A journal-style proof that √3 is irrational is circulated for peer review: it assumes √3 = a/b in lowest terms, deduces a^2 = 3b^2 so that 3 | a, writes a = 3k, and then shows 3 | b, contradicting lowest terms. One referee claims the argument only shows the fraction was not in lowest terms and can a…

Worked solution for TMUA-P2-PRF3-006 is revealed on this page behind the solution details.

Why isn't the answer to TMUA-P2-PRF3-006 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2-PRF3-006 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2-PRF3-006?

No — calculators are banned in the TMUA. TMUA-P2-PRF3-006 is set for a no-calculator target of about 280 seconds; the worked solution shows the fastest by-hand route.

Which TMUA specification point is TMUA-P2-PRF3-006?

TMUA-P2-PRF3-006 is Paper P2 · Proof by contradiction (specification code PRF3), level 3. Drill more of the same topic from this page.

TMUA-P2-PRF3-006 — Paper P2 · Proof by contradiction · Level 3

P2PRF3Level 3~280s target

Spec: PRF3 · No calculator · ~280s

A journal-style proof that

\sqrt{3}

is irrational is circulated for peer review: it assumes

\sqrt{3} = \frac{a}{b}

in lowest terms, deduces

a^2 = 3b^2

so that

3 \mid a

, writes

a = 3k

, and then shows

3 \mid b

, contradicting lowest terms. One referee claims the argument only shows the fraction was not in lowest terms and can always be reduced; another claims the opening assumption is stronger than needed. Which referee report correctly judges the contradiction structure?

Options

The proof assumes $\sqrt{3}$ is rational but should have assumed it is an integer.
The proof shows $3 \mid a$ but $3 \mid b$ does not follow.
The argument proves $\sqrt{3}$ is rational.
The proof is invalid because fractions can always be reduced.
The proof is valid: any rational has a lowest-terms representative, and forcing a common factor of $3$ contradicts minimality, so no representation exists.

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

Hint (check trope)

No written hint yet — check the trope: contradiction-not-negated, assumes-what-proves. Name the trap first, work the stem by hand, then open the solution below.

Show solution

Lowest terms is no extra assumption: every rational has one. The key lemma (

3 \mid a^2 \Rightarrow 3 \mid a

) pushes the factor

3

onto both numerator and denominator, contradicting coprimality. Trap: “just reduce again” misses that infinite descent is impossible — the contradiction kills the assumed rational existence.

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Options

The proof assumes

\sqrt{3}

is rational but should have assumed it is an integer.

The proof shows

3 \mid a

but

3 \mid b

does not follow.

The argument proves

\sqrt{3}

is rational.

The proof is invalid because fractions can always be reduced.

The proof is valid: any rational has a lowest-terms representative, and forcing a common factor of

3

contradicts minimality, so no representation exists.

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-PRF3-011

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

Proof by contradiction

Open TMUA-P2-PRF3-011 →

Same trap: assumes-what-proves

P2ERR1Level 2

TMUA-P2-ERR1-008

A “proof” that the sum of two odds is odd reads: (1) Assume odd m, n with m + n odd. (2) Write m = 2j+1, n = 2k+1. (3) T…

Identifying errors in proofs

Open TMUA-P2-ERR1-008 →

TMUA-P2-PRF3-006 — Paper P2 · Proof by contradiction · Level 3

Options

Keep practising

Master Proof by contradiction

TMUA-P2-PRF3-006 — Paper P2 · Proof by contradiction · Level 3

Options

Keep practising

Master Proof by contradiction