TMUA-P2-PRF3-011 — Paper P2 · Proof by contradiction · Level 3

Options

1. A
   $N$ itself must be prime, and it is not in the list, a contradiction.
2. B
   $N > 1$ has a prime divisor $p$. This $p$ cannot be any $p_i$, since $N$ leaves remainder $1$ on division by each $p_i$; but $p$ must be in the complete list, a contradiction.
3. C
   $N$ is divisible by each $p_i$, so $N + 1$ is prime, a contradiction.
4. D
   Since the list was arbitrary, there must be infinitely many such lists.

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

Keep practising

• Same topic · same level

  P2PRF3Level 3

  TMUA-P2-PRF3-006

  A journal-style proof that √3 is irrational is circulated for peer review: it assumes √3 = a/b in lowest terms, deduces…

  Proof by contradiction

  Open TMUA-P2-PRF3-006 →

• Same trap: single-counterexample

  P2ARG1Level 3

  TMUA-P2-ARG1-009

  A textbook exercise states the following facts about positive integers, where P is “ n is divisible by 6 ” and Q is “ n…

  Statements and connectives

  Open TMUA-P2-ARG1-009 →

• More in this topic

  P2PRF3Level 3

  TMUA-P2-PRF3-012

  A student proves √3 is irrational: assume √3 = a/b in lowest terms, so 3b^2 = a^2. Which next step validly advances the…

  Proof by contradiction

  Open TMUA-P2-PRF3-012 →