Question bank

TMUA-P2-PRF3-001

A proof by contradiction that $\sqrt{2}$ is irrational must begin with which assumption?

TMUA-P2-PRF3-002

To prove by contradiction that the set $\{q \in \mathbb{Q} : q > 0\}$ has no smallest element, which assumption starts t…

TMUA-P2-PRF3-003

A proof by contradiction that “if $n^2$ is even then $n$ is even” starts below. Which opening is correct?

TMUA-P2-PRF3-004

Euclid's proof that there are infinitely many primes assumes finitely many primes $p_1, \dots, p_k$, forms $N = p_1\cdot…

TMUA-P2-PRF3-005

To prove there is no largest integer, a student writes: “Assume the opposite. Then there is a largest integer $N$. But $…

TMUA-P2-PRF3-006

A journal-style proof that $\sqrt{3}$ is irrational is circulated for peer review: it assumes $\sqrt{3} = \frac{a}{b}$ i…

TMUA-P2-PRF3-007

A proof by contradiction that $\sqrt{2}$ is irrational must begin with a correct assumption for contradiction. Which ope…

TMUA-P2-PRF3-008

A student proves by contradiction that there is no greatest integer. Which argument is valid?

TMUA-P2-PRF3-009

Which of the following is a valid proof by contradiction that if $n^2$ is even then $n$ is even?

TMUA-P2-PRF3-010

Which of the following is a valid proof by contradiction that the sum of a rational number and an irrational number is i…

TMUA-P2-PRF3-011

Euclid's proof that there are infinitely many primes begins: suppose, for contradiction, that $p_1, p_2, \dots, p_n$ are…

TMUA-P2-PRF3-012

A student proves $\sqrt{3}$ is irrational: assume $\sqrt{3} = a/b$ in lowest terms, so $3b^2 = a^2$. Which next step val…

TMUA-P2Z-PRF3-001

To prove by contradiction that there is no largest prime — i.e. that for every prime $p$ there is a larger prime — which…

TMUA-P2Z-PRF3-002

A student proves that $\log_2 5$ is irrational by contradiction. Which chain is valid?

TMUA-P2Z-PRF3-003

To prove there is no largest rational number strictly less than $2$, a student assumes the opposite: some largest such r…

TMUA-P2Z-PRF3-004

A journal-style proof that $\sqrt[3]{2}$ is irrational is circulated for peer review: it assumes $\sqrt[3]{2} = \frac{a}…