Animated Solution for Mathematics - Sets and Relations: Let $A=\{1,2,3,\dots ,10\}$ and $B=\{(m,n):m,n\in A,m<n\text{ and }\gcd (m,n)=1\}$. Then $n(B)$ is equal to :

Welcome, fellow traveler of the mathematical landscape! Today, we are going to explore a problem that might seem like a simple counting exercise at first, but it actually opens the door to one of the most elegant concepts in number theory: Euler's Totient Function.

Imagine you have a set $A=\{1,2,3,\dots ,10\}$. We want to form a set $B$ of ordered pairs $(m,n)$ from this set, subject to two specific rules: 1. $m$ must be strictly less than $n$ ($m<n$). 2. $m$ and $n$ must be coprime, meaning their greatest common divisor, $\gcd (m,n)$, must be $1$.

To truly understand this, let us visualize a $10\times 10$ grid. On the horizontal axis, we have $m$, and on the vertical axis, we have $n$.

The condition $m<n$ means we are only interested in the points that lie above the diagonal line $m=n$. Every point in this upper triangular region represents a potential pair $(m,n)$.

The condition $\gcd (m,n)=1$ acts like a filter. It removes all the points where $m$ and $n$ share a common factor greater than $1$. What remains are the coprime pairs.

For any fixed $n$, we are looking for the number of integers $m$ such that $1\le m<n$ and $\gcd (m,n)=1$. This is exactly the definition of Euler's Totient Function, denoted as $\varphi (n)$.

The total number of elements in set $B$, which we call $n(B)$, is simply the sum of $\varphi (n)$ for all possible values of $n$ from $2$ to $10$. Mathematically, we write this as:

Let us calculate these values one by one with precision:

For $n=2$: The only $m<2$ is $1$. Since $\gcd (1,2)=1$, we have $\varphi (2)=1$. For $n=3$: The $m$ values are $1,2$. Both are coprime to $3$, so $\varphi (3)=2$. For $n=4$: The $m$ values are $1,2,3$. Since $\gcd (2,4)=2$, we exclude $2$. We keep $1$ and $3$, so $\varphi (4)=2$. For $n=5$: Since $5$ is prime, all $m<5$ are coprime to it. Thus, $\varphi (5)=4$. For $n=6$: The $m$ values are $1,2,3,4,5$. We exclude $2,3,4$ because they share factors with $6$. We keep $1$ and $5$, so $\varphi (6)=2$. For $n=7$: Since $7$ is prime, all $m<7$ are coprime to it. Thus, $\varphi (7)=6$. For $n=8$: We exclude all even numbers. We keep $1,3,5,7$, so $\varphi (8)=4$. For $n=9$: We exclude multiples of $3$ ($3,6$). We keep $1,2,4,5,7,8$, so $\varphi (9)=6$. * For $n=10$: We exclude multiples of $2$ and $5$. We keep $1,3,7,9$, so $\varphi (10)=4$.

Now, we simply add these values together:

Performing the addition: $1+2=3$ $3+2=5$ $5+4=9$ $9+2=11$ $11+6=17$ $17+4=21$ $21+6=27$ $27+4=31$

The total number of elements in set $B$ is 31. This journey through the grid and the properties of coprime numbers shows us that even a seemingly abstract problem can be broken down into a beautiful, logical sequence.