Why isn't the answer to TMUA-P2Z-PRF3-003 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2Z-PRF3-003 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2Z-PRF3-003?

No — calculators are banned in the TMUA. TMUA-P2Z-PRF3-003 is set for a no-calculator target of about 200 seconds; the worked solution shows the fastest by-hand route.

Which TMUA specification point is TMUA-P2Z-PRF3-003?

TMUA-P2Z-PRF3-003 is Paper P2 · Proof by contradiction (specification code PRF3), level 2. Drill more of the same topic from this page.

TMUA-P2Z-PRF3-003 — Paper P2 · Proof by contradiction · Level 2

P2PRF3Level 2~200s target

Spec: PRF3 · No calculator · ~200s

To prove there is no largest rational number strictly less than

2

, a student assumes the opposite: some largest such rational

r < 2

exists. Which completion gives a valid contradiction?

Options

Then $r+1$ is a rational larger than $r$ — contradiction.
Then $r = 2$ is the largest rational less than $2$ — contradiction.
Then $m = \frac{r+2}{2}$ is rational, satisfies $r < m < 2$ , and contradicts the maximality of $r$ .
Then there are infinitely many rationals, so no largest one exists — contradiction.
Then $r^2$ is rational and larger than $r$ — contradiction.

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, quantifier-negation-flip. Name the trap first, work the stem by hand, then open the solution below.

Show solution

Assume a largest rational

r < 2

exists; then

m = \frac{r+2}{2}

is rational with

r < m < 2

(averaging pulls strictly inside the bound), contradicting maximality. Trap:

r+1

need not stay below

2

(e.g.

r = 1.9

gives

2.9

), so it cannot serve as the contradicting witness; the midpoint is chosen precisely because it stays inside the bound.

Keep practising

Same topic · same level
P2PRF3Level 2
TMUA-P2-PRF3-003
A proof by contradiction that “if n^2 is even then n is even” starts below. Which opening is correct?
Proof by contradiction
Open TMUA-P2-PRF3-003 →
Same trap: quantifier-negation-flip
P2ARG3Level 1
TMUA-P2-ARG3-001
What does the statement “For all integers n, n^2 0 ” assert?
Quantifiers
Open TMUA-P2-ARG3-001 →
Step up a 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 →

Browse more

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Options

Then

r+1

is a rational larger than

r

— contradiction.

Then

r = 2

is the largest rational less than

2

— contradiction.

Then

m = \frac{r+2}{2}

is rational, satisfies

r < m < 2

, and contradicts the maximality of

r

Then there are infinitely many rationals, so no largest one exists — contradiction.

Then

r^2

is rational and larger than

r

— contradiction.

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

Keep practising

Same topic · same level

P2PRF3Level 2

TMUA-P2-PRF3-003

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

Proof by contradiction

Open TMUA-P2-PRF3-003 →

Same trap: quantifier-negation-flip

P2ARG3Level 1

TMUA-P2-ARG3-001

What does the statement “For all integers n, n^2 0 ” assert?

Quantifiers

Open TMUA-P2-ARG3-001 →

Step up a 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 →

TMUA-P2Z-PRF3-003 — Paper P2 · Proof by contradiction · Level 2

Options

Keep practising

Master Proof by contradiction

TMUA-P2Z-PRF3-003 — Paper P2 · Proof by contradiction · Level 2

Options

Keep practising

Master Proof by contradiction