Statements, implication, converse and equivalence | Year 11

01 / LEARN

The idea, step by step

Read the direction of the claim

A mathematical statement has a truth value once its variables and domain are specified. In an implication $P \Rightarrow Q$ , $P$ is the assumption and $Q$ is what must follow whenever $P$ holds. The implication does not say that $P$ always happens.

For example, on the integers, being a multiple of four guarantees being even. It is enough to begin with an arbitrary multiple of four and show that it is twice an integer. Testing a few multiples illustrates the idea but does not prove the universal claim.

Reverse carefully

The converse of $P \Rightarrow Q$ is $Q \Rightarrow P$ . A converse needs its own argument and can be false even when the original statement is true. A counterexample must satisfy the converse assumption while failing its conclusion.

Write $P \Leftrightarrow Q$ only after both directions are established. This means that $P$ is both sufficient and necessary for $Q$ . State the domain: a claim about positive real numbers may fail for negative numbers.

A common trap

Using the same proof to claim the converse automatically.

Instead: Write the converse separately and either prove it or produce a counterexample.

02 / WORKED EXAMPLES

See the thinking

Example 1

A true implication and a false converse

For an integer $n$ , consider: if $n$ is divisible by 4, then $n$ is even. Prove it and test its converse.

Write $n = 4 k$ for an integer $k$ .
Then $n = 2 (2 k)$ , so $n$ is even.
The converse would claim every even integer is divisible by 4. The integer 6 is even but not divisible by 4.

The implication is true; the converse is false.

Example 2

An equivalence

For an integer $n$ , prove that $n$ is odd if and only if $n + 1$ is even.

If $n = 2 k + 1$ , then $n + 1 = 2 (k + 1)$ is even.
Conversely, if $n + 1 = 2 k$ , then $n = 2 (k - 1) + 1$ is odd.
Both directions use integer parameters, so both are valid on the stated domain.

n

odd

\Leftrightarrow n + 1

even.

Check your understanding

State the converse of: if $x > 3$ , then $x > 1$ .

Show answer

If $x > 1$ , then $x > 3$ .

The converse is false, for example at $x = 2$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

State the assumption and conclusion in: if an integer is divisible by 10, it is divisible by 5.
2 marks
Prove that every multiple of 6 is a multiple of 3.
2 marks
Write the converse of the claim in Question 2 and give a counterexample.
2 marks
For real $x$ , test both directions of $x = 2 \Rightarrow x^{2} = 4$ .
2 marks
For real $x$ , prove $x = 0 \Leftrightarrow x^{2} = 0$ .
3 marks
Is “ $n$ is a multiple of 12” sufficient, necessary, both or neither for “ $n$ is a multiple of 4”, on the integers?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

State the assumption and conclusion in: if an integer is divisible by 10, it is divisible by 5.
2 marks
The condition after “if” is assumed. The statement after “then” must be established.
Assumption: divisible by 10. Conclusion: divisible by 5.
Prove that every multiple of 6 is a multiple of 3.
2 marks
Write $n = 6 k = 3 (2 k)$ for $k \in ℤ$ . Since $2 k$ is an integer, $3 ∣ n$ .
True.
Write the converse of the claim in Question 2 and give a counterexample.
2 marks
3 is divisible by 3 but not by 6.
Every multiple of 3 is a multiple of 6; false at 3.
For real $x$ , test both directions of $x = 2 \Rightarrow x^{2} = 4$ .
2 marks
Substitution gives $2^{2} = 4$ .
For the converse, $x = - 2$ has $x^{2} = 4$ but $x \neq 2$ .
The original is true; the converse is false.
For real $x$ , prove $x = 0 \Leftrightarrow x^{2} = 0$ .
3 marks
If $x = 0$ , squaring gives 0.
If $x^{2} = 0$ , then $x \cdot x = 0$ , so the zero-product property gives $x = 0$ .
Both directions hold.
Is “ $n$ is a multiple of 12” sufficient, necessary, both or neither for “ $n$ is a multiple of 4”, on the integers?
3 marks
Every $12 k = 4 (3 k)$ is divisible by 4.
The number 4 is divisible by 4 but not by 12.
Sufficient but not necessary.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: geometric reasoning and real-number/proof material, regrouped into a school chapter; all exercises are original.

Official curriculum source ↗

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