How to show if relation on $\mathbb N\times\mathbb N$ defined $(a,b) \sim (c,d)$ by $ad(b+c)=bc(a+d)$ is transitive?

Question

I can show it is reflexive and symmetric but I don't know how to show transitivity using only the properties of natural numbers (no division).

Answer 1

Rewrite the equivalence as $\;\dfrac1a+\dfrac1d=\dfrac1b+\dfrac1c$.

Answer 2

Suppose $(a,b)\sim (c,d)$ and also that $(c,d)\sim (e,f)$.

By definition of our relation, you have then that $ad(b+c) = bc(a+d)$ and that $cf(d+e)=de(c+f)$

We ask if it follows that $af(b+e)=be(a+f)$ (which would imply that $(a,b)\sim (e,f)$ and that $\sim$ is transitive).

From $(a,b)\sim(c,d)$ we know that $ad(b+c)=bc(a+d)\Leftrightarrow acd - bcd = abc - abd\Leftrightarrow (a-b)cd = (c-d)ab$

Similarly, $(c,d)\sim (e,f)$ implies $(c-d)ef = (e-f)cd$

Then we have $((a-b)ef)cd=((a-b)cd)ef = ((c-d)ab)ef = ((c-d)ef)ab = ((e-f)cd)ab=((e-f)ab)cd$

Since $c\neq 0$ and $d\neq 0$, this implies that $(a-b)ef=(e-f)ab$. This in turn implies that $(a,b)\sim (e,f)$ and the relation is indeed transitive.

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Without using subtraction:

Note that $(a,b)\sim (c,d)\Leftrightarrow ad(b+c) = bc(a+d) \Leftrightarrow abd + acd = abc + bcd~~~$ (eqn 1)

Similarly, $(c,d)\sim (e,f)\Leftrightarrow cdf + cef = cde + def~~~$(eqn 2)

We have then:

$\begin{array}{ll|l} abcdf+abdef+acdef & =abcdf + (abd+acd)ef&\text{rearranging and distributivity}\\ &=abcdf + (abc+bcd)ef&\text{by eqn 1}\\ &=abcdf+abcef+bcdef&\text{rearranging and distributivity}\\ &=ab(cdf+cef) + bcdef&\text{rearranging and distributivity}\\ &=ab(cde+def)+bcdef&\text{by eqn 2}\\ &=abcde + abdef + bcdef&\text{rearranging and distributivity} \end{array}$

So, $abcdf + \color{blue}{abdef} + acdef = abcde + \color{blue}{abdef} + bcdef$

So, $ab\color{blue}{cd}f + a\color{blue}{cd}ef = ab\color{blue}{cd}e + b\color{blue}{cd}ef$.

Since $cd\neq 0$, we have $abf + aef = abe + bef$

This implies then that $af(b+e) = be(a+f)$, showing that $(a,b)\sim (e,f)$