculation
 $2s_n = \left[ \frac11-\frac13 \right] + \left[ \frac12-\frac14 \right] + \ldots + \left[ \frac1n-\frac1{n+2} \right] = 1+\tfrac12 -\frac1{n+1} -\frac1{n+2} \to 1$ as $n\to\infty$.

Here we used the cancellation
 \\[\begin{alignedat}{14}
 &\tfrac11 & & & &- \tfrac13 \\\\
 & & &+ \tfrac12 & & & & - \tfrac14 \\\\
 & & & & &+ \tfrac13 & & & & - \tfrac15 \\\\
 & & & & & & &+ \tfrac14 & & & & - \tfrac16 \\\\
 & & & & & & & & & & & & \hspace{1em}\vdots\hspace{1em} \\\\
 & & & & & & & & & & & & & &+ \tfrac1{n-1} & & & & - \tfrac1{n+1} \\\\
 & & & & & & & & & & & & & & & &+ \tfrac1{n} & & & & - \tfrac1{n+2} = \tfrac11+\tfrac12 - \tfrac1{n+1} - \tfrac1{n+2}.
 \end{alignedat}\\]
 So $\sum\_{k=1}^\infty \frac1{k(k+2)}=\frac12\sum\_{k=1}^\infty \frac2{k(k+2)}=\frac12\sum\_{k=1}^\infty \left[\frac1k-\frac1{k+2}\right]=\frac12\times\frac32=\frac34$.
 
 ---
 ## Telescoping series
 In the same way, we can show that, for any $p>0$, the series $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+p)}$ converges, and calculate its limit.
 
 ---
 ## Telescoping series
 In the same way, we can show that, for any $p>0$, the series $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+p)}$ converges, and calculate its limit.

Or can we?
 
 ---
 ## Telescoping series
 In the same way, we can show that, for any $p>0$, the series $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+p)}$ converges, and calculate its limit.

Or can we?

Not if $p\notin\mathbb N$.
 For example, try to make the telescoping sum argument for $p=\frac32$.
 
 ---
 ## Telescoping series
 In the same way, we can show that, for any $p>0$, the series $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+p)}$ converges, and calculate its limit.

Or can we?

Not if $p\notin\mathbb N$.
            For example, try to make the telescoping sum argument for $p=\frac32$.

You never get the cancellation, because the terms are "missing" each other.

Let's try another approach.
 
 ---
 ## Comparison of series
 We know $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+1)}=\sum\_{k=1}^\infty a_k$ converges.
 Try $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)} = \sum\_{k=1}^\infty b_k$.

Here $0 \leqslant b_k \leqslant a_k$ so we are summing smaller things.

So $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)}$ must also converge.
 And we could replace $2$ with any real $p>1$.
 
 ---
 ## Comparison of series
 We know $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+1)}=\sum\_{k=1}^\infty a_k$ converges.
 Try $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)} = \sum\_{k=1}^\infty b_k$.

Here $0 \leqslant b_k \leqslant a_k$ so we are summing smaller things.

So $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)}$ must also converge.
            And we could replace $2$ with any real $p>1$.

#### Theorem (Comparison test)
 If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty a_k$ converges, then so too does $\displaystyle\sum\_{k=1}^\infty b_k$.
 
 ---
 ## Comparison of series
 We know $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+1)}=\sum\_{k=1}^\infty a_k$ converges.
 Try $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)} = \sum\_{k=1}^\infty b_k$.

Here $0 \leqslant b_k \leqslant a_k$ so we are summing smaller things.

So $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)}$ must also converge.
            And we could replace $2$ with any real $p>1$.

#### Corollary
 If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty b_k$ diverges, then so too does $\displaystyle\sum\_{k=1}^\infty a_k$.
 
 ---
 ## Comparison test
 #### Theorem (Comparison test)
 If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty a_k$ converges, then so too does $\displaystyle\sum\_{k=1}^\infty b_k$.

#### Example
            $\sum\_{k=1}^\infty \frac1{k^2}$ converges.

#### Idea
            Want to use comparison with $\sum\_{k=1}^\infty \frac1{k(k+1)}$.
            But can't because $0 \leqslant \frac1{k^2} \leqslant \frac1{k(k+1)}$ is false!

#### Proof
            $\sum\_{k=1}^\infty \frac1{k^2} = \frac11 + \sum\_{k=2}^\infty \frac1{k^2} = \frac11 + \sum\_{k=1}^\infty \frac1{(k+1)^2}$.
            But $0 \leqslant \frac1{(k+1)^2} \leqslant \frac1{k(k+1)}$ and $\sum\_{k=1}^\infty \frac1{k(k+1)}$ converges.
            Therefore, by the comparison test, so too does $\sum\_{k=1}^\infty \frac1{(k+1)^2}$.

Hence $\sum\_{k=1}^\infty \frac1{k^2}$ converges.   □
 
 ---
 ## Comparison test
 #### Theorem (Comparison test)
 If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty a_k$ converges, then so too does $\displaystyle\sum\_{k=1}^\infty b_k$.

#### Example
 $\displaystyle \sum\_{k=1}^\infty \frac1{k^t}$ converges for all $t\geq2$ and diverges for all $t\leq1$.
 
 ---
 ## Comparison test
 #### Theorem (Comparison test)
 If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty a_k$ converges, then so too does $\displaystyle\sum\_{k=1}^\infty b_k$.

#### Example
            $\displaystyle \sum\_{k=1}^\infty \frac1{k^t}$ converges for all $t\geq2$ and diverges for all $t\leq1$.

This follows by comparison with $\displaystyle\sum\_{k=1}^\infty \frac1{k^2}$ and $\displaystyle\sum\_{k=1}^\infty \frac1k$.

We still don't know about $1<t<2$.
 See later.
 
 ---
 ## Absolute convergence
 #### Definition
 If $\sum\_{n=1}^\infty \lvert a_n \rvert$ converges, then the series $\sum\_{n=1}^\infty a_n$ is said to be *absolutely convergent*.
 #### Theorem
 If, a series is absolutely convergent, then it is convergent
 #### Example
 Because $\sum\_{n=1}^\infty \frac1{n^3}$ converges, $\sum\_{n=1}^\infty \frac{(-1)^n}{n^3}$ is absolutely convergent, hence also converges.

#### Example
            Which of

$\sum\_{n=1}^\infty \frac{(-1)^n}{n},\\; \sum\_{n=1}^\infty \frac{n+3}{2-3n-2n^2},\\; \sum\_{n=1}^\infty \frac{500\sin(\pi n/2)}{n^2},\\; \sum\_{n=1}^\infty \frac{(-1)^n\times3}{n-5\pi},\\; \sum\_{n=1}^\infty \frac{(-1)^nn^2}{n+100}$

can you tell are are absolutely convergent, convergent, or divergent?
 
 ---
 ## Absolute convergence
 #### Definition
 If $\sum\_{n=1}^\infty \lvert a_n \rvert$ converges, then the series $\sum\_{n=1}^\infty a_n$ is said to be *absolutely convergent*.
 #### Theorem
 If, a series is absolutely convergent, then it is convergent
 #### Example
 Because $\sum\_{n=1}^\infty \frac1{n^3}$ converges, $\sum\_{n=1}^\infty \frac{(-1)^n}{n^3}$ is absolutely convergent, hence also converges.

#### Example
            The series $\sum\_{n=1}^\infty \frac{(-1)\^{n+1}}{n}$ is not absolutely convergent, but it is convergent.

A series which is convergent but not absolutely convergent is called *conditionally convergent*.
 A conditionally convergent series will converge to different values if you reorder its terms.
 
 ---
 ## Conditional convergence
 #### Example
 The series $S = 1-1+\frac12-\frac12+\frac13-\frac13+\frac14-\frac14+\ldots$ is not absolutely convergent, but it is convergent.

Indeed $S=0$.

But, if we could reorder the terms without changing $S$, then take two positive terms before each negative term to get

$S = 1+\frac12-1+\frac13+\frac14-\frac12+\ldots$

$S = 1-\frac12+\frac13-\frac14+\ldots = \log(2) = \sum\_{n=1}^\infty \frac{(-1)\^{n+1}}{n}$, as we shall prove later.
 
 ---
 ## Ratio test
 #### Example
 $\sum\_{n=1}^\infty \frac{1}{2^n}$ converges.
 #### Proof
 The sequence of partial sums is $s_n = \frac12+\frac14+ \ldots+ \frac1{2^n}=\frac{1-(\frac12)^n}{1-\frac12} - 1 \to \frac1{1/2}-1=1$, as $n\to\infty$.   □
 
 ---
 ## Ratio test
 #### Example
 $\sum\_{n=1}^\infty \frac{1}{2^n}$ converges.
 #### Proof
 The sequence of partial sums is $s_n = \frac12+\frac14+ \ldots+ \frac1{2^n}=\frac{1-(\frac12)^n}{1-\frac12} - 1 \to \frac1{1/2}-1=1$, as $n\to\infty$.   □

Here the ratio between consecutive terms is $\frac12$ and $0<\frac12<1$.
 The same argument would work for convergence of $\sum\_{n=1}^\infty r^n$ with any $r\in(-1,1)$.
 
 ---
 ## Ratio test
 #### Example
 $\sum\_{n=1}^\infty \frac{1}{2^n}$ converges.
 #### Proof
 The sequence of partial sums is $s_n = \frac12+\frac14+ \ldots+ \frac1{2^n}=\frac{1-(\frac12)^n}{1-\frac12} - 1 \to \frac1{1/2}-1=1$, as $n\to\infty$.   □

Here the ratio between consecutive terms is $\frac12$ and $0<\frac12<1$.
 The same argument would work for convergence of $\sum\_{n=1}^\infty r^n$ with any $r\in(-1,1)$.

But we can do better.

#### Theorem
 The series $\sum\_{n=1}^\infty a_n$ with $\lim\_{n\to\infty} \left\lvert\frac{a\_{n+1}}{a_n}\right\rvert = \ell$ converges if $\ell<1$, but diverges if $\ell>1$.
 
 ---
 ## Ratio test
 #### Theorem
 The series $\sum\_{n=1}^\infty a_n$ with $\lim\_{n\to\infty} \left\lvert\frac{a\_{n+1}}{a_n}\right\rvert = \ell$ converges if $\ell<1$, but diverges if $\ell>1$.
 #### Examples
 $\sum\_{n=1}^\infty \frac1{3^n}$ converges, but $\sum\_{n=1}^\infty 2^n$ diverges.
 
 ---
 ## Ratio test
 #### Theorem
 The series $\sum\_{n=1}^\infty a_n$ with $\lim\_{n\to\infty} \left\lvert\frac{a\_{n+1}}{a_n}\right\rvert = \ell$ converges if $\ell<1$, but diverges if $\ell>1$.
 #### Examples
 $\sum\_{n=1}^\infty \frac1{3^n}$ converges, but $\sum\_{n=1}^\infty 2^n$ diverges.
 #### Examples
 Using the ratio test, and possibly some other tests, decide convergence of:

$\sum\_{n=1}^\infty \frac{2^n}{3^n},\\; \sum\_{n=1}^\infty \frac{1}{n\sqrt n},\\; \sum\_{n=1}^\infty n\^{100}(\frac9{10})^n,\\; \sum\_{n=1}^\infty \frac{n\^{100}}{(\frac9{10})^n},\\; \sum\_{n=1}^\infty \frac{2^n + n^5}{n\sin(n) + 3^n},\\; \sum\_{n=1}^\infty \frac{2^n + 4n}{5 - 2^n}$
 
 ---
 ## Functions
 A function is a machine that takes in things and puts out things.
 Usually, the things are real numbers.

$f:X\to Y$ has

*domain* $X$, the set of inputs,

*codomain* $Y$, the set that all the outputs belong to,

*range* (or *image*) $f(X)=\\{y\in Y$ such that $y=f(x)$ for some $x\in X\\}$.

Can use a table, a rule, a formula to define a function.

| $x$ | \| | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ |
 |-----|-|----|-----|-----|---|----|----|
 | $f(x)$ | \| | $13$ | $\pi$ | $\frac12$ | $1$ | $1$ | $3$ |
 
 ---
 ## Functions
 A function is a machine that takes in things and puts out things.
 Usually, the things are real numbers.

$f:X\to Y$ has

*domain* $X$, the set of inputs,

*codomain* $Y$, the set that all the outputs belong to,

*range* (or *image*) $f(X)=\\{y\in Y$ such that $y=f(x)$ for some $x\in X\\}$.

Can use a table, a rule, a formula to define a function.

$f(x) = x$ if $x$ is an integer, and $f(x) = t$ such that $t$ is the number of pixels used to render the letter $x$ in this font, if $x$ is not an integer.
 
 ---
 ## Functions
 A function is a machine that takes in things and puts out things.
 Usually, the things are real numbers.

$f:X\to Y$ has

*domain* $X$, the set of inputs,

*codomain* $Y$, the set that all the outputs belong to,

*range* (or *image*) $f(X)=\\{y\in Y$ such that $y=f(x)$ for some $x\in X\\}$.

Can use a table, a rule, a formula to define a function.

$f(x) = x^2 - \sin(x)$
 
 ---
 ## Functions
 A function is a machine that takes in things and puts out things.
 Usually, the things are real numbers.

$f:X\to Y$ has

*domain* $X$, the set of inputs,

*codomain* $Y$, the set that all the outputs belong to,

*range* (or *image*) $f(X)=\\{y\in Y$ such that $y=f(x)$ for some $x\in X\\}$.

Can use a table, a rule, a formula to define a function.

$f(x) = x^2 - \sin(x)$

Most of our functions will be defined by formulae.
 
 ---
 ## Functions
 A function is a machine that takes in things and puts out things.
 Usually, the things are real numbers.

A function always gives the same output for the same input.

If $x$ is fixed, then $f(x)$ is fixed.

$f$ is the function, $f(x)$ means "the function $f$ evaluated at the input $x$" means "the output of $f$ which corresponds to the input $x$".
 
 ---
 ## Functions
 A function is a machine that takes in things and puts out things.
 Usually, the things are real numbers.

A function always gives the same output for the same input.

But different inputs could give the same output.

#### Example
            $f:\mathbb R\to\mathbb R$ defined by $f(x) = x^2$ has $f(-x)=f(x)$ for all $x$, so $f$ is not injective.

#### Definition
 A function $f$ is *injective* if $f(x_1)=f(x_2) \implies x_1=x_2$.
 
 ---
 ## Functions: injectivity and surjectivity
 A function is a machine that takes in things and puts out things.
 Usually, the things are real numbers.

A function always gives the same output for the same input.

#### Definition
            A function $f$ is *injective* if $f(x_1)=f(x_2) \implies x_1=x_2$.

A function $f:X\to Y$ is *surjective* if its range is its codomain; if $\\,\forall y \in Y,\\;\exists\\,x \in X$ such that $f(x)=y$.

A function is *bijective* if it is both injective and surjective.
 
 ---
 ## Functions: injectivity and surjectivity

#### Definition
            A function $f$ is *injective* if $f(x_1)=f(x_2) \implies x_1=x_2$.

A function $f:X\to Y$ is *surjective* if its range is its codomain; if $\\,\forall y \in Y,\\;\exists\\,x \in X$ such that $f(x)=y$.

A function is *bijective* if it is both injective and surjective.
            #### Examples
            Which of the following are injective, surjective, bijective on domain $\mathbb R$ and codomain $\mathbb R$?

$f(x) = x^3, \\;\\; g(x) = x^3-x, \\;\\; h(x) = \frac1{\lvert x \rvert +1}, \\;\\; p(x) = \sin(x)$,

$q(x) = 5, \\;\\; r(x) = x+\lceil x \rceil$.

(How) can you restrict their domains and codomains to make them bijective?