TMUA Worked Examples

What you'll get from this section. The whole course put to work. Below are three TMUA-style questions with full solutions, each leaning on a different part of what you've built.

A few habits that make logic questions safer under exam pressure:

• Translate every option back to basics — $\Rightarrow$, necessary, sufficient — before judging it.
• Watch the direction. The converse and the inverse love to pose as the contrapositive.
• The negation of "if $P$ then $Q$" is "$P$ and not $Q$" — not another if–then.
• Check the edge cases, including vacuous ones, before committing.

Example 1 — Necessary and sufficient

Let $n$ be a positive integer. Which of the following is both necessary and sufficient for $n$ to be divisible by $12$?

• A. $n$ is divisible by $6$
• B. $n$ is divisible by $24$
• C. $n$ is divisible by $3$ and by $4$
• D. $n$ is divisible by $2$ and by $3$
• E. $n$ is divisible by $4$

Solution. We need a condition $C$ for which both directions hold: divisible by $12 \Rightarrow C$ (necessary) and $C \Rightarrow$ divisible by $12$ (sufficient).

Test the options. Divisibility by $6$, by $4$, or by "$2$ and $3$" (which is just divisibility by $6$) is necessary — every multiple of $12$ has these — but not sufficient: $n = 6$ is divisible by $6$, by $2$ and by $3$, yet not by $12$. So A, D, E fail sufficiency.

Divisibility by $24$ is sufficient (every multiple of $24$ is a multiple of $12$) but not necessary: $n = 12$ is divisible by $12$ but not by $24$. So B fails necessity.

That leaves C. Since $3$ and $4$ share no common factor, being divisible by both forces divisibility by their product, $3 \times 4 = 12$ — so the condition is sufficient. And every multiple of $12$ is certainly divisible by $3$ and by $4$ — so it is necessary. Both directions hold.

(The trap is D: $2$ and $3$ only force divisibility by $6$, because $2$ and $3$ multiply to $6$, not $12$. You need two factors whose product is $12$.)

The answer is C.

Example 2 — Negating a quantified statement

Consider the statement:

For every integer $n$, if $n^2$ is odd then $n$ is odd.

Which of the following is its negation?

• A. For every integer $n$, if $n^2$ is odd then $n$ is even.
• B. There exists an integer $n$ such that $n^2$ is odd and $n$ is even.
• C. There exists an integer $n$ such that if $n^2$ is odd then $n$ is even.
• D. For every integer $n$, $n^2$ is odd and $n$ is even.
• E. There exists an integer $n$ such that $n^2$ is even and $n$ is odd.

Solution. Negate in two steps. First the quantifier: the negation of "for every $n$, $\dots$" is "there exists an $n$ such that not $\dots$" — so $\forall$ becomes $\exists$ (Section 3).

Then the inside. The negation of an implication "if $P$ then $Q$" is "$P$ and not $Q$" (Section 7) — not another implication. Here $P$ is "$n^2$ is odd" and $Q$ is "$n$ is odd", so the negation of the inside is "$n^2$ is odd and $n$ is not odd", i.e. "$n^2$ is odd and $n$ is even".

Putting the pieces together: there exists an integer $n$ such that $n^2$ is odd and $n$ is even — which is B.

Why the others fail: C keeps an if–then inside, but the negation of an implication is never an implication. D keeps "for every" instead of switching to "there exists". E negates the wrong parts, swapping which of $n^2$ and $n$ is odd. A negates only the conclusion and leaves $\forall$ and the if–then untouched.

(As an aside: the original statement is actually true — it's the contrapositive of "$n$ even $\Rightarrow n^2$ even" from Section 5 — so its negation B is false. But the question asks only for the form of the negation, and B is that form.)

The answer is B.

Example 3 — Deduction by contrapositive

You are told that both of the following are true:

(i) If a function is differentiable, then it is continuous. (ii) The function $g$ is not continuous.

Which of the following can be validly deduced?

• A. $g$ is differentiable.
• B. $g$ is not differentiable.
• C. $g$ is continuous.
• D. Nothing can be deduced about whether $g$ is differentiable.
• E. $g$ is both differentiable and not continuous.

Solution. Statement (i) is an implication: differentiable $\Rightarrow$ continuous. Its contrapositive (Section 4) is the equivalent statement

Statement (ii) tells us $g$ is not continuous. Feeding that into the contrapositive, we deduce that $g$ is not differentiable.

The others don't follow: C directly contradicts (ii); A and E assert $g$ is differentiable, which the contrapositive rules out; and D is wrong precisely because the contrapositive does let us conclude. This is the difference between a dead end and a valid deduction — running the implication backwards (the converse) would give nothing, but the contrapositive gives everything.

The answer is B.

Where to go from here

That completes the course. You've now got the full toolkit: implication and its many phrasings, necessary and sufficient conditions, negation and counter-examples, the contrapositive, proof by contradiction, equivalence, and truth tables. Every TMUA reasoning question is some combination of these moves — so the most valuable next step is to take that toolkit into full practice papers and watch the same ideas reappear under exam conditions.