Logic & Proof

10 methods · 48 questions

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A multi-line proof or solution is offered and you must name the first faulty step. Do not re-derive from scratch: read down the lines and stop at the first operation that is not a genuine equivalence, then confirm by testing a value that exposes it. A handful of moves cause almost every TMUA trap.

Trigger: a numbered argument with options "the first error is on line ...".

Instances:
(i) squaring both sides, or replacing a value by a square root, can add or drop a sign and create extraneous roots;
(ii) dividing both sides by an expression that might be zero silently loses a solution;
(iii) applying a decreasing function (such as a log with base below 1) without reversing the inequality;
(iv) taking only the positive square root and forgetting the negative branch;
(v) reading "product equals a constant" as forcing each factor to a value.

Linked questions (3)

TMUA 2017 Paper 2 · Q18

Consider the following problem: Solve the inequality

(\frac{1}{4})^{n} < (\frac{1}{32})^{10}

, where

n

is a positive integer. A student produces the following argument:

(\frac{1}{4})^{n} < (\frac{1}{32})^{10} (I)

lo g_{\frac{1}{2}} (\frac{1}{4})^{n} < lo g_{\frac{1}{2}} (\frac{1}{32})^{10} (II)

n lo g_{\frac{1}{2}} (\frac{1}{4}) < 10 lo g_{\frac{1}{2}} (\frac{1}{32}) (III)

n < \frac{10 lo g _{\frac{1}{2}} ( \frac{1}{32} )}{lo g _{\frac{1}{2}} ( \frac{1}{4} )} (IV)

n < \frac{10 \times 5}{2} = 25 (V)

1 ⩽ n ⩽ 24

Which step (if any) in the argument is invalid?

AThere are no invalid steps; the argument is correct
BOnly step (I) is invalid; the rest are correct
COnly step (II) is invalid; the rest are correct
DOnly step (III) is invalid; the rest are correct
EOnly step (IV) is invalid; the rest are correct
FOnly step (V) is invalid; the rest are correct

TMUA 2018 Paper 2 · Q13

The following is an attempted proof of the conjecture:

[[B]] i f [[/ B]] tan θ > 0, [[B]] t h e n [[/ B]] sin θ + cos θ > 1.

Suppose

tan θ > 0

, so in particular

cos θ \neq = 0

.

Since

tan θ = \frac{sin θ}{cos θ}

, then

sin θ cos θ = tan θ cos^{2} θ > 0

. (I)

It follows that

1 + 2 sin θ cos θ > 1

. (II)

Therefore

sin^{2} θ + 2 sin θ cos θ + cos^{2} θ > 1

, (III)

which factorises to give

(sin θ + cos θ)^{2} > 1

. (IV)

Therefore

sin θ + cos θ > 1

. (V)

Which one of the following is the case?

AThe proof is correct.
BThe proof is incorrect, and the first error occurs in line (I).
CThe proof is incorrect, and the first error occurs in line (II).
DThe proof is incorrect, and the first error occurs in line (III).
EThe proof is incorrect, and the first error occurs in line (IV).
FThe proof is incorrect, and the first error occurs in line (V).

TMUA 2018 Paper 2 · Q9

Consider the following attempt to solve the equation

4 x 2 x - 1 = 10 x - 5

4 x 2 x - 1 = 10 x - 5 ↓ (I)

4 x 2 x - 1 = 5 (2 x - 1) ↓ (II)

16 x^{2} (2 x - 1) = 25 (2 x - 1)^{2} ↓ (III)

16 x^{2} = 25 (2 x - 1) ↓ (IV)

16 x^{2} - 50 x + 25 = 0 ↓ (V)

(8 x - 5) (2 x - 5) = 0 ↓ (VI)

The solutions of the original equation are

x = \frac{5}{8}

and

x = \frac{5}{2}

.

Which one of the following is true?

AThe solution is correct.
BOnly one of $x = \frac{5}{8}$ and $x = \frac{5}{2}$ is correct and the error arises as a result of step (II).
COnly one of $x = \frac{5}{8}$ and $x = \frac{5}{2}$ is correct and the error arises as a result of step (III).
DOnly one of $x = \frac{5}{8}$ and $x = \frac{5}{2}$ is correct and the error arises as a result of step (IV).
EThere is another value of $x$ that satisfies the original equation and the error arises as a result of step (II).
FThere is another value of $x$ that satisfies the original equation and the error arises as a result of step (III).
GThere is another value of $x$ that satisfies the original equation and the error arises as a result of step (IV).