1.01bPureCore

Logical Connectives and Mathematical Arguments

A calculation may be correct while the argument around it is false. Learn to read implication, equivalence and negation precisely, so that solving an equation and proving a statement both preserve the logic.

35 min Proof

The big picture

A proof is a sequence of justified implications. The direction matters: every square has four sides, but a shape with four sides need not be a square. The same issue appears when you square an equation, cancel a factor or take a square root. Understanding the language lets you spot extra solutions, missing cases and unjustified conclusions.

What you'll be able to do

Distinguish $P\Rightarrow Q$ from its converse and from $P\Leftrightarrow Q$
Use necessary and sufficient conditions accurately
Negate statements involving all, some, inequalities, and conjunctions
Identify steps that introduce or lose solutions

Implication and the converse

$P\Rightarrow Q$ means that whenever $P$ is true, $Q$ must be true. It makes no claim that $Q$ can happen only through $P$ . The converse is $Q\Rightarrow P$ , and it needs a separate justification.

In $P\Rightarrow Q$ , $P$ is sufficient for $Q$ , while $Q$ is necessary for $P$ . Being divisible by 4 is sufficient for being even; being even is necessary for being divisible by 4.

Example 1Is the converse of “if

n

is a multiple of 6, then

n

is a multiple of 3” true for integers?

1The converse says: if

n

is a multiple of 3, then it is a multiple of 6.

2Choose

n=3

. It satisfies the premise but not the conclusion.

3One counterexample disproves the converse.

AnswerThe original implication is true; its converse is false.

Tip — To disprove an implication, find a case with a true premise and a false conclusion. A case with a false premise tells you nothing.

Equivalence needs both directions

$P\Leftrightarrow Q$ means both $P\Rightarrow Q$ and $Q\Rightarrow P$ . In words: $P$ if and only if $Q$ . Use this symbol only when every step is reversible.

For real numbers, $x=3\Rightarrow x^2=9$ , but $x^2=9\not\Rightarrow x=3$ : $x=-3$ is also possible. The correct equivalence is $x^2=9\Leftrightarrow (x=3\text{ or }x=-3)$ .

Example 1Solve

\sqrt{x+2}=x

and explain why checking is necessary.

1The original equation requires

x\geq0

. Squaring gives

x+2=x^2

2Factorise:

(x-2)(x+1)=0

, so the squared equation has candidates

2

and

-1

3Check

x=2

\sqrt4=2

. Check

x=-1

\sqrt1\neq-1

. Squaring introduced an extra candidate.

AnswerOnly

x=2

solves the original equation.

Negation and the scope of a claim

The negation of “all integers have property $P$ ” is “there exists an integer without property $P$ ”. It is not “no integers have property $P$ ”.

The negation of $x>2$ is $x\leq2$ , including the boundary. The negation of “ $P$ and $Q$ ” is “not $P$ or not $Q$ ”; the negation of “ $P$ or $Q$ ” is “not $P$ and not $Q$ ”. Mathematical “or” normally includes the possibility of both.

In contradiction proofs, negate the entire statement precisely before starting the algebra. In counterexample questions, the example must lie in the domain named by the claim.

Example 1Negate: “for every real

x

x^2>x

”. Then disprove it.

1Negation: there exists a real

x

for which

x^2\leq x

2Take

x=\tfrac12

. Then

x^2=\tfrac14\leq\tfrac12

3The example is real and violates the claimed strict inequality.

AnswerThe universal claim is false.

Audit an algebraic proof

Division by a variable can discard a case. From $x(x-2)=0$ , dividing by $x$ is permitted only if $x\neq0$ . Instead use the zero-product rule to keep both $x=0$ and $x=2$ .

An identity holds throughout its stated domain. Simplifying $\dfrac{x^2-1}{x-1}$ to $x+1$ does not restore the excluded value $x=1$ : the original expression is undefined there.

When asked to show or prove, give the logical links, not only the final expression. State integer assumptions, denominator restrictions and reasons for rejecting candidates.

Example 1A student writes

a=b

, then

a^2=ab

, then

(a-b)(a+b)=b(a-b)

, then

a+b=b

. Locate the flaw.

1Because

a=b

, the factor

a-b

is zero.

2Cancelling

a-b

divides by zero, so that step is invalid.

3All previous equalities may be true; the invalid cancellation breaks the proof.

AnswerThe apparent contradiction comes from division by zero.

Think like an examiner

Use implication arrows in the direction justified by the working.
State restrictions before dividing by an expression that may be zero.
A counterexample must satisfy the premise and domain of the statement.

Common misconceptions

Assuming a true statement has a true converse.

Test or prove the converse separately.

Negating “all” as “none”.

The negation of “all” is “at least one not”.

Key results to remember

P\Leftrightarrow Q\quad\text{means}\quad(P\Rightarrow Q)\text{ and }(Q\Rightarrow P)

Equivalence requires both implications.

Stretch yourself

Assess “for every real $x$ , $x^2>4$ if and only if $x>2$ ”. Correct it.

Hint — Consider negative values of large magnitude.

Key takeaways

An implication does not automatically give its converse.
Equivalence requires two valid directions.
Negation must include the boundary and correct quantifier.
Squaring can add candidates; cancellation can lose cases.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

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