Constructing the singlet state by orthogonality Every set of notes I can find says that the singlet state can be found by requiring that it be orthogonal to the triplet state with $S_z=0$ but they never explain how you actually do it. I can sort of see that $|ud-du>$ looks orthogonal to $|ud+du>$ but I need to be able to do it and understand it myself, so that I can derive the 3 particle states later. If I do the maths:
$ <ud+du|ud-du>=<ud|ud>-<ud|du>+<du|ud>-<du|du>$ 
$= 1-<ud|du>+<du|ud>-1 = <ud|du>+<du|ud> $
Both of these terms look like they could be 0 but I don't know how to do an inner product when you have two particles in there. Are these two inner products trivially 0 or is one the negative of the other?
Thanks if you can explain
 A: You can split up the two particle states like this:
$$|\psi\phi\rangle = |\psi\rangle\otimes|\phi\rangle$$
Then the two particle inner product becomes trivial:
$$\langle\alpha\beta|\psi\phi\rangle = (\langle\alpha|\otimes\langle\beta|)(|\psi\rangle\otimes|\phi\rangle) = \langle\alpha|\psi\rangle\langle\beta|\phi\rangle$$
This also generalises in the obvious way to n particle states.
A: Since the two particles don't interact, you can separate, as or1426 said, a state of type $|ψϕ⟩$ into the direct product of the two single particle states, $|ψ⟩⊗|ϕ⟩$. Thus your last line becomes first Finding the singlet state may be
$= 1-<ud|du>+<du|ud>-1 = <u|d><d|u>+<d|u><u|d> $,
each inner product here being between states of a single particle. And, as you know, $<u|d> = <d|u> = 0$. 
$$ $$
But it seems to me very strange the statement that "the singlet state can be found by requiring that it be orthogonal to the triplet state with $S_z = 0$ ". 
It's like travelling from Rome to Napoli by going around the Earth, because you first have to find the triplet state with $S_z = 0$. 
The singlet state of, e.g. 2 particles with spin, should be the eigenvector of the operator 
$$(1) \ (S_1 + S_2)^2 = (S_{1,x} + S_{2,x})^2 + (S_{1,y} + S_{2,y})^2 + (S_{1,z} + S_{2,z})^2$$ 
for the eigenvalue zero.
For a solution, try first this
$$|\psi> \ \ \ = \ \ \ (|up>_1 + |down>_1)⊗(|up>_2 - |down>_2)$$
$$ = (|up>_1⊗|up>_2 - |down>_1⊗|down>_2) - (|up>_1|⊗|down>_2) - |down>_1⊗|up>_2).$$
Check this solution first of all on two identical fermions of spin $\hbar /2$. You can immediately see that the part between the 1st pair of parentheses should be discarded, because it is symmetrical in the two fermions, i.e. interchanging the spins of the fermions nothing happens. The wave-function of two identical fermions should be antisymmetrical (change sign) at the interchanging of the spin of the fermions. So, we remain with
$$(2) \ |Singlet> = (|up>_1⊗|down>_2) - |down>_1⊗|up>_2).$$
which satisfies the antisymmetry requirement. Note that this state should be normalized, for which purpose divide it by $\sqrt (2)$.
