Why isn't the answer to TMUA-P2-ERR1-016 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2-ERR1-016 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2-ERR1-016?

No — calculators are banned in the TMUA. TMUA-P2-ERR1-016 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-ERR1-016?

TMUA-P2-ERR1-016 is Paper P2 · Identifying errors in proofs (specification code ERR1), level 3. Drill more of the same topic from this page.

TMUA-P2-ERR1-016 — Paper P2 · Identifying errors in proofs · Level 3

P2ERR1Level 3~280s target

Spec: ERR1 · No calculator · ~280s

A student imitates the

\sqrt{2}

proof: (1) Assume

\sqrt{2} = p/q

for some integers

p

q

with

q \ne 0

. (2) Then

2q^2 = p^2

, so

p^2

is even, so

p

is even, say

p = 2k

. (3) Then

2q^2 = 4k^2

, so

q^2 = 2k^2

, so

q

is even. (4) Both

p

and

q

are even, a contradiction. Hence

\sqrt{2}

is irrational. Which line contains the first fault?

Options

Line (1): it omits “in lowest terms”, so “both even” in line (4) is no contradiction (e.g. $4/6$ has both terms even yet is a valid fraction).
Line (2): from $p^2$ even it does not follow that $p$ is even.
Line (3): the algebra $2q^2 = 4k^2$ giving $q^2 = 2k^2$ is wrong.
There is no fault; the proof is complete as written.

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: line-fault-missing-assumption, contradiction-setup. Name the trap first, work the stem by hand, then open the solution below.

Show solution

Without the lowest-terms condition, deducing that

p

and

q

share the factor

2

contradicts nothing — reducible fractions exist. The classic proof assumes

p/q

in lowest terms, so a common factor of

2

is impossible. Trap: the algebra and the “even square gives even root” lemma are both correct; the fault is the missing quantifier-free setup in line (1), not the computations.

Keep practising

Same topic · same level
P2ERR1Level 3
TMUA-P2-ERR1-009
A six-line algebra proof submitted for a scholarship problem claims to show that x = 5 follows from the equation (x - 2)…
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Open TMUA-P2-ERR1-009 →
Same trap: contradiction-setup
P2PRF3Level 1
TMUA-P2-PRF3-007
A proof by contradiction that √2 is irrational must begin with a correct assumption for contradiction. Which opening is…
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Open TMUA-P2-PRF3-007 →
More in this topic
P2ERR1Level 3
TMUA-P2-ERR1-010
A flawed lemma in a student journal claims that from the true statement “For every real x there exists a real y with y >…
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Open TMUA-P2-ERR1-010 →

Browse more

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Options

Line (1): it omits “in lowest terms”, so “both even” in line (4) is no contradiction (e.g.

4/6

has both terms even yet is a valid fraction).

Line (2): from

p^2

even it does not follow that

p

is even.

Line (3): the algebra

2q^2 = 4k^2

giving

q^2 = 2k^2

is wrong.

There is no fault; the proof is complete as written.

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

Keep practising

Same topic · same level

P2ERR1Level 3

TMUA-P2-ERR1-009

A six-line algebra proof submitted for a scholarship problem claims to show that x = 5 follows from the equation (x - 2)…

Identifying errors in proofs

Open TMUA-P2-ERR1-009 →

Same trap: contradiction-setup

P2PRF3Level 1

TMUA-P2-PRF3-007

A proof by contradiction that √2 is irrational must begin with a correct assumption for contradiction. Which opening is…

Proof by contradiction

Open TMUA-P2-PRF3-007 →

TMUA-P2-ERR1-016 — Paper P2 · Identifying errors in proofs · Level 3

Options

Keep practising

Master Identifying errors in proofs

TMUA-P2-ERR1-016 — Paper P2 · Identifying errors in proofs · Level 3

Options

Keep practising

Master Identifying errors in proofs