Trying to figure out whether the following relation is an equivalence relation

Question

LEt $R$ be a relation on $\mathbb{N}$ given by $m R n$ iff $m$ and $n$ have the same digit in the tens place. What does it mean to have the same in digit in the tens place?

Answer 1

If $x=100a+10b+c$ for $a\in\mathbb N$ and $b,c\in\{0,1,...,9\}$, then $b$ is the digit of $x$ in the tens place. Then you need to check that $R$ is reflexive, symmetric and transitive, which it is.

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Another way to write the digit $b$ in the tens place of $x$ is $b=\lfloor x/10\rfloor\pmod{10}$.

Answer 2

Like $\mathbf{1}2$ and $3\mathbf{1}8$, $900\mathbf{7}1$ and $\mathbf{7}0$.

But not $381\mathbf{0}2$ and $381\mathbf{1}2$.

Answer 3

Not really an answer to your question, but too much for a comment.

In the context of proving that the relation is an equivalence it is irrelevant what is meant by "have the same in digit in the tens place". Relevant is only the existence of a function $f$ with domain $\mathbb N$ that sends $n\in\mathbb N$ to this digit (whatever it is).

Then:$$mRn\iff f(m)=f(n)$$ and this makes it easy to prove that $R$ is an equivalence relation:

• $f(m)=f(m)$ so reflexive.
• $f(m)=f(n)\implies f(n)=f(m)$ so symmetric.
• $f(m)=f(n)\wedge f(n)=f(k)\implies f(m)=f(k)$ so transitive.