MAT102 Lecture 07B

• When you see negations, like prove there are no $x$. It's likely a contradiction or contrapositive proof.
• It can only be a contrapositive if there are implications to flip.
• $x\sim y$
  • $x\sim x$
    • $x^2+x=x+x^2$
    • True
  • $x\sim y$, $y\sim x$
    • $x^2+y=y^2+x$
    • $y^2+x=x^2+y$
    • True
    • Just flip the vars from equals sign
    • $x^2+y=y^2+x\parallel y^2+x=x^2+y$
  • $x\sim y$ and $y\sim z$
    • $x\sim z$
    • $x^2+y=y^2+x$
    • $y^2+z=z^2+y$
    • $x^2+z=z^2+x$

    ---

    • Assume $x\sim y$ and $y\sim z$
    • Show that $x\sim z$
      • By assumption
        • $x^2+y=y^2+x$
        • $y^2+z=z^2+y$
      • Show that $x^2+z=z^2+x$
        • Manipulate the left hand side to become the right hand side
        • $x^2+y=y^2+x$
        • $y^2-y=x^2-x$
        • $y^2-y=z^2-z$
        • Equate
          • $x^2-x=z^2-z$
          • $x^2+z=z^2+x$
          • Proven
• $\left[\frac{1}{3}\right]$
  • ${\frac{2}{3}}^2+\frac{1}{3}={\frac{1}{3}}^2+\frac{2}{3}$
  • To find what's in this equivalence class
  • Set $x=\frac{1}{3}$
  • $x\sim y$
  • ${\left(\frac{1}{3}\right)}^2+y=y^2+\left(\frac{1}{3}\right)$
  • $\left(\frac{1}{9}\right)+y=y^2+\left(\frac{1}{3}\right)$
  • $y^2-y-\frac{2}{9}=0$
  • $y=\frac{1\pm \sqrt{1-4\left(\frac{2}{9}\right)}}{2}$
  • $y=\frac{1\pm \sqrt{\frac{1}{3}}}{2}$
  • $y=\frac{1}{3},\frac{2}{3}$
  • Possible to prove every equivalence class that has at most 2 elements in it.
  • There is only one equivalence class that has one element here.
• Can the sets $S_1=\{0,1,2,3,\dots \}$ and $S_2=\{0,-1,-2,\dots \}$ be equivalence classes for some equivalence relation $\sim$ on $\mathbb{Z}$
  • No because they're non-disjoint and non-equal
  • Equivalence classes partition a set.
  • $B_i$ partitions the space.
  • Claim:
    • If $S_1$ and $S_2$ are equivalence classes with $x\in S_1\cap S_2$ then $S_1=S_2$
    • Proof: $\subseteq$
      • Let $y\in S_1$ so that $x\cap y$, but then $y\in S_2$
      • If $y$ is related to $x$ and $x$ is in $S_1$ and $S_2$, then $y$ must be in $S_2$.
    • $\supseteq$ the same thing.