Let
$\qquad \mathrm{DISJOINT} = \{ \langle M_1,M_2 \rangle : M_1, M_2 \text{ are TMs and } L(M_1) \cap L(M_2) = \emptyset\}$.
How do I know if this language is decidable or not? And How do I prove my answer?
Let
$\qquad \mathrm{DISJOINT} = \{ \langle M_1,M_2 \rangle : M_1, M_2 \text{ are TMs and } L(M_1) \cap L(M_2) = \emptyset\}$.
How do I know if this language is decidable or not? And How do I prove my answer?
Assuming you have a decider $R$ for DISJOINT, you could use this to make a decider D for $E_\text{TM} = \{\langle M\rangle\mid L(M)=\emptyset\}$ as follows:
D(<M>) =
return R(<M>, <A>)
where $A$ was a TM, selected in such a way that $\langle M\rangle\in E_\text{TM}$ if and only if $(\langle M\rangle, \langle A\rangle)\in\text{ DISJOINT}$. All that's left for you is to find the $A$ and show that it satisfies the needed conditions. (There are a couple of ways to make this choice.)
Hint: If DISJOINT were decidable then even the special case in which $M_1$ is some fixed machine which accepts all inputs is decidable. This is the language $\{ \langle M \rangle : L(M) = \emptyset \}$, which you might be more familiar with.