# Associativity is not always given

Many of my students tends to believe that associativity of a given operation is always true.
We have seen in the previous post that addition is associative so that we always have $(a+b)+c=a+(b+c)$. This actually hold true for all integers (positives and negatives).

But clearly you would agree that subtraction is not associative. $(a-b)-c\ne a-(b-c)$.

This has obviously to do with the way subtraction is define. One way to think of it is to write $-a$ as the additive inverse (or opposite) of $a$, such that: $a+(-a)=0$. The subtraction $b-a$ is then defined as $b+(-a)$. The rule is that we can replace $+(-a)$ directly by $-a$.

It follows that the additive inverse of $b+c$ is $(-b)+(-c)$ as $b+c+(-b)+(-c)=0$. Therefore we chose to write the additive inverse of $b+c$ as $-(b+c)$.

Let us now have a look at $(a-b)-c=(a+(-b))+(-c)=a+((-b)+(-c))=a+(-(b+c))$ which we write $a-(b+c)$ so that clearly subtraction is not associative

$(a-b)-c=a-(b+c)$.

# Showing that addition is associative and commutative using Mathematical Induction

I recently found a book left on the AMRC coffee table “Fundamental Concepts of Mathematics” by R.L. Goodstein. Inside his book Goodstein propose to use mathematical induction to show that addition is commutative.

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We first need to define the addition of two number as satisfying the following property: $m+(n+1)=(m+n)+1$ (this is associativity with $1$ under addition). Let us refer this property as $A_0$.

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We can first show that associativity is always true. We write $A_1$ the property $m+(n+p)=(m+n)+p$.

The proof by induction is:

1) we know $A_1$ to be true for $p=1$ as it give $A_0$

2) let us assume it exists an integers $k$ satisfying $A_1$ so that we can write $m+(n+k)=(m+n)+k$.

It follows that:

using $A_0$    $m+(n+(k+1))=m+((n+k)+1)$

using $A_0$    $m+(n+(k+1))=(m+(n+k))+1$

since $A_1$ is true for $k$    $m+(n+(k+1))=((m+n)+k)+1$

using $A_0$    $m+(n+(k+1))=(m+n)+(k+1)$

We now see that if $k$ satisfies $A_1$ then the next integer $k+1$ satisfies $A_1$. Since $A_1$  is true for $k=1$ it is true for all positive integers.

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From here we can show commutativity with $1$. We refer to this property as $C_0$: $a+1=1+a$ for all $a$ (positive) integers.

The proof by induction is the following:

1) we know that $C_0$ is true for $a=1$, as we simply have $1+1=1+1$

2) let us assume that it exists an integer $k$ for which $C_0$ is true. Then we can write $k+1=1+k$.

Since $k$ satisfies $C_0$    $(k+1)+1=(1+k)+1$

using $A_0$    $(k+1)+1=1+(k+1)$

We now see that if $k$ satisfy $C_0$ then the next integer $k+1$ satisfy $C_0$. Since $C_0$  is true for $k=1$ it is true for all positive integers.

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Next we can show the property $C_1$: commutativity with any integer $a+b=b+a$.

The proof by induction is:

1) we know $C_1$ is true for $b=1$ as it gives $a+1=1+a$ which simply is $C_0$ which we now know to be true.

2) let us assume that $C_1$ is true for some integer $k$ then we can write $a+k=k+a$.

It follows that

using $A_0$   $a+(k+1)=(a+k)+1$

since $C_1$ is true for $k$    $a+(k+1)=(k+a)+1$

using $A_0$    $a+(k+1)=k+(a+1)$

using $C_0$    $a+(k+1)=k+(1+a)$

using $A_1$    $a+(k+1)=(k+1)+a$

So that we see that if $k$ satisfy $C_1$ then the next integer $k+1$ satisfy $C_1$. Therefore $C_1$ is true for all integers.

# Martin Gardner – Problem 1 – The Returning Explorer

In “My Best Mathematical and Logic Puzzles” Martin Gardner’s first problem is titled “The Returning Explorer” in reference of the following riddle.

“”An explorer walks one mile due south, turns and walks one mile due east, turns again and walks one mile due north. He finds himself back where he started. He then shoot [or see (depending of the riddle you wish to tell)] a bear. What color is the bear ?””

The problem proposed by M. Gardner is then to find all points on the globe from which you can walk one mile south, east and north and finally end up at your starting point.

There is an infinite number of points which satisfies this property. Can you find them and give the latitude at which you need to start your journey ?

The answer is the following.  Writing $R$ the earth radius and $\phi_n$ the latitude ($n$ is integer), all latitude on which you can start are given by:

$\phi_n=\frac{1}{R}-arccos\left(\frac{1}{2\pi R n}\right).$

Which at the first order in $1/R$ leads to

$\phi_n=-\frac{\pi}{2}+\frac{1}{R}\frac{2\pi n+1}{2\pi n}$.

Now, starting from any other latitude $\theta_0$, you would not be able to “complete a triangle” by walking south for $X$ miles, east for $Y$ miles and back north for $X$ miles. But how far away from the starting point would you be exactly ?

Well if you were to walk back west from the finish to the starting point, you would have to walk for a distance $\Delta$

$\Delta =Y\frac{\cos(\theta_0)}{\cos(\theta_0-X/R)}.$

Assuming $X/R\ll1$ we have

$\Delta\simeq Y\frac{\cos(\theta_0)}{\cos(\theta_0)+\frac{X}{R}\sin(\theta_0)}.$

However, this is not the shortest distance. The fact is that the shortest path between two points $A$ and $B$ on a sphere is along the circle of radius $R$ passing by $A$ and $B$. Can we find out what the shortest path is ?

# Visiting Scholar at U Del

Since Feb 1st 2015, I am a visiting scholar at University of Delaware. I am working with Professor Abhuydai Singh in the Electronical and Engineering Department.

I also had the chance to meet with Professor Pak Wing Fok. In addition of sharing a car for our regular trips between philadelphia and Newyark, both of us are sharing similar research interests in bio-physics and stochastic processes.

So far I had the change to give two presentation of my work:

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February 2015

Department of Mathematical Sciences, University of Delaware :

Analytical Approaches for simple models of Gene Expression

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March 2015

Department of Physics, Virginia Tech, Society of Physics Students :

Birth and Death on Lineland or a purely geometrical model of creation and annihilation of particles

# Amanda Goodall: Why we need expert leaders in universities, hospitals & other settings

The works of Amanda Goodall is focused on the correlations between “manager’s expertise” and “success”. If we admit that “good management leads to better performance”, we surely should investigate what does define good management ?

Amanda Goodall will prove to you (with data to back it up) that “Experts, not managers, make the best leaders”. In her book Amanda shows that the performance of universities improves when led by presidents who are outstanding scholars. See here. Amanda explains why it is important for experts to be guided by expert leaders. Amanda raises the same statement in other sectors as well. One should mention that more and more hospitals are lead by managers with no expertise in medicine. In this field (but others too) it is of capital importance for managers to understand (best experience) what challenges are facing doctors and nurses everyday.

We will follow up on a related subject : How stress or well being affect your productivity. This leads to the work of Andrew Oswald (warwick university).

# 16 is a special number

In one of the Numberphile video –  we can learn that 16 is a very special number because it is the only integer number which can be  written as $2^4$ or $4^2$.

But how do we show that this is actually true ? Well let us look for all integers $a$ and $b$ such that $a^b=b^a$. Since we are not looking for the trivial solution for which $a=b$ let us impose $a>b$.

It follows that $(a/b)^b=b^{a-b}$. And since $a-b>0$ we know that $b^{a-b}$ is an integer, so that $(a/b)^b$ is an integer and therefore $a/b$ is an integer too.

In other words $a=kb$ for some integer $k$. Replacing $a$ by $kb$ in the equation we started with we get

$(kb)^b=b^{kb}$ which leads to

$kb=b^k$ or $k=b^{k-1}$.

From here you can:

1 – observe that $kb$ is always smaller than $b^k$ expect for $k=2$

2 – observe that $k^{1/(k-1)}$ is never integer but for $k=2$

both impose $b=2$ and $a=4$, which ends the proof.