TMUA-P2-PRF3-012 — Paper P2 · Proof by contradiction · Level 3

Options

1. A
   Since $3$ divides $a^2$ and $3$ is prime, $3$ divides $a$ (if $a \equiv 1$ or $2 \pmod 3$ then $a^2 \equiv 1 \pmod 3$). Write $a = 3k$; then $3b^2 = 9k^2$, so $b^2 = 3k^2$ and $3$ divides $b$ too, contradicting lowest terms.
2. B
   Since $3$ divides $a^2$, $9$ divides $a$, so write $a = 9k$ and cancel immediately.
3. C
   Since $a^2 = 3b^2$, $a = 3b$, so $\sqrt{3} = 3$, a contradiction.
4. D
   Check $a = 3$, $b = 2$: $(3/2)^2 = 9/4 \ne 3$, so no fraction works.

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