Equivalence Relations on Integers with a Unique Property

doggie_Walkes
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This is a question from A consise introduction to pure mathematics (Martin Liebeck)

Hi guys, just stuck on one problem was wondering if someone could lend me hand.

Let ~ be an equivalence relation on all intergers with the property that for all "m" is an element of the set of intergers , we have,

m ~ m +5
and also m ~ m+8

Prove that m~ n for all m, n is an element of intergers.

This is on page 161 of Martin Liebeck's book, number 7.

Im really stuck!
 
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didn't i see this posted a few days ago?

anyways, noinking the solution form those posts, use repeated applications of the equivalence relation to show for any n:
n ~ n+1,

then you're pretty much done, maybe with induction implicit, but it should be reasonably easy to see that any n is equivalent to any m, witr repeated application of the above
 
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Hint: can you show m~m+15 and m~m+16? Then you are almost there.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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