The relation R defined on Z by aRb if and only if a=(2^n)b prove/disprove symmetry, transitive and reflexive

by Eva   Last Updated August 02, 2018 21:20 PM

The relation $\mathbf R$ defined on $\mathbb Z$ by $a\mathbf{R} b$ if and only if $\exists n, a=(2^n)b$

prove/disprove symmetry

prove/disprove transitive

prove/disprove reflexive

This was a question I had on my math test previously and it is driving me crazy. I was trying to use an induction proof but couldn't figure out what to put as my base case.

Answers 1

This is an order relation. It is transitive and reflexive, but asymmetric.

  • Transitive since if $a=(2^n)b$ and $b=(2^m)c$ then $a=(2^{n+m})c$
  • Reflexive since $a=2^0a$
  • Assymetric since if $a=(2^n)b\wedge b=(2^m)a$ then $a=(2^{n+m})a$ so $n=m=0$ and therefore $a=b$
Ahmed Lazhar
Ahmed Lazhar
August 02, 2018 21:06 PM

Related Questions

How does this operation make a relation?

Updated June 04, 2017 17:20 PM

Understanding Equivalence Classes?

Updated August 25, 2017 21:20 PM

recurrence solution of placing tiles

Updated March 06, 2016 01:08 AM