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Proving transitivity on relation: $aRb=7|(|a-b|)$, so $aRb\, \wedge bRc\implies aRc$

What I tried:

$$7k =|a-b| $$ $$7l=|b-c|$$ $$l,k\in\mathbb{Z}$$

Now I squared the two equations and subtracted the top one from the bottom one: $$49(l^2-k^2)=(b-c)^2-(a-b)^2 =c^2-2b(a-c)-a^2\neq|a-c|^2$$

I see that this approach does not work, since I can't get the square root of the distance between $a$ and $c$, so that $\sqrt{49(l^2-k^2)}$ would be a rational number for all $l$ and $k$ (for instance we could have $l=3$ and $k= 2,$ and we get $7\cdot\sqrt{5}\notin\mathbb{Q}$), so my equation above does not imply that transitivity does not exist.

My question is what would be the best way to prove if it does or doesn't exist?

Bernard
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VLC
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2 Answers2

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You want to see that $aRc$, that is $7$ divides $|a-c|$. By hypothesis $7k=a-b$ and $7l=b-c$ (you can do this choosing $k,l$ from $\mathbb{Z}$). Then: $$a-c=a-b+b-c=7k+7l=7(k+l)\rightarrow |a-c|=7|k+l|$$

Marcos
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Alternative approach:

You want to prove that

$$\{ ~( ~7 ~| ~|a-b| ~) ~~\wedge~~ ~( ~7 ~| ~|b-c| ~) ~\} ~~\implies ~~ ~( ~7 ~| ~|a-c| ~).$$

Note that

  • Either $~~~~~|a - b| = (a-b)~~~~~$ or $~~~~~|a - b| = (-1) \times (a-b).$
  • Either $~~~~~|b - c| = (b-c)~~~~~$ or $~~~~~|b - c| = (-1) \times (b-c).$
  • Either $~~~~~|a - c| = (a-c)~~~~~$ or $~~~~~|a - c| = (-1) \times (a-c).$

By assumption, there exists $r,s \in \Bbb{Z}$ such that

  • $7r = |a - b|.$
  • $7s = |b - c|.$

Define $R$ so that:

  • $R = (r)~~$ if $~~|a - b| = (a-b)$.
  • $R = (-r)~~$ Otherwise.

Define $S$ so that:

  • $S = (s)~~$ if $~~|b - c| = (b-c)$.
  • $S = (-s)~~$ Otherwise.

Then,

$$\{ ~[ ~7R = (a-b) ~] ~~\wedge~~ ~[ ~7S = (b-c) ~] ~\} ~~\implies$$

$$[ ~7(R+S) = (a-c) ~: ~(R+S) \in \Bbb{Z} ~].$$

Define $T$ so that:

  • $T = (R+S)~~$ if $~~|a - c| = (a-c)$.
  • $T = [-(R+S)]~~$ Otherwise.

Then, $~~T \in \Bbb{Z}~~$ and $~~7T = |a - c|.$

Therefore, $~7 ~| ~|a - c|.$

user2661923
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  • Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 18 '21 at 03:39
  • @BillDubuque My answer, which occurred after Marcos posted his answer, was in response to what I considered an analytical error in Marcos' answer. That is, he confused (for example) $|a-b| + |b-c|$ with $(a-b) + (b-c)$. I think that you and I have debated this issue before. When an answer is posted that I disagree with, then I think it is important to post a rebuttal answer, directly, rather than linking to a different article. – user2661923 Nov 18 '21 at 03:47
  • You can comment (and vote) on incorrect answers. That's not a good excuse to post further duplicate answers (this topic is a big FAQ) - as should be clear to any experienced user. – Bill Dubuque Nov 18 '21 at 03:55
  • @BillDubuque For one thing, the situation is fairly rare, so it will not present itself very often. For another, you are looking at the situation from the viewpoint of site maintenance, while I am looking at the situation from the viewpoint of the effect on the OP. Again, labeling the situation as a vanilla dupe, is, in my mind, iffy. Besides that, we already have one dead horse pending. – user2661923 Nov 18 '21 at 04:01
  • Rampant duplication is one of the biggest problems the site faces. Keep in mind that most of your answers will have limited value if no one can find them in the future because they are buried among hundreds (or thousands) of (mostly) useless dupes. – Bill Dubuque Nov 18 '21 at 04:04