Relations, Equivalence class

Question

Define the relation $R$ on the set $\Bbb Z^+$ of all positive integers by: for all $a, b \in \Bbb Z^+$, $aRb$ if and only if the largest digit of a is equal to the largest digit of $b$. For example, $\,271\,R\, 770\,$ because the largest digit of $271$ is 7 which is also the largest digit of $770$.

1. Prove that $R$ is an equivalence relation on $Z^+$.
2. Find the number of equivalence classes of $R$. Explain.
3. Find and simplify the number of positive integers between $100$ and $1000$ which are in the equivalence class $[271]$. Explain.

My attempt to answer:

1. A relation $R$ is on a set $Z^+$ is equivalence if and only if, $R$ is reflexive, and $R$ is symmetric and $R$ is transitive.

   Here, $R$ is reflexive because, for all $a \in \Bbb Z\ aRa$.
   $R$ is symmetric as well. For all $a, b \in \Bbb Z$, if $aRb$ then $bRa$.
   $R$ is transitive as well, because, for all $a, b, c \in \Bbb Z$, if $aRb$ and $bRc$, then $aRc$.
   Therefore, $R$ is equivalence.

2. Number of equivalence classes of $R$, $\ [R]$ = $\left\{\,x \mid x \in\; \equiv \,\right\}$ so we have 9 equivalence classes.

I am not sure about this. Can someone explain this part to me please?

Answer 1

Hint:

If you investigate the questions like: "is $R$ and equivalence relation on set $A$?" then often (even stronger: almost always) it is very handsome to look for a function that has $A$ as domain and satisfies $$aRb\iff f(a)=f(b)\tag1$$

If you have found such a function then you are allowed to conclude:

• $R$ is an equivalence relation on $A$.
• The equivalence classes are the fibres of the function $f$, so take the form $[a]:=\{b\in A\mid f(a)=f(b)\}$

It is clear also that the number of equivalence classes is the cardinality of the range of function $f$.

You can do it with the function $f:\mathbb Z^+\rightarrow\{1,2,\cdots,9\}$ prescribed by: $$n\mapsto\text{largest digit of } n$$

Why is it so that you can conclude immediately that $R$ is an equivalence relation? Well:

• $f(a)=f(a)$ for each $a\in A$ (reflexive)
• $f(a)=f(b)\implies f(b)=f(a)$ for each $a,b\in A$ (symmetric)
• $f(a)=f(b)\wedge f(b)=f(c)\implies f(a)=f(c)$ for each $a,b,c\in A$ (transitive)

It is clear as crystal that these things are true for any function $f$ and $(1)$ makes it legal to replace expressions like $f(a)=f(b)$ by $aRb$.

Answer 2

Well, the equivalence class of $n$ (say the largest digit of $n$ is $k$) would be the positive integers whose largest digit is $k$. Now $k$ can be any number from 1 to 9 (you must omit 0 because a positive integer must have a positive digit). So for Question 3, you just need to find the number of integers between 100 and 1000 whose largest digit is 7.