Math Challenge - June 2023

[Infrared]

I would prefer that you do it by hand :)

With that being said, do you see an easy to way to check that $\sqrt{5}+\sqrt[3]{7}+\sqrt[5]{3}$ is an algebraic integer (I think you mean this as opposed to algebraic number, which just means that it is a root of an integer, not necessarily monic, polynomial) without explicitly writing down its minimal polynomial?

 

[julian]

The sum of algebraic integers is an algebraic integer and each of $\sqrt{5}$, $\sqrt[3]{7}$, and $\sqrt[5]{3}$ are algebraic integers. Do I need to use that theorem? Or is there another way?

It is getting quite late here. Tomorrow!

 

[Infrared]

That theorem should be fair game to use and I trust that you can do the arithmetic to verify that $\sqrt{5}+\sqrt[3]{7}+\sqrt[5]{3}$ is not an integer, so I'll count it as solved.

I'll share the solution I had in mind. If the sum is rational, then $\sqrt{5}\in\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{3}).$ The field on the righthand side contains $\mathbb{Q}(\sqrt[3]{7})$ and $\mathbb{Q}(\sqrt[5]{3})$ as subfields so its degree over $\mathbb{Q}$ is a multiple of both 3 and 5, hence a multiple of 15. Also, its degree over $\mathbb{Q}$ is at most 15 because in general $[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha)]\cdot [K(\alpha):K]\leq [K(\beta):K]\cdot [K(\alpha):K].$ So $[\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{3}):\mathbb{Q}]=15$ and it cannot have a degree 2 element like $\sqrt{5}.$

One interesting thing to note is that our methods are good for different generalizations of the original problem. If I instead asked to prove that the numbers $\sqrt{5},\sqrt[3]{7},\sqrt[5]{3}$ are rationally independent, then your method seems quite difficult to use since after clearing denominators you would have to explain why no integer combination of these numbers could be an integer, which doesn't simplify the problem.

Instead, if some of the degrees of the roots were not coprime, then my method could fail because the degree of the composite field might not be the product of degrees of the base fields (you could get a new prime factor in the degree!) and also last sentence could fail to be a contradiction if there was the needed divisibility.

 

[julian]

Now I'm more awake, I can look into it and find that the theorem isn't difficult to prove. It is easy to establish that:

\begin{align*}
& 2.2 < \sqrt{5} < 2.3
\nonumber \\
& 1.9 < \sqrt[3]{7} < 2
\nonumber \\
& 1.2 < \sqrt[5]{3} < 1.3
\nonumber \\
\end{align*}

So that

\begin{align*}
5.3 < \sqrt{5} + \sqrt[3]{7} + \sqrt[5]{3} < 5.6
\end{align*}

 

[fresh_42]

@Infrared I have a question regarding problem #2. It looks as if you expected to use the intermediate value theorem. I think a minimality argument is sufficient, but I may have overlooked something in my proof:

Set $$P(f;T):=\left\{nT\,|\,n\in \mathbb{Z}\right\}\quad \text{ for some } T\in \mathbb{R}$$
Since $0\in P(f)$ in any case, we have $P(f)=\{0\}=P(f;0)$ for non-periodic functions $f.$ We may therefore assume that $f$ is periodic and $T_0$ the minimal, positive, non-zero element of $P(f).$

Edit: If no such minimal period $T$ existed, and $\{0\}\subsetneq P(f)$ then we had a sequence $(T_n)_{n\in \mathbb{N}}$ with $\lim_{n \to \infty}T_n=0.$ In every neighborhood of $x$ was an element $|T_n|<\varepsilon$ such that $f(x)=f(x+T_n)$. As $f$ is continuous, $f(x)$ was constant for all $|x|<\varepsilon,$ i.e. on an interval of positive length. Therefore, $f(x)$ would be constant everywhere, i.e. $T=0$ and $P(f)=\{0\}.$

Or: The sets $A(x_0)=\{x_0\}+\{T\in P(f)\,|\,f(x_0)=f(x_0+T)\}$ are closed by the continuity of $f.$ If no minimal $T_0$ existed at some point $x_0$ then we could construct a zero-sequence $(T_n)$ such that $x_0\not\in A(x_0)$ which contradicts closeness.

If such an element was negative then (by using the for-all-quantifier in the definition of $P(f)$)
\begin{align*}
f(x+|T_0|)=f((x+|T_0|)+T_0)=f((x+|T_0|)-|T_0|)=f(x)
\end{align*}
and we can choose $|T_0|$ instead. By the same argument, we will find per induction in both directions that $P(f;T_0)\subseteq P(f).$ Now assume $0\neq T\in P(f).$ Then we have to show that $T=nT_0$ for some $n\in \mathbb{Z}.$ We assume that $T>T_0$, since there is nothing to show for $T=T_0$ and negative values, can be turned into positive by the above argument. Let $T\in [\,nT_0,(n+1)T_0\,).$ Then
$$
f(x+(T-nT_0))=f((x-nT_0)+T)\stackrel{T\in P(f)}{=}f(x-nT_0)\stackrel{P(f;T_0)\subseteq P(f)}{=}f(x)
$$
contradicting the minimality of $T_0$ since
$$
T-nT_0 < (n+1)T_0-nT_0=T_0.
$$
except $T-nT_0=0.$

I may have overlooked something, which leads me to my question: What do you mean by counterexample? The function
$$
f(x)=\begin{cases}|\sin(x)|&\text{ if }x\not \in \mathbb{Z}\pi \\ -2 &\text{ if }x \in \mathbb{Z}\pi\end{cases}
$$
is still periodic with $P(f)=P(f;\pi)$ but not continuous anymore.

 

[Infrared]

@fresh_42 The error in your proof is that you assume there is a minimal positive element of $P(f)$ when $f$ is periodic. Why can it not contain elements arbitrarily close to 0? If there is a minimal positive element of $P(f),$ call it $T_0$ then I think a simpler way of writing your argument is to say that for any other $T\in P(f)$, you can perform long division $T=nT_0+r$ for integer $n$ and $0\leq r<T_0.$ Since $P(f)$ is an additive group containing $T$ and $T_0$, you have $r\in P(f)$ which contradicts the minimality of $T_0$ unless $r=0$, in which case $T=nT_0$ as desired.

I may have overlooked something, which leads me to my question: What do you mean by counterexample?

I mean that the statement "If $f$ is continuous and not constant, then $P(f)=T\mathbb{Z}$ for some $T\in\mathbb{R}$" is false without the condition that $f$ is continuous. @pasmith has already done this in post 27. Since you didn't actually use the fact that $f$ is continuous in your argument as far as I can see, that's a sign it can't be right since that assumption is necessary.

 

[fresh_42]

@fresh_42 The error in your proof is that you assume there is a minimal positive element of $P(f)$ when $f$ is periodic. Why can it not contain elements arbitrarily close to 0?

Because then it would be zero as $0\in P(f).$ That's the only place I use continuity, admitted.

If there is a minimal positive element of $P(f),$ call it $T_0$ then I think a simpler way of writing your argument is to say that for any other $T\in P(f)$, you can perform long division $T=nT_0+r$ for integer $n$ and $0\leq r<T_0.$ Since $P(f)$ is an additive group containing $T$ and $T_0$, you have $r\in P(f)$ which contradicts the minimality of $T_0$ unless $r=0$, in which case $T=nT_0$ as desired.

This is equivalent to my minimality argument, only that minimality is hidden in the remainder term of the Euclidean algorithms. A matter of taste.

I mean that the statement "If $f$ is continuous and not constant, then $P(f)=T\mathbb{Z}$ for some $T\in\mathbb{R}$" is false without the condition that $f$ is continuous.

Have you looked at my example with the sine? It is clearly not continuous and periodic ($\pi$) and not constant.

@pasmith has already done this in post 27. Since you didn't actually use the fact that $f$ is continuous in your argument as far as I can see, that's a sign it can't be right since that assumption is necessary.

Oh, you meant a counterexample to $P(f)=T\mathbb{Z}$ and not to continuity.

 

[Infrared]

Because then it would be zero as $0\in P(f).$ That's the only place I use continuity, admitted.

Great, so as soon as you include a proof that $P(f)$ has a minimal positive element when $f$ is continuous (assuming of course that $P(f)\neq\{0\}$), your solution will be correct :)

 

[fresh_42]

Great, so as soon as you include a proof that $P(f)$ has a minimal positive element when $f$ is continuous (assuming of course that $P(f)\neq\{0\}$), your solution will be correct :)

Added to the proof as "Edit" comment.

 

[Infrared]

Added to the proof as "Edit" comment.

Edit: Sorry, I actually don't quite follow the first argument. $f(x)=f(x+T_n)$ for a sequence $T_n\to 0$ does not imply that $f$ is constant in a neighborhood of $x.$ This would be satisfied by any function continuous at $x.$ I'm probably misreading your argument?

I'm also having a hard time reading the second argument, though. For exmaple, the set $\{T\in P(f):f(x_0)=f(x_0+T)\}$ is just $P(f)$ since the condition imposed $f(x_0)=f(x_0)+T$ is already implied by the definition of $P(f).$

 

[fresh_42]

Edit: Sorry, I actually don't quite follow the first argument. $f(x)=f(x+T_n)$ for a sequence $T_n\to 0$ does not imply that $f$ is constant in a neighborhood of $x.$ This would be satisfied by any function continuous at $x.$ I'm probably misreading your argument?

I'm also having a hard time reading the second argument, though. For exmaple, the set $\{T\in P(f):f(x_0)=f(x_0+T)\}$ is just $P(f)$ since the condition imposed $f(x_0)=f(x_0)+T$ is already implied by the definition of $P(f).$

Yes, please forget it. It was too quick (and very dirty), too late, too hot and I was too tired. I will have to think about it in daylight. Sorry for your inconvenience.

 

[nuuskur]

Effectively, can we have uncountable chains in the lattice \mathcal P(\mathbb N)?

Spoiler: Problem 8

Dedekind cuts. For any r\in\mathbb R let S_r := \{q\in\mathbb Q \mid q&lt;r\}. It follows that \mathbb R\to\mathcal P(\mathbb Q), r\mapsto S_r, is injective. Let f:\mathbb Q\to\mathbb N be a bijection, which induces F:\mathcal P(\mathbb Q)\to\mathcal P(\mathbb N), A\mapsto \{f(a)\mid a\in A\}. Clearly, this is bijective.

Let r&lt;s in \mathbb R. Note that
<br /> F(S_r) = \{f(q) \mid q\in S_r\} = \{f(q) \mid q&lt;r\} \subseteq \{f(q) \mid q&lt;s\} = F(S_s).<br />

So F(S_r), r\in \mathbb R, is a linearly ordered subset of \mathcal P(\mathbb N).

 

[Infrared]

@nuuskur Nice job! A friend gave me this one and it took quite a bit longer for me to solve than it should have. My instinct was that there was no such family and I spent a while trying to prove it before I considered that it might be possible.

 

[nuuskur]

Initially, I thought it was impossible too, because I couldn't somehow "constructively" make a chain with R-many elements when thinking explicitly about subsets of N. Then I started thinking about it for Q and the answer was more or less apparent. The black box is the bijection $f$, since its existence can be justified if an injection \mathbb Q\to\mathbb N is found. I am not even going to try and understand what this chain looks like

 

[Throwaway_for_June]

@nuuskur Nice job! A friend gave me this one and it took quite a bit longer for me to solve than it should have. My instinct was that there was no such family and I spent a while trying to prove it before I considered that it might be possible.

I don't see a problem with constructing an uncountable sequence like that using Dedekind cuts, but it bugs me because it also seems like we can also construct an injection from a sequence like that into $\mathbb{N}$: Let's say that $A \in \mathcal{F}$ is some element of the chain. Then $\Delta A = A - \bigcup_{B \in \mathcal{F}, B \subsetneq A} B$ should be a non-empty set since all the $B$s are proper subsets of $A$ and that union is equal to one of them [Edit: this isn't true for infinite unions] and since $\Delta A$ is a set of natural numbers and the natural numbers are well-ordered, it should have a smallest element $\min \Delta A$. Then for any particular chain $\mathcal{F}$, $A \rightarrow \min \Delta A$ should be an injection. Clearly $\min \Delta A$ can't be in any set $B \subsetneq A$ in chain because those elements were subtracted out so $\min \Delta B \neq \min \Delta A$, and similarly for any $C \supsetneq A$ the elements of $A$ can't be in $\Delta C$ so $\min \Delta C \neq \min \Delta A$.

Am I missing something obvious? Edit: Ah, right $\mathbb{N} = \bigcup \{ 1 \}, \{1,2\}, \{1,2,3\} \ldots$ even if all of the individual elements of the union are finite and included in each other.

 

[Throwaway_for_June]

It's not immediately obvious to me why $\Delta A$ should be nonempty. You seem to claim something like $[0,1-1/n)$ does not converge to $[0,1]$ because all the individual sets are proper subsets.

Well, it's that $\bigcup_{n=2}^{\infty} [0,1-1/n) = [0,1)$. The union won't contain 1 so it doesn't converge to the closed interval.

 

[Infrared]

@Throwaway_for_June @nuuskur Are you responding to the correct thread? I don't see the notation $\Delta A$ anywhere in this thread.

 

[Throwaway_for_June]

@Throwaway_for_June @nuuskur Are you responding to the correct thread? I don't see the notation $\Delta A$ anywhere in this thread.

Yeah, we were discussing one of the answers. Mods are reviewing my edits.

 

[nuuskur]

Part of his or her argument was the following. Let $\mathcal A$ be a chain of $\mathcal P(\mathbb N)$. Take $A\in\mathcal A$ and consider
<br /> \Delta A := A\setminus \bigcup _{B\subset A}B.<br />
Let's assume $\Delta A$ is nonempty subset of $\mathbb N$, hence it has smallest element $m_A$. So, we have $\mathcal A\to\mathbb N, A\mapsto m_A$. They also had some argument to show it's injective, but the problem is if it was injective, then clearly we can't have uncountable chains.

I think it's reasonable to claim nonemptiness, though that is also not obvious to me. But I don't think injectivity follows (I mean, in hindsight, it couldn't).

 

[Infrared]

It is also not necessarily true that $\Delta A$ is nonempty. If so, it would also be the case if you replaced $\mathbb{N}$ with $\mathbb{Q}$ but with if you use the Dedekind cuts $S_a=\{q\in\mathbb{Q}:q<a\}$ for $a\in\mathbb{R},$ you see that any $S_a\setminus\bigcup_{b<a}S_b$ is empty.

Also, please don't have conversation in threads where you delete the messages after. It makes it impossible for anyone else who's reading to follow.

 

[Throwaway_for_June]

...
I think it's reasonable to claim nonemptiness, though that is also not obvious to me. But I don't think injectivity follows (I mean, in hindsight, it couldn't).

If the $\Delta A$s are all non-empty and disjoint, then taking the least element of each $\Delta A$ will produce an injection. The $\Delta$s have to be disjoint because of the way the chain is ordered, so at most countably many of the elements in one of these chains can have a non-empty $\Delta$ . Infinities are weird.

 

[projective]

4)

Spoiler

From the definition of the commutator $[S_7, S_3] \leq S_7 \cap S_3$ but $S_7 \cap S_3 = 1$ and so $[S_7, S_3] = 1$ but $S_7$ and $S_3$ are abelian and therefore $G$ also is

 

[fresh_42]

@Infrared Again for #2. I think I have a proof for uniformly continuous functions. My proof does not work for only continuous functions because I get a circle argumentation with the quantifiers if continuity depends on the location. Are you sure it isn't necessary? It may not play a role in $P(f)$ but it does in $\mathbb{R}\backslash P(f).$ I even tried tricks with closed intervals but I cannot get rid of the location.

 

[mathwonk]

#2: We may assume P(f) contains more than just 0. Since f is continuous, P(f) is closed, and if P(f) contained an arbitrarily small positive element, it would also be dense. Since f is constant on P(f) with value f(0), this would imply f is constant with value f(0). Thus if f is continuous and non constant, and P(f) ≠ {0}, there is a smallest positive element T of P(f). Since P(f) is closed under addition and subtraction, P(f) = Z.T.

 

[fresh_42]

Here is my proof for #2 under the assumption that $f$ is uniformly continuous.

Spoiler: 2

Lemma: $P(f)$ is a group under addition.

Proof: Clearly, $0\in P(f).$ By the all quantifier $(*)$ in the definition of $P(f)$
\begin{align*}
T\in P(f)&\Longrightarrow f(x)=f(x+T-T)=f((x-T)+T)\stackrel{(*)}{=}f(x-T)\\&\Longrightarrow -T\in P(f)\\
S,T\in P(f)&\Longrightarrow f(x)=f(x+T)\stackrel{(*)}{=}f((x+T)+S)=f(x+(T+S))\\&\Longrightarrow T+S \in P(f)
\end{align*}
Since $0\in P(f)$ in any case, we have $P(f)=\{0\}=0\cdot \mathbb{Z}$ for non-periodic functions $f.$ We may therefore assume that $f$ is periodic and in particular not constant.

If $P(f)$ contains an interval of positive length, then $f$ is constant. If $P(f)$ has a minimal positive element $T_0$ then $T_0\mathbb{Z}\subseteq P(f)$ since $P(f)$ is an additive group. Now assume $0\neq T\in P(f).$ Then we have to show that $T=nT_0$ for some $n\in \mathbb{Z}.$ We assume that $T>T_0$, since there is nothing to show for $T=T_0$ and negative values, can be turned into positive by the above argument. Let $T\in [\,nT_0,(n+1)T_0\,).$ Then
$$
f(x+(T-nT_0))=f((x-nT_0)+T)\stackrel{T\in P(f)}{=}f(x-nT_0)\stackrel{T_0\mathbb{Z}\subseteq P(f)}{=}f(x)
$$
contradicting the minimality of $T_0$ since
$$
T-nT_0 < (n+1)T_0-nT_0=T_0
$$
except for $T=nT_0.$ Equivalently you can write any $T\in P(f)$ as $P(f)\ni T-qT_0=r$ by the Euclidean algorithm. Since $0<r<T_0$ is impossible, we have $r=0$ and $T=qT_0.$

$P(f)$ is closed since $f$ is continuous: Let $(T_n)\subset P(f)$ be a converging sequence. Then
$$
f(x+\lim_{n \to \infty}T_n)=f(\lim_{n \to \infty}(x+T_n))=\lim_{n \to \infty}f(x+T_n)=f(x)
$$
i.e. $\displaystyle{\lim_{n \to \infty}T_n\in P(f).}$ The set $U:=\mathbb{R}\backslash P(f)$ is therefore open and dense as $P(f)$ does not contain an interval of positive length. Assume there is no minimal, positive element in $P(f).$ This means there is a strictly monotone decreasing sequence $(T_n)\subset P(f)$ with $\displaystyle{\lim_{n \to \infty}T_n}=0$ and $T_n>0.$

Let $u\in U$ and $x_u\in \mathbb{R}$ such that $|f(x_u+u)-f(x_u)|=d>0$ and $\varepsilon =d/2.$ We can choose $T_n$ for sufficiently large $n$ and $u$ sufficiently close to $0$ such that $|T_n-u|<\delta(\varepsilon )$ if we want to get $|f(T_n)-f(u)|<\varepsilon .$
Then $|(T_n+x_u)-(u+x_u)|=|T_n-u|<\delta (\varepsilon )$ and if $f$ is uniformly continuous at $T_n$ and $T_n+x_u$ then
$$
d=|f(x_u+u)-f(x_u)|=|f(x_u+u)-f(x_u+T_n)|<\varepsilon =d/2.
$$
This is a contradiction and means that there is no such sequence $(T_n),$ i.e. $P(f)$ has a minimal element.

 

[mathwonk]

#10: If A is path connected then B cannot be connected, hence also not path connected, essentially by the Jordan curve theorem. It is "obvious" but already tricky to prove even that they cannot both be path connected. I do not see whether both can be connected. In fact I only know one connected set that is not path connected, the topologist's sine curve plus (0,0), (or the full closure of the topologist's sine curve).

here is a little more detail on why 3), hence nor 2), cannot be true: if there is a path in A connecting (0,0) to (1,1), then there is a simple such path (every path connected Hausdorff space is "simple path"-connected). Then add on 3/4 of the circle with center (1,0) and radius one (all but the NW quarter), to form a simple closed curve containing (1,0) but not (0,1) in its interior. Then Jordan says the interior and exterior of this curve are disjoint open sets. These open sets will then disconnect any subset B of the square which is disjoint from A and contains both (0,1) and (1,0). QED.

For a slick argument just that A and B cannot both be path connected, one wants to show that any path connecting (0,0) to (1,1) within the square, meets any path connecting (0,1) to (1,0) within the square. But intersection number is a homotopy invariant, so we can replace these 2 paths by the 2 diagonals, which clearly intersect once transversely. QED.

Rmk: One can also do this directly, arguing via various projections within the square, essentially a homotopy argument. In his beautiful book on ODE, Arnol'd uses this fact to solve the following problem: if two cars manage to drive between two points A and B along two different non intersecting roads, while joined by a rope of length < 2L, can two circular wagons of radius L drive along those roads in opposite directions (with their center on the road) without colliding? (He constructs a "phase space" for this problem which is identical to the setup in problem #10. 2).)

 

[mathwonk]

#9: the preimage of [1:0] is the circle consisting of all (z,w) such that w = 0 and |z|^2 = 1, and the preimage of [0:1] is the circle consisting of all (z,w) such that z=0 and |w|^2 = 1. Their linking number is the intersection number of one loop with a disc in S^3 capping off the other loop. Such a disc in S^3 capping off the loop {(z,0): |z|^2 = 1} is the set {(z,t): |z|^2 + t^2 = 1, and 0≤t≤1, t real}. This disc intersects {(0,w): |w|^2 = 1} transversely in the point (0,1). Hence the loops are linked.

 

[Infrared]

@fresh_42 A continous function which is periodic is uniformly continuous.

@mathwonk Everything you said it right, but very concise. Perhaps you would like to elaborate on how the Jordan curve theorem shows that a path from $(0,0)$ to $(1,1)$ disconnects the square? I had a solution in mind without the Jordan curve theorem but perhaps what you have is good too.

@projective This is right, but quite brief, so just to fill in the gaps for anyone else: $S_7$ and $S_3$ are the (Sylow) subgroups of orders $7$ and $9$ respectively. If $x\in S_7$ and $y\in S_3$ because if $x\in S_7$ and $y\in S_3$ then $xyx^{-1}y^{-1}=(xyx^{-1})y^{-1}=x(yx^{-1}y^{-1})$ shows it to be in $S_3\cap S_7=\{1\}$ since both subgroups are normal.

So, every element of $S_3$ commutes with every element of $S_7.$ Together, these subgroups generate the whole group because the map $S_3\times S_7\to G, (x,y)\mapsto xy$ is injective and hence surjective. Since every group of order $9$ and $7$ is already abelian, this means that $G$ is abelian.

 

[fresh_42]

@fresh_42 A continous function which is periodic is uniformly continuous.

I suspected this but couldn't see an immediate argument for locations whose distance isn't in $P(f).$

I was already happy that I recognized that I had to use it. People usually simply write $\forall \varepsilon \exists \delta$ where they actually use $\forall \varepsilon \exists\delta (x_0;\varepsilon ).$ This is a nice example for my mantra to always keep track of dependencies like $\forall \varepsilon \exists N(\varepsilon )$ instead of $N.$

 

[Throwaway_for_June]

Here is my proof for #2 under the assumption that $f$ is uniformly continuous.
...

Seems like you're using some unnecessary big guns...

Spoiler: 2

Suppose for the sake of contradiction that $f$ is not constant, that $f$ is continuous, and that $P(f)$ has arbitrarily small strictly positive elements.

Since $f$ is not constant, it must take on at least two different values. Let those values be $a=f(x_a)$ and $b=f(x_b)$ so $b-a \neq 0$

Since $f$ is continuous everywhere, it's continuous at $0$. So for $\forall \epsilon > 0 \exists \delta>0 : |x| < \delta \implies \left| f(x) - f(0) \right| < \epsilon$. In particular, there's some $\delta_{ab} > 0$ corresponding to $\epsilon = \left| \frac{ b-a }{2} \right|$.

Since $P(f)$ has arbitrarily small strictly positive elements it contains some element $\iota$ with $0 < \iota < \delta_{ab}$.

Now we have $\delta_{ab} > \iota > \left| x_a - \lfloor \frac{x_a}{\iota} \rfloor \iota \right|$, $\delta_{ab} > \iota > \left| x_b - \lfloor \frac{x_b}{\iota} \rfloor \iota \right$, so $\left| b-a \right| > \left| f(x_a - \lfloor \frac{x_a}{\iota} \rfloor \iota) - f(0) \right| + \left| f(x_b - \lfloor \frac{x_b}{\iota} \rfloor \iota) - f(0) \right| = \left| f(x_a) - f(0) \right| + \left|f(x_b - f(0) \right| \geq \left| f(x_b) - f(x_a) \right| = \left| b-a \right|$, but $\left| b-a \right|$ can't be bigger than itself so we have a contradiction.