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For all $a \in\mathbb{Z}$, $a \sim a + 3$;

For all $a \in\mathbb{Z}$, $a \sim a + 7$;

Show that $a ∼ b$ for all $a,b \in\mathbb{Z}$.

I found the equivalence above online yet I don't comprehend how this relation is equivalent for $a\sim b$.

Thank you for any guidance or help given.

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schrodingerscat1950
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1 Answers1

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Hint 1: For every $a\in\mathbb Z$,

$$a+1 = a+7-3-3\\ a-1=a+3+3-7$$

Hint 2:

For every $a$, you can also show that $a\sim a-7$ and $a\sim a-3$

5xum
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  • Thank you for your guidance I still don't think my question is an equivalence relation – schrodingerscat1950 Nov 07 '18 at 12:44
  • @schrodingerscat1950 Where did I say the relation is an equivalence relation? – 5xum Nov 07 '18 at 12:46
  • Sorry I meant to say that the problem doesn't seem to be an equivalence relation. – schrodingerscat1950 Nov 07 '18 at 12:51
  • @schrodingerscat1950 I never said it was an equivalence relation. My hints are still useful if you think about them for a second. Try to use them, for example, to prove that $1\sim 2$. Then use them to prove that $1\sim 3$... and so on. – 5xum Nov 07 '18 at 12:53
  • I'm sorry but I am still confused with how to show 1~2 even though for every a, a~a-7 and a~a-3 – schrodingerscat1950 Nov 07 '18 at 13:02
  • @schrodingerscat1950 My bad. I think you have to assume that the relation is an equivalence relation. Otherwise, I don't think it's possible to solve the task, since, for example, the relation $\leq$ also satisfies the two conditions, but clearly, you cannot prove that $a\leq b$ for all $a,b$. – 5xum Nov 07 '18 at 13:06
  • Assuming it is an equivalence relation, do I treat each property case by case ? – schrodingerscat1950 Nov 07 '18 at 13:16
  • @schrodingerscat1950 Assuming it is an equivalence relation, you don't need to treat the cases. Just think about how you would, for example, prove that $10\sim 16$. – 5xum Nov 07 '18 at 13:23
  • I am finding it difficult to show 10~16. – schrodingerscat1950 Nov 07 '18 at 13:50
  • @schrodingerscat1950 Can you prove $10\sim 13$? – 5xum Nov 07 '18 at 13:52
  • set a=10, then a~a+3 <=> 10+3=13 thus 10~13 – schrodingerscat1950 Nov 07 '18 at 13:54
  • @schrodingerscat1950 OK. Can you prove that $13\sim 16$? – 5xum Nov 07 '18 at 13:57
  • Ok I got it thank you for your help ! – schrodingerscat1950 Nov 07 '18 at 13:59
  • will I need to do the same for a~a+7 ? – schrodingerscat1950 Nov 07 '18 at 14:00
  • @schrodingerscat1950 now that you got it, use my hints to prove that $a\sim a+1$. – 5xum Nov 07 '18 at 14:01
  • Would you mind showing me the proof so my mind can rest easy ? – schrodingerscat1950 Nov 07 '18 at 14:13
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    @schrodingerscat1950 No. I will not to all the work for you. I gave you two hints, I lead you step by step to proving that $10\sim 16$, now it's your turn to do some work. If I give you the solution to this problem, then the moment you encounter a similar problem, you will freeze again. Mathematics is not learnt by watching others doing it. Mathematics is learnt by sitting down to grueling sessions of hard work, sometimes for hours with no success, until finally figuring it out. – 5xum Nov 07 '18 at 14:17