You want to prove that the additive identity in a group is unique. You have already shown that 0 satisfies the identity property (0 + a = a for all a). What is the correct next step in a uniqueness proof?
AShow by construction that no other element can satisfy the identity property
BAssume there is a second element 0' that also satisfies the identity property, then derive 0 = 0'
CUse induction to eliminate all candidate identity elements one by one
DArgue that the construction of 0 was the only logically possible one, so uniqueness follows from existence
The canonical uniqueness proof template is: assume two objects a and b both satisfy the property P, then prove a = b. Here, you let 0' also be an additive identity. Since 0 is an identity: 0 + 0' = 0'. Since 0' is an identity: 0 + 0' = 0. Therefore 0 = 0'. Both hypotheses were used simultaneously — that is the structural signature of a correct uniqueness proof.
Question 2 Multiple Choice
A student proves constructively that √2 exists as a real number and concludes: 'Since I've shown it exists, it must be unique — there's only one real square root of 2 equal to √2.' Is this reasoning valid?
AYes — existence and uniqueness are the same claim for irrational numbers
BNo — existence and uniqueness are separate claims; a separate argument must show no other value satisfies the same defining property
CYes — constructive existence proofs automatically establish uniqueness because the construction is specific
DNo — uniqueness proofs only apply to algebraic objects, not real numbers
Existence and uniqueness are logically independent claims. 'An x with property P exists' and 'at most one x with property P exists' must be argued separately. A constructive proof exhibits one object satisfying P but does nothing to rule out others. The uniqueness argument must go further: assume two objects both satisfy P, and derive they are equal. Conflating these two steps is the most common error when working with ∃!x P(x) claims.
Question 3 True / False
In a uniqueness proof, you use both hypotheses simultaneously — the fact that a satisfies P and the fact that b satisfies P — to derive a = b.
TTrue
FFalse
Answer: True
This simultaneous use of both conditions is the structural signature of a correct uniqueness proof. In the additive identity example: 'e is an identity' gives e + e' = e'; 'e' is an identity' gives e + e' = e. Both hypotheses are needed to conclude e = e'. If you only used one, you couldn't establish equality. This is also why the template is stated with exactly two objects — you need two to derive equality, and two is sufficient.
Question 4 True / False
Proving that any two elements satisfying property P should be equal primarily establishes pairwise uniqueness — a separate argument is still needed to rule out three or more distinct elements most satisfying P.
TTrue
FFalse
Answer: False
Pairwise equality is sufficient for full uniqueness. If a = b whenever any two objects both satisfy P, then for any third object c also satisfying P, applying the same argument to (a, c) gives a = c, and to (b, c) gives b = c. So all three collapse to the same value. The pairwise template captures the general case: you cannot have two distinct elements both satisfying P, let alone three, four, or any number.
Question 5 Short Answer
Why is the proof template 'assume a and b both satisfy P, then prove a = b' sufficient to establish that at most one element satisfies P?
Think about your answer, then reveal below.
Model answer: If any two elements satisfying P must be equal, then you cannot have two distinct elements satisfying P. For any further element c satisfying P, the same argument applied to (a, c) gives a = c, and to (b, c) gives b = c — so all elements satisfying P are identical. Pairwise equality is the minimal, general structure: proving it for any two objects covers all possible cases.
The elegance of the template is that 'at most two objects' is all you ever need to consider. The proof structure is minimal in the same way that proving a statement for arbitrary n is more general than listing cases — assuming two arbitrary objects satisfying P and showing they're equal captures the full content of uniqueness.