1. Interpret the relations carefully

The question defines relations on the set of real numbers $\mathbb{R}$:

$R_1 = \{(a,b) \in \mathbb{R}^2 : a^2+b^2 \in \mathbb{Q}\}$

and

$R_2 = \{(a,b) \in \mathbb{R}^2 : a^2+b^2 \notin \mathbb{Q}\}.$

We must check whether each relation is transitive.

A relation $R$ on a set is transitive if

$(a,b) \in R \text{ and } (b,c) \in R \implies (a,c) \in R.$

2. Test transitivity of $R_1$

We need to see whether

$a^2+b^2 \in \mathbb{Q}, \quad b^2+c^2 \in \mathbb{Q}$

necessarily implies

$a^2+c^2 \in \mathbb{Q}.$

A good way is to look for a counterexample.

Take

$a=1, \quad b=0, \quad c=\sqrt{2}.$

Then:

• For $(a,b)=(1,0)$, $a^2+b^2 = 1^2+0^2=1 \in \mathbb{Q},$ so $(1,0) \in R_1$.

• For $(b,c)=(0,\sqrt{2})$, $b^2+c^2 = 0^2+(\sqrt{2})^2=2 \in \mathbb{Q},$ so $(0,\sqrt{2}) \in R_1$.

• But for $(a,c)=(1,\sqrt{2})$, $a^2+c^2 = 1+2=3 \in \mathbb{Q}.$

This still belongs to $R_1$, so this is not a counterexample.

Let us try a more careful choice.

Take

$a=\sqrt{\pi}, \quad b=\sqrt{1-\pi},$

which is not real since $1-\pi<0$, so not valid. We need real numbers whose squares combine rationally.

Instead choose:

$a=\sqrt{\alpha},\quad b=\sqrt{q-\alpha},$

with irrational $\alpha$ and rational $q$ such that both are nonnegative. Similarly for $b,c$.

Let us construct explicitly:

Take

$b=0,\quad a=\sqrt{2},\quad c=\sqrt{3}.$

Then:

$a^2+b^2=2 \in \mathbb{Q}, \qquad b^2+c^2=3 \in \mathbb{Q},$

so $(a,b)\in R_1$ and $(b,c)\in R_1$.

But

$a^2+c^2=2+3=5 \in \mathbb{Q}.$

Still not a counterexample.

So let us examine algebraically.

If

$a^2+b^2 \in \mathbb{Q}, \qquad b^2+c^2 \in \mathbb{Q},$

then subtracting,

$a^2-c^2 = (a^2+b^2)-(b^2+c^2) \in \mathbb{Q}.$

But this does not force $a^2+c^2$ to be rational.

Now choose irrational $b^2$ and rational numbers $r,s$ such that

$a^2=r-b^2, \qquad c^2=s-b^2$

are nonnegative. Then

$a^2+c^2 = r+s-2b^2,$

which can be irrational.

Take

$b^2=\frac{\sqrt{2}}{2}, \quad r=1, \quad s=1.$

Then

$a^2=1-\frac{\sqrt{2}}{2} >0, \qquad c^2=1-\frac{\sqrt{2}}{2}>0,$

so real $a,c$ exist.

Now:

$a^2+b^2 = \left(1-\frac{\sqrt{2}}{2}\right)+\frac{\sqrt{2}}{2}=1 \in \mathbb{Q},$

$b^2+c^2 = \frac{\sqrt{2}}{2}+\left(1-\frac{\sqrt{2}}{2}\right)=1 \in \mathbb{Q}.$

Hence $(a,b)\in R_1$ and $(b,c)\in R_1$.

But

$a^2+c^2 = 2\left(1-\frac{\sqrt{2}}{2}\right)=2-\sqrt{2} \notin \mathbb{Q}.$

So $(a,c)\notin R_1$.

Therefore, $R_1$ is not transitive.

3. Test transitivity of $R_2$

Now check whether

$a^2+b^2 \notin \mathbb{Q}, \quad b^2+c^2 \notin \mathbb{Q}$

implies

$a^2+c^2 \notin \mathbb{Q}.$

Again, we look for a counterexample.

Take

$a=1,\quad b=\sqrt{2},\quad c=1.$

Then:

$a^2+b^2 = 1+2=3 \in \mathbb{Q},$

so this does not belong to $R_2$. Not useful.

We want both first two sums irrational, but the third rational.

Take

$a^2=1, \quad b^2=\sqrt{2}, \quad c^2=1.$

Then

$a^2+b^2 = 1+\sqrt{2} \notin \mathbb{Q},$

$b^2+c^2 = \sqrt{2}+1 \notin \mathbb{Q},$

but

$a^2+c^2 = 1+1=2 \in \mathbb{Q}.$

Thus if we choose

$a=1, \quad b=\sqrt[4]{2}, \quad c=1,$

then

• $(a,b) \in R_2$,
• $(b,c) \in R_2$,
• but $(a,c) \notin R_2$.

Therefore, $R_2$ is not transitive.

4. Conclusion

• $R_1$ is not transitive.
• $R_2$ is not transitive.

So the correct option is

$\boxed{\text{A: Neither } R_1 \text{ nor } R_2 \text{ is transitive.}}$

5. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A, so they agree.