To prove that for a relation $R$ on $\mathbb{R} \times \mathbb{R}$ such as $(a,b)R(c,d)$ exists if and only if $\exists k \in \mathbb{N}$ so that :
$k \le \sqrt{a^2+b^2} < k+1$
$k \le \sqrt{c^2+d^2} < k+1$
- I understand that to prove I need to show that it is reflexive, symetric and transitive. However, the proof just seems silly because the equation on both sides are the same. Would this be an acceptable proof?
Reflexive & Symmetric: $(a,b)\, \mathscr{R}\, (a,b) \iff k \le \sqrt{a^2+b^2} < k+1 \iff k \le \sqrt{a^2+b^2} < k+1$ $(c,d)\, \mathscr{R}\, (a,b) \iff k \le \sqrt{c^2+d^2} < k+1 \iff k \le \sqrt{a^2+b^2} < k+1$
Because we can substitute $(a,b)$ for $(c,d)$ and obtain the exact same equation on both sides.
Alternatively I was also thinking that we could prove that if:
$\sqrt{a^2+b^2} = k$ and $\sqrt{c^2+d^2} = k$ then:
$\sqrt{a^2+b^2} - \sqrt{a^2+b^2} = 0$
$\sqrt{c^2+d^2} - \sqrt{a^2+b^2} = 0$
Transitive:
$(a,b)R(c,d) \rightarrow (c,d)R(e,f) \rightarrow (a,b)R(e,f)$
$k \le \sqrt{a^2+b^2} < k+1 \iff k \le \sqrt{c^2+d^2} < k+1 \iff k \le \sqrt{e^2+f^2} < k+1$
Because all three will reside between a value $k \le (x,y) \le k+1$
I am not sure how to go about proving this, I've seen other examples in the book but they really rely on one condition that links both $a,b$ and $c,d$. Given that in this example they are seperated I am confused as to how I should be tackling the problem. I feel as if what I wrote doesn't prove much, but rather just shows the definition of the given properties.
- Also how would you show the equivalence classes graphically?
Thanks,
Sometimes proving equivalence relations IS trivial.
For example if the equivalence is based adding or multiplication of two numbers symmetry and reflexivity are automatic as addition and multiplication is commutative.
However this one, although very easy, is not as trivial as you think. Note $(a,b) R (c,d)$ if there exists some $k$ so that $k \le \sqrt{a^2+ b^2}, \sqrt{c^2 + d^2} < k+1$ but that if $(a,b) R (cd)$ and $(e,f)R (g,h)$ then the $k$ so that $k \le \sqrt{a^2+ b^2}, \sqrt{c^2 + d^2} < k+1$ need not be the same naturall number $j$ so that $j \le \sqrt{e^2+ f^2}, \sqrt{g^2 + h^2} < j+1$.
But it is still almost trivial.
Obviously as $\sqrt{a^2 + b^2} = \sqrt{a^2 + b^2}$ Reflexivity holds. You can spell it out if you want to (and you probably should if 1) you are teaching a student learning the stuff 2) you are a student learning the stuff or 3) you don't know how fussy you teacher is and s/he might assume you don't really get this stuff if you don't) by writing the absurdly obvious: $k \le \sqrt{a^2 + b^2} < k+1 \iff k \le \sqrt{a^2 + b^2} < k+1$.
Oh, we should also point out the obvious: that i) As $a^2 + b^2 \ge 0$, $\sqrt{a^2 + b^2}$ always exists and ii) by the archimedian principal there always exist a $k \in \mathbb Z$ so that $k \le \sqrt{a^2 + b^2} < k+1$ and iii) as $\sqrt{a^2 + b^2} \ge 0$ then the $k$ in ii) is a natural number (we're including zero I presume), then all ordered pairs are actually in an equivalence class.
And as $k = k$ and $k +1 = k+1$ Symmetry and Transitivity hold. I.E. If $k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{c^2 + d^2} < k+1$ and $k\le \sqrt{e^2 + f^2} < k+1$ then .... welll, $k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{c^2 + d^2} < k+1$ and $k\le \sqrt{e^2 + f^2} < k+1$, duh.
But we can spell them out specifically:
Symmetry:($k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{c^2 + d^2} < k+1$) $\iff$ ($k\le \sqrt{c^2 + d^2} < k+1$ and $k\le \sqrt{a^2 + b^2} < k+1$)
And Transitivity: ($k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{c^2 + d^2}<k+1)$ AND ($k\le \sqrt{c^2 + d^2} < k+1$ and $k\le \sqrt{e^2 + f^2})$\implies ($k\le \sqrt{a^2 + b^2} < k+1$ and $k\le \sqrt{e^2 + f^2})$
....
So, yes, the proof WAS that trivial.