Is the relation $a^2 +b^2 =0$ transitive?

Question

A relation on the real numbers is given by $a^2 +b^2 =0$. I know this is not reflexive, I believe it is symmetric(although there is only one solution) but I'm not sure whether it can be called transitive as the only solution is $a=b(=0)$.

Answer 1

The definition of the "the relation is transitive" is

For all $a, b, c$ it holds that if $a\sim b$ and $b\sim c$, then $a\sim c$.

The "if" and "then" here are supposed to be understood in the particular mathematical jargon (known as a "material conditional") where their meaning is that whenever you have a situation where [$a\sim b$ and $b\sim c$] is false, that alone will cause the whole if-then claim to be considered true.

That is completely the case for your relation. Some examples:

$$ \begin{array}{|c|c|}\hline a & b & c & a \sim b\text{ and }b\sim c? & a \sim c? & \text{so, are we happy?} \\ \hline 0 & 0 & 0 & \text{yes} & \text{yes} & \text{yes} \\ \hline 0 & 0 & 1 & \text{no: }b\sim c\text{ is false} & \text{no} & \text{yes} \\ \hline 2 & 2 & 2 & \text{no, neither} & \text{no} & \text{yes} \\ \hline 0 & 5 & 0 & \text{no, neither} & \text{yes} & \text{yes} \\ \hline \end{array} $$

The one combination that can't possibly arise is $$ \begin{array}{|c|c|}\hline a & b & c & a \sim b\text{ and }b\sim c? & a \sim c? & \text{so, are we happy?} \\ \hline ?? & ?? & ?? & \text{yes} & \text{no} & \textbf{no} \\ \hline \end{array} $$ and since that is impossible, your relation is transitive.

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For what's worth, there's a similar variety of intermediate results when we check whether $<$ is transitive -- which it had better be, since it's one of the ur-examples of transitive relations:

$$ \begin{array}{|c|c|}\hline a & b & c & a < b\text{ and }b < c? & a < c? & \text{so, are we happy?} \\ \hline 1 & 2 & 3 & \text{yes} & \text{yes} & \text{yes} \\ \hline 3 & 2 & 1 & \text{no, neither} & \text{no} & \text{yes} \\ \hline 1 & 1 & 2 & \text{no: }a<b\text{ is false} & \text{yes} & \text{yes} \\ \hline \end{array} $$

But there's no requirement that all three situations in fact arise. For example the relation that never holds is transitive, since the only situation we can get is

$$ \begin{array}{|c|c|}\hline a & b & c & a \sim b\text{ and }b\sim c? & a \sim c? & \text{so, are we happy?} \\ \hline \text{whatever} & \text{whatever} & \text{whatever} & \text{no, neither} & \text{no} & \text{yes} \\ \hline \end{array} $$

And the relation that relates everything to everything is also transitive since for it we end up with $$ \begin{array}{|c|c|}\hline a & b & c & a \sim b\text{ and }b\sim c? & a \sim c? & \text{so, are we happy?} \\ \hline \text{whatever} & \text{whatever} & \text{whatever} & \text{yes} & \text{yes} & \text{yes} \\ \hline \end{array} $$