TMUA Worked Examples

What you will 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 this course.

Example 1

Given exactly one of the following statements is true for all real $x$, regarding a real-valued function $f$ and a real number $a$, which is it?

• A. $f(x) \leq 0$ if $x > a$.
• B. $x \leq a$ is equivalent to $f(x) > 0$.
• C. $f(x) > 0$ only if $x\leq a$.
• D. $x>a$ and $f(x) > 0$ cannot both be true.
• E. $f(x)\leq 0$ is a necessary condition for $x>a$.
• F. $f(x) > 0$ is a sufficient condition for $x\leq a$.
• G. $f(x) < 0$ and $x\leq a$ cannot both be true.
• H. $x > a$ implies $f(x) \leq 0$.

Solution

Rewrite each statement in the form: condition on $x$ and $a$ implies condition on $f(x)$.

• A. $x > a \Rightarrow f(x) \leq 0$.
• B. $x > a \Leftrightarrow f(x) \leq 0$.
• C. $x > a \Rightarrow f(x) \leq 0$.
• D. $x > a \Rightarrow f(x) \leq 0$.
• E. $x > a \Rightarrow f(x) \leq 0$.
• F. $x > a \Rightarrow f(x) \leq 0$.
• G. $x \leq a \Rightarrow f(x) \geq 0$.
• H. $x > a \Rightarrow f(x) \leq 0$.

Therefore all except B and G are equivalent. Since exactly one statement is true, this group of equivalent statements must be false.

Also, B implies A, so B cannot be the unique true statement.

Thus, if exactly one statement is true, it must be G.

Example 2

Suppose $P$ is a statement that depends on $x$. Given exactly one of the following is true about $P$, which is it?

• A. If $|x| < 1$, then $P$ is true.
• B. $P$ is false and $|x| \leq 1$ cannot both be true.
• C. $0.5< x < 2.5$ is necessary for $P$ to be true.
• D. $|x| < 2$ only if $P$ is true.
• E. $0.5< x < 2.5$ if $P$ is true.
• F. $P$ is false if and only if $|x|\geq 1$.

Solution.

Rewrite each statement with the condition on $x$ first, using the contrapositive where needed.

• A. $|x| < 1 \Rightarrow P$.
• B. $|x| \leq 1 \Rightarrow P$.
• C. $x \leq 0.5 \text{ or } x \geq 2.5 \Rightarrow \neg P$.
• D. $|x| < 2 \Rightarrow P$.
• E. $x \leq 0.5 \text{ or } x \geq 2.5 \Rightarrow \neg P$.
• F. $|x| \geq 1 \Leftrightarrow \neg P$.

Statements C and E are equivalent, so neither can be the only true statement.

Statement D implies B, since $|x| \leq 1$ is contained inside $|x| < 2$. Another way to understand this is: D says whenever $x$ is a number between -2 and 2, $P$ is true, given this is the case, B is also true, hence D implies B.

Statement B implies A, since $|x| < 1$ is contained inside $|x| \leq 1$. Therefore B and D cannot be the only true statement.

Statement F implies $|x| < 1 \Rightarrow P$, so F implies A. Therefore F cannot be the only true statement.

Thus the only statement which can be true by itself is A.

The answer is A.

Example 3

Let $f$ be a differentiable real-valued function.

We say that $f$ is positively increasing if and only if, for every real number $a$, there exists a real number $h>0$ such that, for every real number $x$ with $a-h\leq x\leq a+h$, if $f(x)>0$, then $f'(x)>0$.

Which of the following means that $f$ is not positively increasing?

Choices

• A. For every real number $a$, for every real number $h>0$, there exists a real number $x$ with $a-h\leq x\leq a+h$ such that $f(x)>0$ and $f'(x)\leq 0$.

• B. There exists a real number $a$ such that there exists a real number $h>0$ such that, for every real number $x$ with $a-h\leq x\leq a+h$, if $f(x)>0$, then $f'(x)\leq 0$.

• C. There exists a real number $a$ such that, for every real number $h>0$, there exists a real number $x$ with $a-h\leq x\leq a+h$ such that $f(x)>0$ and $f'(x)\leq 0$.

• D. There exists a real number $a$ such that, for every real number $h>0$, for every real number $x$ with $a-h\leq x\leq a+h$, we have $f(x)>0$ and $f'(x)\leq 0$.

• E. There exists a real number $a$ such that, for every real number $h>0$, there exists a real number $x$ with $x<a-h$ or $x>a+h$ such that $f(x)>0$ and $f'(x)\leq 0$.

• F. There exists a real number $a$ such that, for every real number $h>0$, there exists a real number $x$ with $a-h\leq x\leq a+h$ such that $f(x)\leq 0$ and $f'(x)\leq 0$.

Solution.

$f$ is not positively increasing is the negation of $f$ is positively increasing. So we just follow our negation rules: reverse each quantifier from the outermost quantifier, and negate the inner statement. Do this recursively.

So we first reverse the outermost quantifier. The negation of "for every real number $a$" is "there exists a real number $a$".

Next, reverse the next quantifier. The negation of "there exists a real number $h>0$" is "for every real number $h>0$".

Next, reverse the next quantifier. The negation of "for every real number $x$ with $a-h\leq x\leq a+h$" is "there exists a real number $x$ with $a-h\leq x\leq a+h$".

Finally, negate the inner implication:

The negation of this is:

Therefore $f$ is not positively increasing means:

There exists $a$ such that for every $h>0$, there exists $x$ with $a-h\leq x\leq a+h$ such that $f(x)>0$ and $f'(x)\leq 0$.

This is statement C, so the answer is C.

Worksheets

Put the whole toolkit to work on more TMUA-style reasoning questions — implication, necessary and sufficient conditions, negation, quantifiers, the contrapositive, and proof by contradictions. You may find this last worksheet particularly interesting!

• Worksheet 1 — Questions · Solutions

Where to go from here

That completes the course, well done! You've now got the full toolkit: implication and its many phrasings, necessary and sufficient conditions, negation and counterexamples, 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. Best of luck!