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# Guest post: Sequence Numbers

This is a guest post, sent in by David, who’s discovered an interesting property of numbers, and is looking for collaborators to take it further.

Mathematics is not a careful march down a well-cleared highway, but a journey into a strange wilderness, where the explorers often get lost. Rigour should be a signal to the historian that the maps have been made, and the real explorers have gone elsewhere.

– W. S. Anglin

A few years ago I saw a post on a website that showed that the inverse of 998,001 produces a decimal expansion that counts, using three digit strings, from 000 to 997 without error.

$\frac{1}{998,\!001} = 0.0000010020030040050060070080090100110120130\ldots$

I immediately thought that this had to be a hoax. I decided to work it out to prove it was a hoax – after all some people put anything they want on the web whether it is true or not.

My calculator cannot handle more than eight digits. And I really did not want to work it out by hand. The arithmetic is simple enough, I just did not want to spend that much time on it (we all like instant everything – personally I get impatient waiting on the microwave). So I took this problem to a wonderful website, Wolfram|Alpha. It did the calculation for me, and I checked it out – digit by digit. I did not like what I found out – the equation was right, and I was wrong.

But I did start wondering – I wondered if there were other numbers that had similar properties as this one. Are there other integers that, when you turn the number into a unit fraction and then convert it to a decimal, produce an interesting sequence of numbers (like counting, or multiplying, or whatever).

So I looked on the web for answers – but the internet did not have any answers to this question (stupid internet!). Could it be true that the example I found in a blog post was the only example of this special property?

I contacted a few other mathematicians by e-mail to see if they knew anything about this kind of a mathematics – most did not – but a few gave me so hints on where I might find out more. But those websites only gave me more hints.

By this time I realized two things – I was not going to find an answer out there – and if I wanted to know the answer I was going to have to find it myself. And I really wanted to know the answer!

Things were slow at first, and at second, and at third… but I found a few more hints, and eventually found another number with a similar properties.

Then I found a few more.

But I did not understand why some numbers worked and others did not. So I kept working on the problem, and as I found more of these special numbers I started to see patterns, and I learned how to fill in some blank spots in these patterns.

I decided this group of numbers needed a special name – after all I could not keep calling them just “special numbers. “Mr. B’s Super-duper-fantastic Arithmetically Accurate and Mathematically Mystical Sequence Numbers”. This name is really kind of long and it did not catch on. Now I just call them “Sequence Numbers” because they produce interesting sequences.

I define sequence numbers as integers that produce interesting sequences when you take their inverses and convert them into decimal form. I prefer that the “interesting sequence” is listed in the Online Encyclopedia of Integer Sequences, but if the sequence is an obvious or well known sequence I can accept that too.

So what are some of these other numbers? Let me tell you about some of my new buddies:

• 9,899 produces the first few terms of the well known Fibonacci sequence.
• 196,020 produces a list of the multiples of five.
• 998,999 produces the terms of the Fibonacci sequence, using three digit strings.
$\frac{1}{998999} = 0.000 \; 001 \; 002 \; 003 \; 005 \; 008 \; 013 \; 021\ldots$
• 998,998 produces the terms of the Jacobsthal sequence, using three digit strings.
$\frac{1}{998998} = 0.000 \; 001 \; 001 \; 003 \; 005 \; 011 \; 021 \; 043 \; 085 \; 171\ldots$
• 76,922,923,077 counts by multiples of 13. (Imagine that! How far can you count by multiples of 13?)
• 999,999,700,000,029,999,999 shows a list of all of the triangular numbers with 6 digits or less, and even most of the triangular numbers with seven
• 999,999,999,997 lists all of the powers of two ($2^0$, $2^1$, $2^2$, $2^3$, etc.) up to 12 digits long, except for the largest 12-digit power of 2. And it writes them all in 12-digit strings.
• 999,999,998,999,999,998,999,999,998,999,999,998,999,999,998, 999,999,998,999,999,998,999,999,998,999,999,998,999,999,998, 999,999,998,999,999,998,999,999,999 shows the terms of the tridecanacci sequence, defined as $a(n) = \begin{cases} 0 & 0 \leq n \lt 12, \\ 1 & n=12, \\ \sum_{i=1}^{13}a(n-i) & n \gt 12 \end{cases}$
• 999,999,999,876,543,210 – this one shows the 123,456,789 times table (all of the multiples of 123,456,789) up to, but not including the last 24 digit multiple of 123,456,789.

There are a lot of these special numbers out there, and I know I have not found them all.

I hope some of you don’t believe me. Well let me rephrase that – I hope that some of you don’t believe me AND have the gumption to try to prove me wrong. If you do you will learn how to find these numbers, and even how to design some to do specific things.

If you find something you want to show me, or you want me to post for others, please email me at: mbiom.edu@gmail.com. Check out my website at: SequenceNumbers.blogspot.com.

Don’t be afraid of the big numbers. That may be where all the really interesting and fun stuff starts.

## About the author

### 4 Responses to “Guest post: Sequence Numbers”

1. Thomas Oléron Evans

These are rather neat.

The fraction that lists powers of 2 can be derived using the formula for the sum of a geometric series (though the one you’ve listed actually gives powers of 3, I think; 1/(10^k – n), gives the powers of n up to k digits, in general).

Also, Colin Beveridge has a clear explanation of how 1/999999999998999999999999 generates the Fibonacci sequence here:

http://www.flyingcoloursmaths.co.uk/ask-uncle-colin-is-the-fibonacci-series-witchcraft/

Reply
2. Stephen Miller

The original example generalises nicely. Say we want to list (almost) all k digit numbers; then we first need to know that
$\frac{1}{10^k-1}=0.\dot 0…\dot 0 \dot 1$
where the recurring term is $k-1$ zeroes followed by a one. Then we need to add a few of those up – well, I won’t spoil ALL the fun, but we get the nice sequence number
$\frac{1}{(10^k-1)^2}$

Reply
3. Studen691

If you use generating functions then this becomes trivial. For your first example consider
$f(x) = \frac{x^2}{(1-x)^2} = 0x + 1x^2 + 2x^3 + 3x^4 + \dots \,.$
Then it is $f(\frac{1}{1000}) = \frac{1}{998001}$. So take the generating functions of your favourite number and just plug in $10^{-d}$ to get a number which has the sequence at their decimalplaces. The $d$ just gives the number of “space” they have.

Reply


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# The Inverse Image of an Ideal by a Ring Homomorphism is an Ideal

## Problem 411

Let $f:R\to R’$ be a ring homomorphism. Let $I’$ be an ideal of $R’$ and let $I=f^{-1}(I)$ be the preimage of $I$ by $f$. Prove that $I$ is an ideal of the ring $R$.

Contents

## Proof.

To prove $I=f^{-1}(I’)$ is an ideal of $R$, we need to check the following two conditions:

1. For any $a, b\in I$, we have $a-b\in I$.
2. For any $a\in I$ and $r\in R$, we have $ra\in I$.

Let us first prove condition 1. Let $a, b\in I$. Then it follows from the definition of $I$ that $f(a), f(b)\in I’$.
Since $I’$ is an ideal (and hence an additive abelian group) we have $f(a)-f(b)\in I’$.
Since $f$ is a ring homomorphism, it yields that
$f(a-b)=f(a)-f(b)\in I’.$ Thus we have $a-b \in I$, and condition 1 is met.
This implies that $I$ is an additive abelian group of $R$.

Next, we check condition 2. Let $a\in I$ and $r\in R$. Since $a\in I$, we have $f(a)\in I’$.
Since $I’$ is an ideal of $R’$ and $f(r)\in R’$, we have $f(r)f(a)\in I’$.
Since $f$ is a ring homomorphism, it follows that
\begin{align*}
f(ra)=f(r)f(a)\in I’,
\end{align*}
and hence $ra\in I$. So condition 2 is also met and we conclude that $I$ is an ideal of $R$.

## Comment.

Instead of condition 1, we could have used

Condition 1′: For any $a, b\in I$, we have $a+b\in I$.

The reason is that condition 2 guarantee the existence of the additive inverses, and hence condition 1 and 2 are equivalent to condition 1′ and 2.

### 2 Responses

1. 05/15/2017

[…] $f^{-1}(I_k)$ of the ideal $I_k$ by a ring homomorphism is an ideal of $R$. (See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a […]

2. 08/11/2017

[…] preimage of an ideal by a ring homomorphism is an ideal. (See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a […]

##### Polynomial $(x-1)(x-2)\cdots (x-n)-1$ is Irreducible Over the Ring of Integers $\Z$

For each positive integer $n$, prove that the polynomial $(x-1)(x-2)\cdots (x-n)-1$ is irreducible over the ring of integers $\Z$.

Close

format: plain


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# MO Diagrams¶

• As bonding occurs and atomic orbitals (AOs) combine to form molecules, new molecular orbitals (MOs) are formed in the process
• Two types of bonding can occur

#### Sigma bonding¶

The AOs meet end to end and have a significant overlap

#### Pi bonding¶

The AOs meet adjacent to each other and have lesser overlap

• Pi bonds are weaker than sigma bonds, however pi bonds are usually observed as a part of a double or triple bond and thus have the combined strength of pi and sigma bonds

## The quantum¶

• As two AOs combine as wavefunctions, there is both constructive and destructive interference, that causes two distinct waveforms/MOs to form:
• Bonding orbitals
• Have a lower energy than the two AOs that make them up
• Are primarily located between the two nuclei that make formed the bond
• Antibonding orbitals
• Have a higher energy than the two AOs that make them up
• Are primarily located on the outside of the two nuclei that make formed the bond
• Have one more node than the parent orbitals

## The rules¶

1. The total number of MOs is equal to the total number of AOs
2. Bonding MOs are lower energy than the AOs
3. Antibonding MOs and higher energy than the AOs
4. Electrons follow the Aufbau principle - they fill from lowest to highest energy
5. They follow the Pauli exclusion principle - they will remain unpaired until no equivalent MOs are left
6. MOs are most stable when formed of AOs of like energy

## Bond order (BO)¶

• A measure of stability
• Uses the equation:
$\frac{1}{2}(\text{bonding}\:e^− −\text{antibonding}\:e^−)$
• A molecule with a bond order $$>0$$ is likely to exist and be more stable
• The higher the bond order the more stable the bond

## How to draw MO diagrams¶

1. Only consider your valence shells, as core shells will always have a BO of 0 and will not contribute to bonding
2. Draw out the contributing valence AOs in order of energy

1. Connect the AOs with lines to show how the MOs will form in terms of energy
1. Pi orbitals will form into $$1\sigma$$ and $$2\pi$$ bonding orbitals and $$1\sigma^∗$$ and $$2\pi^∗$$ antibonding orbitals
2. In atoms up to N, the $$\pi$$ bonding MO will be lower energy than the $$\sigma$$ bonding MO $$\sigma^∗$$ always comes after $$\pi^∗$$
3. If an atom cannot contribute to a particular MO formation, they’re considered non bonding MOs and you do not connect them with a line

4. Assign electrons according to Aufbau’s, Pauli’s and Hund’s rules 1. Don’t forget to assign extra or fewer electrons if the atom has a charge

5. Don’t forget to number your MOs, including the core orbitals 1. Molecule will be paramagnetic if any of the MOs have unpaired electrons * If all the electrons are paired, it is considered diamagnetic


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# Help with branches

I'm trying to do something like this, but I doon't know how to creat [+silábico] or organize "NUC" in an equal high. Thanks!!

Here is what I used: (I've tried "[{[-silábico]},before drawing Vocálico={y=-1.0cm,x=0.8cm},anchor= south west]", but it didn't work :/).

\documentclass[12 pt, a4paper, doubleside]{article}%showtrims
\usepackage[top=3cm, bottom=2cm, left=3cm, right=2cm]{geometry}
\usepackage{forest}
\usepackage{phonrule}
\usepackage[brazil]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[left,modulo]{lineno}
\usepackage{multicol,textcomp,amsmath,amssymb}
\usepackage{textcase,graphicx,lastpage,covington}
\usepackage[makeindex,split]{splitidx}
\usepackage[tone,extra]{tipa}
\usepackage{tikz}
\usepackage{tikz-qtree}
\usepackage{mathtools}

\begin{document}

\begin{forest}
[Nuc
[x
[Raiz
[Vocálico
[Altura
[\textsc{[+{alto}]}
]
]
]
]
]
]
]
\end{forest} \hspace{0.4cm}

\begin{forest}
[\textsc{[+{ac}]}
[Nuc
[x
[Raiz
]
]
]
]
\end{forest} \hspace{0.2cm} $\rightarrow$ \hspace{0.2cm}
\begin{forest}
[Nuc
[x
[Raiz
]
]
]
\end{forest} \hspace{0.4cm}

\begin{forest}
[\textsc{[+{ac}]}
[Nuc
[x
[Raiz
]
]
]
]
\end{forest}

\end{document}

Expected output

• Are you fully convinced that you want to use forest for that? Why not just a tikz chain or matrix? – user121799 Nov 6 '18 at 20:04

Even though I really love forest and continue to be stunned by what it can do, in this case I'd like to argue that it is easier to employ other tools. An arguably rather simple way is to use tikz-cd for that.

\documentclass[12 pt, a4paper, doubleside]{article}%showtrims
\usepackage[top=3cm, bottom=2cm, left=3cm, right=2cm]{geometry}
\usepackage{mathtools}
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}[column sep=1mm,
/tikz/column 4/.append style={column sep=0.7cm}]
& & {[+\text{ac}]}\arrow[d,dash] & & & & {[+\text{ac}]}\arrow[d,dash] &\\
\text{x}\arrow[d,dash] & & \text{x}\arrow[d,dash] & &
\text{x}\arrow[d,dash,"="below=-2pt] & & \text{x}\arrow[d,dash] & \\
\text{Nuc}\arrow[d,dash] & & \text{Nuc}\arrow[d,dash] & &
\text{Nuc}\arrow[d,dash] & & \text{Nuc}\arrow[d,dash] & \\

• @Amanda I added it, as well as the missing line of xs. Note that your diagram is a bit wide for the page geometry. One possible way to slightly decrease the width of the diagram and make the horizontal lines longer is to replace column sep=1mm by column sep=1.5mm,font=\small,. – user121799 Nov 6 '18 at 20:37


text:
m
• E

F Nous contacter

0

# Documents  Voloch, José Felipe | enregistrements trouvés : 4

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## Generators of elliptic curves over finite fields Voloch, José Felipe | CIRM H

Multi angle

Research talks;Algebraic and Complex Geometry;Number Theory

We will discuss some problems and results connected with finding generators for the group of rational points of elliptic curves over finite fields and connect this with the analogue for elliptic curves over function fields of Artin’s conjecture for primitive roots.

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## Differential descent obstructions Voloch, José Felipe | CIRM H

Multi angle

Research talks;Algebraic and Complex Geometry;Number Theory

We will discuss a new obstruction to the existence of rational and integral points on algebraic varieties over function fields obtained by considering covers described by differential equations.

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## Maps between curves and diophantine obstructions Voloch, José Felipe | CIRM H

Multi angle

Research talks;Algebraic and Complex Geometry;Number Theory

Given two algebraic curves $X$, $Y$ over a finite field we might want to know if there is a rational map from $Y$ to $X$. This has been looked at from a number of perspectives and we will look at it from the point of view of diophantine geometry by viewing the set of maps as $X(K)$ where $K$ is the function field of $Y$. We will review some of the known obstructions to the existence of rational points on curves over global fields, apply them to this situation and present some results and conjectures that arise. Given two algebraic curves $X$, $Y$ over a finite field we might want to know if there is a rational map from $Y$ to $X$. This has been looked at from a number of perspectives and we will look at it from the point of view of diophantine geometry by viewing the set of maps as $X(K)$ where $K$ is the function field of $Y$. We will review some of the known obstructions to the existence of rational points on curves over global fields, apply them to ...

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## Códigos corretores de erros16e colóquio brasileiro de matemática Voloch, José Felipe | Instituto de Matemática Pura e Aplicada

Ouvrage

Localisation : Salle de manutention

code MDS # code correcteur d'erreur # code cyclique # code de Goppa # géométrie finie

#### Filtrer

##### Codes MSC

Ressources Electroniques (Depuis le CIRM)

Books & Print journals

Recherche avancée

0
Z


text:
NormalizeRegularChain - Maple Help

RegularChains[ChainTools]

 NormalizeRegularChain
 decompose a regular chain into normalized regular chains

 Calling Sequence NormalizeRegularChain(rc, R) NormalizeRegularChain(rc, R, 'normalized'='strongly')

Parameters

 rc - regular chain of R R - polynomial ring 'normalized'='strongly' - (optional) boolean flag

Description

 • The command NormalizeRegularChain(rc, R) returns a triangular decomposition of rc into regular chains in the following sense. The intersection of the saturated ideals of the output regular chains has the same radical as the saturated ideal of rc. Moreover, each output regular chain is normalized.
 • In addition, if NormalizeRegularChain(rc, R) returns only one regular chain, then this regular chain has the same saturated ideal as rc.
 • More generally, if a saturated ideal $I$ of an output regular chain has the same dimension as the saturated ideal $J$ of rc, then $I$ is radical whenever $J$ is radical.
 • If 'normalized'='strongly' is provided, then the output regular chains are strongly normalized.
 • The notion of a (strongly) normalized regular chain is defined in the ChainTools subpackage overview page.
 • The notion of a saturated ideal (of a regular chain) is defined in the RegularChains package overview page.

Examples

 > $\mathrm{with}\left(\mathrm{RegularChains}\right):$$\mathrm{with}\left(\mathrm{ChainTools}\right):$

We define a ring of polynomials.

 > $R≔\mathrm{PolynomialRing}\left(\left[x,y,z\right]\right)$
 ${R}{≔}{\mathrm{polynomial_ring}}$ (1)

We define a regular chain.

 > $\mathrm{rc}≔\mathrm{Empty}\left(R\right)$
 ${\mathrm{rc}}{≔}{\mathrm{regular_chain}}$ (2)
 > $\mathrm{rc}≔\mathrm{Chain}\left(\left[{y}^{2}-z,yx+z\right],\mathrm{rc},R\right)$
 ${\mathrm{rc}}{≔}{\mathrm{regular_chain}}$ (3)
 > $\mathrm{Equations}\left(\mathrm{rc},R\right)$
 $\left[{y}{}{x}{+}{z}{,}{{y}}^{{2}}{-}{z}\right]$ (4)

We apply RegularChains[NormalizeRegularChain].

 > $\mathrm{lrc}≔\mathrm{NormalizeRegularChain}\left(\mathrm{rc},R\right)$
 ${\mathrm{lrc}}{≔}\left[{\mathrm{regular_chain}}\right]$ (5)
 > $\mathrm{map}\left(\mathrm{Equations},\mathrm{lrc},R\right)$
 $\left[\left[{x}{+}{y}{,}{{y}}^{{2}}{-}{z}\right]\right]$ (6)

We check that both chains have the same saturated ideal.

 > $\mathrm{EqualSaturatedIdeals}\left(\mathrm{rc},\mathrm{lrc}\left[1\right],R\right)$
 ${\mathrm{true}}$ (7)

We check that the new regular chain is strongly normalized.

 > $\mathrm{IsStronglyNormalized}\left(\mathrm{lrc}\left[1\right],R\right)$
 ${\mathrm{true}}$ (8)


text:
• No products in the cart.

# Rank and Trace of Idempotent matrix (TIFR 2013 problem 39)

Question:

True/False?

If $$A$$ is a complex nxn matrix with $$A^2=A$$, then rank$$A$$=trace$$A$$.

Hint:

What are the eigenvalues of $$A$$? What is trace in terms of eigenvalues?

Discussion:

If $$v$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda$$ then $$Av=\lambda v$$, therefore $$\lambda v=Av=A^2v=\lambda Av =\lambda^2 v$$. Therefore, since any eigenvector is non-zero, $$\lambda =0 or 1$$.

Sum of eigenvalues is trace of the matrix. So, trace$$A$$= number of non-zero eigenvalues= total number of eigenvalues – number of 0 eigenvalues

Since $$A$$ satisfies the polynomial $$x^2-x$$, the minimal polynomial is either $$x$$ or $$x-1$$ or $$x(x-1)$$. This means the minimal polynomial breaks into distinct linear factors, so $$A$$ is diagonalizable. Therefore, the algebraic multiplicity of an eigenvalue is same as its geometric multiplicity.

In total there are n eigenvalues (for A is nxn) and the number of 0-eigenvalues is the algebraic multiplicity of 0, which is same as the geometric multiplicity of 0, i.e, the dimension of the kernel of A.

Therefore, trace$$A$$$$=n-$$nullity$$A$$.

By the rank-nullity theorem, the right hand side of the above equation is rank$$A$$.

September 22, 2017


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# Interpreting a weak model proceeding from there

I originally posted this question and since then I've posted another more specific branch-off question.

I've followed the advice I received in response to my questions, and I've done some statistical analysis. Now I need help analyzing the results and determining my next move. Here's my original explanation of the situation:

My data is a list of records, each one representing an educational seminar event. I have a continuous variable that represents the revenue brought in by each seminar, which is the response variable in my regression. I also have a number of categorical variables which are acting as factors/IVs.

To add a little more detail, these factors include things like day of the week the seminar was held, topic, speaker, etc.

My main goal is to build a model that A) can help to explain what factors most influence revenue and how, and B) has some predictive power.

I performed a multiple regression in R, but much to my dismay, the adjusted $R^2$ value was a mere 0.2188. I know this doesn't mean the factors have no predictive power at all, but I'm wary of making any major strategic decisions based on such a poorly fit model--am I right to feel that way?

I also performed an ANOVA test, and certain factors seemed to be labelled as more significant than others.

My questions are:

1. How should I proceed given the weak fit of my model? Could it be that the data simply isn't helpful and that either random chance or factors I haven't considered are at play here?

2. Though ANOVA tells me which factors appear significant, it doesn't give me any idea of how they are significant. How can I determine what effect each significant factor has from a practical standpoint?

3. Are there any other tests that would help me understand how all these variables are related?

Bear in mind that I am new to R and rusty on statistical methods (I took an intro-level course a few years ago).

-

I would suggest you the following steps of further analysis.

1. Perform sensitivity analysis (global or possibly local if your data has complex behavior) in order to find out most important variables. I think this R package would be suitable for your needs. On this step you will have a sorted list of variables according to their impact on response variable.
2. Drop unimportant (it is generally your choice how many) variables from the further analysis. Just try several numbers of dropped variables in order to get the best results in next steps.
3. Build your model. Do not forget to use dummy variables for categorical factors! If your are not obliged to use linear model try polynomial or other nonlinear (that is of course if you have enough observation). Although multiple ridge regression is a good choice (but you need to carefully select the parameters).
4. Now you need to estimate accuracy of the model build on the previous steps. $R^2$ is one of the ways although not very illustrative. ANOVA accounts only for linear dependence. Another and more general way is to perform cross validation and calculate mean error of prediction.

Hope this helps! And good luck in your analysis ;)

EDIT: By the way about you last question. You can visualize internal relations in your regressors using e.g. matrix scatterplots. It is a graphical representation of correlation matrix. And if you see straight line somewhere it tells you that the corresponding variables are perfectly correlated and you can drop one of them. As I understand in your case all factors are categorical. So it is not straightforward how to perform correlation analysis. Although R provides some specific tools.

-

I think statement 1 is probably right. The total R square can be partitioned into contributions due to each of the factors. This shows in the ANOVA table (though the effect of the facotrs is influenced by the other factors in the model. You could also look at how big R square is by trying separate 1 factor ANOVAs. So keep what looks to be the most imprtant and see if you can think about other factors that could explain the response better.

-


text:
### Pilot testing of a dry HCl abatement process for geothermal steam

Pubblicato dalla rivista Chemical Engineering Transactions edita dall’AIDIC – Associazione Italiana Di Ingegneria Chimica, l’articolo “Pilot testing of a dry HCl abatement process for geothermal steam“.

La pubblicazione è frutto della collaborazione di CPTM con Enel Green Power e Dipartimento di Ingegneria Civile e Industriale dell’Università di Pisa.

Abstract
In many geothermal areas, an abatement of hydrochloric acid present in the steam is essential in order to produce energy before turbine expansion. In fact, concentrations of acid higher than 5 ppm risk seriously corroding the turbine, significantly increasing the costs of this energy source. In 2018, Enel Green Power decided to support a research project, called Georockwash, and aimed at identifying a new process for the abatement of HCl in geothermal steam with the collaboration of Magona Technological Pole, as a centre specialized in the experimentation of industrial processes on a pilot scale.
The project has developed a new abatement process based on the use of dry CaCO3, as agent for the neutralization of acid. The process has been tested initially on a laboratory scale and then on a pilot scale with the construction of a 500 kg/h steam plant, skid mounted and transportable.
The experimentation campaign on the pilot plant was conducted first at the facilities of the Magona Pole and then at an Enel Green Power geothermal power plant. In the experimentation the effects of the main operating parameters on the abatement efficiency were evaluated and correlated with a semi-empirical equation.
Finally, the experimentation showed abatement efficiencies often higher than 95%, confirming the suitability of the new process for geothermal use with low investment and operating costs.


text:
# Pythagoras Puzzler

### Why do this problem?

This problem provides the basis for a proof of Pythagoras' theorem using a dissection. In constructing a proof there is the opportunity for rich discussion about what constitutes a proof and why it is not enough to say that the pieces seem to form two squares, it has to be proved that lines match up and corners are right angles.

### Possible approach

In pairs, students construct a square and choose a point on one side. Make sure everyone understands that they need to mark each point the same distance from each corner as shown in the diagram. Different pairs could choose different sized squares and mark at different distances from the edge to give a variety of different dissections.

Cutting out and reassembling the pieces is really only the first part of this challenge. If anyone really struggles to reassemble it, there's a diagram in the Hint which might help. Once the two joined squares are formed, pairs should come up with arguments to convince themselves that the angles which appear to be $90^{\circ}$ actually are, and that lines which match up are equal in length.

Pairs could produce a poster or a presentation to show the stages of their proof, and then present their proofs to the rest of the class. Encourage students to be critical of each step in their own and others' proofs to make sure they are mathematically sound.

### Key questions

Which angles are $90^{\circ}$? How can you be sure?
How do the side lengths of the squares formed relate to the side lengths of the right angled triangles?
Is this enough to prove Pythagoras' Theorem? What do we need to do to make it a proof rather than a demonstration?

### Possible extension

Explore other proofs of Pythagoras' Theorem in this problem.

### Possible support

Tilted squares offers another way to think about proving Pythagoras' Theorem.