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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?

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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.

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

String
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  • @crash: That approach seems less general to me. Why make it more detailed about the higher places, when we do not need them anyway? – String Mar 11 '15 at 09:18
  • @crash If $N=1087$ then $N= 100(100)+8(10)+7$, and $100\in \Bbb N$. We only need to be concerned about the 'tens' place. – Graham Kemp Mar 11 '15 at 09:20
  • @crash: Then $N=1327=100\cdot 13+10\cdot 2+7$ and so on. More generally $1000a+100b+10c+d=100(10a+b)+10c+d$. – String Mar 11 '15 at 09:21
  • $1327 = 100\times 13+10\times 2+7$, $10293785576 = 100\times 102937855+ 10\times 7+6$, et cetera – Graham Kemp Mar 11 '15 at 09:23
  • @String Well, it's a very weird way of representing base-10 numbers IMO, but I can see what you mean, and only the tens place is being considered anyway. I'll remove my comments. – Daniel W. Farlow Mar 11 '15 at 09:23
  • @crash: Please do not remove them - it is a valid question, a valid concern IMO. – String Mar 11 '15 at 09:24
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    @String Sorry. I already removed them. :( I'll briefly express here again what I mentioned so others will see what I commented on -- Usually a base-10 number $N$ is represented as $$ N= a_nb^n+a_{n-1}b^{n-1}+\cdots+a_2b^2+a_1b+a_0, $$ where $0\leq a_i<b, i=0,1,\ldots,n$. My quibble had to do with the following: Our own numeral system is an example of a positional numeral system with base $10$. For such a system, after the base $b$ has been selected, basic symbols are adopted for $0,1,2,\ldots,b-1$. Thus, there are $b$ basic symbols, frequently called the digits of the system. – Daniel W. Farlow Mar 11 '15 at 09:29
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    [cont'd] We then represent the number $N$ (as done above) to base $b$ by the sequence of basic symbols $$ a_na_{n-1}\ldots a_2a_1a_0. $$ Thus, a basic symbol in any given numeral represents a multiple of some power of the base, the power depending on the position in which the basic symbol occurs. – Daniel W. Farlow Mar 11 '15 at 09:31
  • @crash: Nice context and details in those last two comments! I find it relevant, since others may very well think along the same lines. – String Mar 11 '15 at 09:34
  • @String Thanks. Hopefully now people won't think I'm an idiot haha. In view of our petty misunderstanding, good answer. +1 – Daniel W. Farlow Mar 11 '15 at 09:37
  • I Just show it is an equivalence relation. But just to understand this relation better, I want to compute equivalence classes of some elements. For instance, take in [106], 200 and 30 belong to this equivalence class, correct ? –  Mar 11 '15 at 09:39
  • @ProbabilityGuy: The equivalence classes are infinitely huge. You are correct that $200\in[106]$, but not $30\in[106]$ since $3\neq 0$ which are the tens place of $30$ and $106$ respectively. – String Mar 11 '15 at 09:47
  • how about $0$ ? –  Mar 11 '15 at 09:48
  • @ProbabilityGuy: In the number $30$, the digit $0$ sits in the $1$'s place, not the $10$'s. – String Mar 11 '15 at 09:48
  • I mean, the number $0$ is trivially in this equivalence class, correct? –  Mar 11 '15 at 09:49
  • @ProbabilityGuy: OK, I get what you mean. Yes, $0$ belongs to $[106]$, but I do not find that any more trivial than testing whether any other $x\neq 0$ belongs to $[106]$. – String Mar 11 '15 at 09:52
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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.
drhab
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Like $\mathbf{1}2$ and $3\mathbf{1}8$, $900\mathbf{7}1$ and $\mathbf{7}0$.

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

Eli Rose
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