Irrational sum to integers? Is it possible for $(a-b)k + bf$ to be an integer if $k,f$ are irrational numbers and $a,b$ are integers? What about  $(a-b)k - bf$?
 A: $$(0-1)\sqrt2+1\cdot\sqrt2=0$$
$$(2-1)\sqrt2-1\cdot\sqrt2=0$$

In general this is possible, if and only if the set $\{1,k,f\}$ is linearly dependent over $\mathbb{Q}$. This is close to being a tautology, as seen by the following argument.
Assume the set $\{1,k,f\}$ is linearly dependent over $\mathbb{Q}$, then by clearing the denominators we get a linear dependency relation with integral coefficients $c_i,i=1,2,3$:
$$
c_1k+c_2f=c_3.
$$
Set $c_2=\pm b$ (the sign depending on the case you are interested in) and solve $a$ from the equation $a-b=c_1\implies a=b+c_1$.
On the other hand, if
$$
(a-b)k\pm bf=c
$$
with integers $a,b,c$, then clearly $\{1,k,h\}$ is linearly dependent over $\mathbb{Z}$, and hence also over the rationals.

So for example you can do this with $k=1+\sqrt2$ and $f=\sqrt2$. Or $k=45+17\sqrt{17}$ and $f=-4+\sqrt{17}$. OTOH you cannot do it with $k=\sqrt2$ and $f=\sqrt3$.
Linear dependence over the rationals may not sound like a very useful tool here unless you are familiar with elementary theory of field extensions. That theory will give you useful criteria for deducing linear independence over $\mathbb{Q}$.
