6 mistake patterns students fall for. 2 high-frequency traps appear in almost every exam.
Missing reflexive check for all elements
Very CommonCONCEPT
Students check reflexivity for only some elements of the set and conclude the relation is reflexive, missing elements where (a, a) is absent.
Why: Reflexivity requires (a, a) for EVERY element a in the set. Students check a few elements and assume it holds for all, or confuse 'some pairs are reflexive' with 'the relation is reflexive'.
WRONG:On A={1,2,3}, relation R={(1,1),(2,2),(1,3)}: claiming R is reflexive because (1,1) and (2,2) are present, missing that (3,3) is absent.
RIGHT:R is NOT reflexive because (3,3)∈/R. Reflexivity requires (a,a)∈R for every a∈A, which means we need (1,1), (2,2), AND (3,3).
Students mix up the definitions of injective and surjective, checking the wrong property when asked about one of them.
Why: Both concepts deal with the relationship between domain and codomain, and students memorize them as a pair without clearly distinguishing which is which.
WRONG:To check if f(x)=x2 on R→R is one-one, checking if every positive real has a preimage (that is onto, not one-one).
RIGHT:One-one: f(a)=f(b)⇒a=b. Here f(2)=f(−2)=4, so f is NOT one-one. Onto: is every y∈R achieved? No, f(x)=x2≥0, so f is NOT onto either.
Students write A is a proper subset of B when A could equal B, or use subset notation when proper subset is required.
Why: The symbols for subset and proper subset look similar, and textbooks sometimes use them inconsistently. Students forget that a set is a subset of itself but not a proper subset of itself.
WRONG:Claiming {1,2} is a proper subset of {1,2}.
RIGHT:{1,2}⊆{1,2} is true (subset), but {1,2}⊂{1,2} as a proper subset is false. A proper subset must be strictly smaller.
Students compute the domain of g(f(x)) by only considering where the final expression is defined, ignoring the constraint that f(x) must lie in the domain of g.
Why: Students substitute and simplify first, then find the domain of the simplified expression. But cancellations or simplifications can hide domain restrictions.
WRONG:For f(x)=x2 and g(x)=x, writing domain of g(f(x))=x2=∣x∣ as all of R, without noting f(x)=x2≥0 so g(f(x)) is indeed defined for all x. But reversing: domain of f(g(x))=(x)2=x seems to be all of R, but g(x)=x requires x≥0.
RIGHT:Domain of f∘g: first x must be in dom(g), so x≥0. Then g(x)=x≥0 which is always in dom(f). So domain of f∘g=[0,∞), not R.
Forgetting all three conditions for equivalence relation
OccasionalCONCEPT
Students verify only two of the three properties (reflexive, symmetric, transitive) and conclude the relation is an equivalence relation.
Why: After checking two properties, students assume the third follows automatically. In reality, there exist relations that are symmetric and transitive but not reflexive, or reflexive and symmetric but not transitive.
WRONG:Relation R={(1,1),(1,2),(2,1),(2,2)} on {1,2,3}: claiming it is an equivalence relation because it is reflexive for 1 and 2, symmetric, and transitive, forgetting that (3,3)∈/R.
RIGHT:R is symmetric and transitive, but NOT reflexive (missing (3,3)). Therefore R is NOT an equivalence relation. All three properties must hold for every relevant element.