Consider two drums of sufficiently large capacity, one of them empty and the other full of liquid.

a) Determine if it is possible to place exactly one liter of liquid from the full drum into the empty one, using two buckets, one with a capacity of 5 liters and the other with a capacity of 7 liters.

b) Determine if it is possible to place exactly one liter of liquid from one of the drums into the other using two buckets, one with a capacity of $2-\sqrt{2}$ liters and the other with a capacity of $\sqrt{2}$ liters.
Solution

a) It is enough to fill the empty drum with 15 liters ($3 \times 5$ liters) using the 5-liter bucket three times, and then remove 14 liters ($2 \times 7$ liters) using the 7-liter bucket twice. This way, we transport $3 \times 5 - 2 \times 7 = 1$ liter.
b) The amount $a$ that we can transport from the full drum to the empty one is of the form $k(2-\sqrt{2}) + l(\sqrt{2})$ liters, where $k$ and $l$ are integers indicating how many times we remove or add liquid using each of the buckets. If $l - k \neq 0$, we can write:

$$
\begin{aligned}
a & = k(2-\sqrt{2}) + l \sqrt{2} \\
a - 2k & = \sqrt{2}(l - k) \\
\frac{a - 2k}{l - k} & = \sqrt{2}
\end{aligned}
$$

Thus, the number $\sqrt{2}$ would be the quotient of two integers, which would result in a rational number. We know that this cannot happen because $\sqrt{2}$ is irrational. We still need to analyze what happens when $l = k$. The equation transforms into:

$$
\begin{aligned}
a & = k(2-\sqrt{2}) + l \sqrt{2} \\
& = k(2-\sqrt{2}) + k \sqrt{2} \\
& = 2k
\end{aligned}
$$

Notice that $2k$ is even, and thus we cannot carry an odd value like $a = 1$. In any case, it is not possible to place exactly 1 liter using the buckets with the given capacities in this item.