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Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.

Problem 1

Let $\ast$ denote a binary operation on a set $S$ with the property that

$$ (w \ast x) \ast (y \ast z) = w \ast z \text{ for all } w, x, y, z \in S $$

Show:

  1. If $a \ast b = c$, then $c \ast c = c$.
  2. If $a \ast b = c$, then $a \ast x = c \ast x$ for all $x \in S$.

Hints

Hint for part (a)

Can you use the given equation with some creative substitutions?

Hint for part (b)

If you’ve already tried using the given equation with some substitutions, but can’t quite prove the equation, maybe reusing the result from part (a) would help?

Note 1: the reason that you can reuse it is that they both have the same given assumption, namely that $a \ast b = c$, which in this case means that part (b) can depend on part (a), as part (a) can be proved without using the result from part (b).

Note 2: it’s important to note that both of the parts can’t depend on each other, because that would be circular reasoning.

Solution

Solution for part (a) of the problem. Try solving it first or see the hint above!

Part (a)

Use the given equation, substitute, and simplify:

$$ \begin{align*} (w \ast x) \ast (y \ast z) & = w \ast z & \quad \text{given} \\ (a \ast b) \ast (a \ast b) & = a \ast b & \quad \text{substitute} \ w = a, x = b, y = a, z = b \\ c \ast c & = c & \quad \text{simplify using} \ a \ast b = c \\ \end{align*} $$

This proves that if $a \ast b = c$, then $c \ast c = c$. $\quad \blacksquare$

Solution for part (b) of the problem. Try solving it first or see the hint above!

Part (b)

Use the given equation with some substitutions and utilize the result from part $1$:

$$ \begin{align*} (w \ast x) \ast (y \ast z) & = w \ast z & \quad \text{given} \\ (c \ast c) \ast (x \ast x) & = c \ast x & \quad \text{substitute} \ w = c, x = c, y = x, z = x \\ c \ast (x \ast x) & = c \ast x & \quad \text{simplify using} \ c \ast c = c \ \text{from part (a)} \\ (a \ast b) \ast (x \ast x) & = c \ast x & \quad \text{expand using} \ a \ast b = c \\ a \ast x & = c \ast x & \quad \text{simplify using given equation} \\ \end{align*} $$

This proves that if $a \ast b = c$, then $a \ast x = c \ast x$ for all $x \in S$. $\quad \blacksquare$