Worked examples · Sheet 2

Necessary and sufficient conditions

Theory used: Implication Necessary condition and sufficient condition

Exercise 1

Determine whether, for real $x$ , the condition " $x > 3$ " is necessary, sufficient, or necessary and sufficient for " $x > 1$ ".

x > 3 ⟹ x > 1 true

x > 1 ⟹ x > 3 false: for x = 2 one has 2 > 1 and 2 \leq 3

The condition " $x > 3$ " is sufficient and not necessary for " $x > 1$ ".

The condition under examination is sufficient when the implication from the condition to the target is true, and it is necessary when the implication in the opposite direction, from the target to the condition, is true.

The first implication is true: from $x > 3$ and $3 > 1$ it follows, by the transitive property of inequality, that $x > 1$ . Therefore the condition is sufficient. The second implication is false, because $x = 2$ satisfies $x > 1$ while the comparison $2 \leq 3$ negates $x > 3$ ; the value $x = 2$ is therefore a counterexample. The condition is hence sufficient and not necessary.

Exercise 2

Determine whether, for real $x$ , the condition " $x^{2} = 9$ " is necessary, sufficient, or necessary and sufficient for " $x = 3$ ".

x = 3 ⟹ x^{2} = 9 true

x^{2} = 9 ⟹ x = 3 false: for x = - 3 one has (- 3)^{2} = 9 and - 3 \neq = 3

The condition " $x^{2} = 9$ " is necessary and not sufficient for " $x = 3$ ".

For necessity one looks at the implication from " $x = 3$ " to " $x^{2} = 9$ ": substituting $x = 3$ gives $3^{2} = 9$ , therefore the implication is true and the condition is necessary, because whenever $x = 3$ holds, $x^{2} = 9$ also holds.

For sufficiency one looks at the opposite implication: from $x^{2} = 9$ follow the two possibilities $x = 3$ and $x = - 3$ , because the equation $x^{2} = 9$ has two real solutions. The value $x = - 3$ satisfies $x^{2} = 9$ and negates $x = 3$ , therefore the implication is false. The condition is hence necessary and not sufficient.

Exercise 3

Determine whether, for real $x$ , the condition " $x^{3} \geq 8$ " is necessary, sufficient, or necessary and sufficient for " $x \geq 2$ ".

x \geq 2 ⟹ x^{3} \geq 8 true

x^{3} \geq 8 ⟹ x \geq 2 true (by contrapositive)

x \geq 2 ⟺ x^{3} \geq 8

The condition " $x^{3} \geq 8$ " is necessary and sufficient for " $x \geq 2$ ".

The first implication is proved by multiplying inequalities: from $x \geq 2$ , with both sides positive, it follows that $x^{2} \geq 4$ and, multiplying again by $x \geq 2$ , it follows that $x^{3} \geq 8$ . Therefore the condition is necessary, because it is true whenever $x \geq 2$ is true.

The second implication is proved by contrapositive: from $x < 2$ it follows that $x^{3} < 8$ . There are two cases. If $x \leq 0$ , then $x^{3} \leq 0 < 8$ , because the cube preserves the sign. If $0 < x < 2$ , multiplying $x < 2$ by $x > 0$ gives $x^{2} < 2 x < 4$ and, multiplying again by $x$ , gives $x^{3} < 4 x < 8$ . In both cases $x^{3} < 8$ holds, therefore the contrapositive is true and so is the implication. Both implications hold, hence the two conditions are equivalent and each is necessary and sufficient for the other.