Paper 2 mathematical reasoning

For an integer $n$, let $P$ be '$4$ divides $n$' and $Q$ be '$2$ divides $n$. Every multiple of $4$ is even.

The condition $P$ is sufficient for $Q$, while $Q$ is necessary for $P$. The converse fails at $n=2$.

Why this works: The implication fixes both labels. One allowed value that breaks the converse is enough to rule out equivalence.

Why the tempting route fails: Saying that evenness is sufficient for divisibility by $4$ reverses the implication and ignores the case $n=2$.

Claim: for every integer $n$, the number $n^2+n+1$ is even. Test a small allowed value.

$3$ is odd, so the universal claim is false. There is no need to classify every integer after one valid counterexample.

Why this works: The chosen value is an integer in the stated domain and produces a result that directly contradicts the claim.

Why the tempting route fails: Checking only $n=0$ or several even values can suggest a pattern without proving it for every integer.

A purported proof starts with $ab=ac$, cancels $a$ and concludes $b=c$. The cancellation is valid only when $a$ is not zero.

For $a=0$, both sides are zero for every pair $b,c$. The first error is the unqualified cancellation, not the final numerical contradiction.

Why this works: The audit identifies the earliest line that loses a condition. Later lines cannot repair a broken implication.

Why the tempting route fails: Checking only the final answer misses the point of a first-error question and can overlook a hidden zero case.