Three statements P, Q, R satisfy P ⇒ Q, Q ⇒ R and R ⇒ P. A study group offers five verdicts about the trio. Which is correct?

Worked solution for TMUA-P2-PRF5-011 is revealed on this page behind the solution details.

Why isn't the answer to TMUA-P2-PRF5-011 shown upfront?

The options are lettered without marking the correct one so you attempt TMUA-P2-PRF5-011 first — TMUA questions reward spotting the trap before you check the worked solution below.

Do I need a calculator for TMUA-P2-PRF5-011?

No — calculators are banned in the TMUA. TMUA-P2-PRF5-011 is set for a no-calculator target of about 270 seconds; the worked solution shows the fastest by-hand route.

Which TMUA specification point is TMUA-P2-PRF5-011?

TMUA-P2-PRF5-011 is Paper P2 · Following proofs (specification code PRF5), level 3. Drill more of the same topic from this page.

TMUA-P2-PRF5-011 — Paper P2 · Following proofs · Level 3

P2PRF5Level 3~270s target

Spec: PRF5 · No calculator · ~270s

Three statements

P

Q

R

satisfy

P \Rightarrow Q

Q \Rightarrow R

and

R \Rightarrow P

. A study group offers five verdicts about the trio. Which is correct?

Options

Only $P \Rightarrow R$ follows; the reverse directions need separate proof.
The three arrows are inconsistent, so the hypotheses are contradictory.
$P$ , $Q$ and $R$ are pairwise equivalent: each implies each of the others.
Nothing follows without checking concrete cases.

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: chain-transitivity, iff-split. Name the trap first, work the stem by hand, then open the solution below.

Show solution

The arrows form a directed cycle, so any statement reaches any other: e.g.

Q \Rightarrow P

via

Q \Rightarrow R \Rightarrow P

, and

R \Rightarrow Q

via

R \Rightarrow P \Rightarrow Q

. Hence

P \iff Q \iff R

. Trap: reading only the forward chain

P \Rightarrow Q \Rightarrow R

and missing that the closing arrow

R \Rightarrow P

turns the chain into a cycle of equivalences.

Keep practising

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Browse more

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Options

Only

P \Rightarrow R

follows; the reverse directions need separate proof.

The three arrows are inconsistent, so the hypotheses are contradictory.

P

Q

and

R

are pairwise equivalent: each implies each of the others.

Nothing follows without checking concrete cases.

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

Keep practising

Same topic · same level

P2PRF5Level 3

TMUA-P2-PRF5-005

A revision guide prints a short proof without a theorem statement: suppose x and y are rational, write x = a/b and y = c…

Following proofs

Open TMUA-P2-PRF5-005 →

Same trap: iff-split

P2ARG1Level 1

TMUA-P2-ARG1-002

In mathematics “ P or Q ” uses the inclusive or. The statement “ P or Q ” is FALSE in exactly which case?

Statements and connectives

Open TMUA-P2-ARG1-002 →

TMUA-P2-PRF5-011 — Paper P2 · Following proofs · Level 3

Options

Keep practising

Master Following proofs

TMUA-P2-PRF5-011 — Paper P2 · Following proofs · Level 3

Options

Keep practising

Master Following proofs