TMUA-P2Z-PRF2-003 — Paper P2 · Proof by cases · Level 2

Options

1. A
   Split into $x \ge 0$ and $x < 0$, since the sign of $y$ plays no role.
2. B
   Split into $x$ rational and $x$ irrational, using $|t| = t$ for rational $t$.
3. C
   Split into $x = y$ and $x \ne y$, checking $|x^2| = x^2$ in the first case.
4. D
   No cases are needed since $|x|$ and $|y|$ are non-negative, so $|xy| = |x||y|$ follows at once.
5. E
   Split into $x \ge 0, y \ge 0$; $x \ge 0, y < 0$; $x < 0, y \ge 0$; and $x < 0, y < 0$, using $|t| = t$ for $t \ge 0$ and $|t| = -t$ for $t < 0$ in each case.

Letters only — the answer key is hidden. Work it out, then check the solution below.

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