Definitions as Proof Tools

IB Year 5 | Grade 11

Translate words into algebra

Definitions are not background vocabulary; they are the legal algebraic moves in a proof. Replace a property with its defining form as early as possible, then expose the same form for the conclusion.

Property	Operational definition	Form to expose at the end
\(d\) divides \(a\)	\(a=dq\) for some \(q\in\mathbb Z\)	required expression \(=d\times\text{integer}\)
\(n\) is even	\(n=2q\) for some \(q\in\mathbb Z\)	required expression \(=2\times\text{integer}\)
\(n\) is odd	\(n=2q+1\) for some \(q\in\mathbb Z\)	required expression \(=2\times\text{integer}+1\)
\(x\) is rational	\(x=\dfrac{p}{q}\), where \(p,q\in\mathbb Z\) and \(q\ne0\)	a quotient of integers with non-zero denominator
\(p\) is prime	\(p\ge2\) and its only positive divisors are \(1\) and \(p\)	use a prime-divisibility fact only after naming it

The integer witness matters

A phrase such as “for some integer \(q\)” is essential. Without it, \(a=dq\) is merely an equation with an unspecified real number and does not establish divisibility.

Close the definition loop

To prove that an expression is even, do not finish at “it looks even”. Factor out \(2\) and show that the remaining factor is an integer. The same rule applies to any divisibility claim.

Representation choices

For parity, decide whether one representation handles the whole proof or whether even/odd cases are cleaner.
For rationality, check the denominator is non-zero and that every numerator and denominator used is an integer.
For irrationality by contradiction, choose the fraction in lowest terms before deriving a common factor.
For primality, do not use the false converse “if a prime divides a product, it divides both factors”.

Class check

Suppose \(6\mid a\) and \(6\mid b\). Write the two integer witnesses, then predict the factorised form needed to prove \(6\mid(4a-7b)\).

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