A student argues: 'The Archimedean Property must be added as a separate axiom when defining ℝ, because the completeness axiom only guarantees that bounded sets have suprema — it says nothing about whether infinitely large elements exist.' What is wrong with this argument?
AThe student is correct — completeness and the Archimedean Property are logically independent
BThe Archimedean Property is already built into the definition of the natural numbers, so it requires no axiom at all
CThe Archimedean Property is a theorem: the assumption that ℕ is bounded above in ℝ leads to a contradiction via the least upper bound property
DCompleteness only applies to Cauchy sequences, not to properties of the natural numbers
The Archimedean Property is a theorem, not an axiom — it follows from the completeness of ℝ (the least upper bound property). If ℕ were bounded above, it would have a supremum M by completeness. But then some n ∈ ℕ exceeds M − 1, so n + 1 > M, contradicting M being an upper bound. Therefore ℕ is unbounded, which is precisely the Archimedean Property. The student's error is treating completeness as too weak to imply it — completeness is exactly what rules out infinitely large elements.
Question 2 Multiple Choice
In an ε-δ proof, you need to find N ∈ ℕ such that 1/N < ε for an arbitrary ε > 0. Which fact from the Archimedean Property directly guarantees this choice is always possible?
AThe natural numbers are a subset of the real numbers
BFor any positive real ε, there exists n ∈ ℕ with n > 1/ε, so 1/n < ε
CThe sequence 1/n is bounded below by 0
Dℝ is an ordered field, so division is always well-defined
The Archimedean Property (with a = 1, b = 1/ε) guarantees there exists n ∈ ℕ with n·1 > 1/ε, i.e., n > 1/ε, which gives 1/n < ε. This is the corollary used in virtually every ε-N argument. Options A and C are true but do not imply the existence of such N. Option D guarantees 1/ε is well-defined as a real number but not that a natural number exceeds it — that requires the Archimedean Property.
Question 3 True / False
The Archimedean Property should be included as an explicit axiom when defining the real numbers, because the completeness axiom alone does not prevent ℝ from containing infinitely large elements.
TTrue
FFalse
Answer: False
The Archimedean Property is a theorem proved from completeness, not an additional axiom. The proof: if ℕ were bounded above in ℝ, completeness would give it a supremum M. But then some n > M − 1, so n + 1 > M, contradicting M being an upper bound. Ordered fields that lack the Archimedean Property (containing infinitely large elements) are therefore necessarily incomplete — they violate the least upper bound property. Completeness is sufficient to rule out infinite elements.
Question 4 True / False
In any ordered field that fails the Archimedean Property, there must exist a positive element smaller than 1/n for every natural number n.
TTrue
FFalse
Answer: True
If a field contains an infinitely large element b (one larger than every natural number), then 1/b is a positive element smaller than 1/n for every n ∈ ℕ — a positive infinitesimal. The two pathologies are dual: the existence of an infinitely large element implies the existence of a positive infinitesimal (its reciprocal), and vice versa. This is precisely what the Archimedean Property rules out on both ends: no element is infinitely large relative to 1, and no positive element is smaller than all terms of the sequence 1/n.
Question 5 Short Answer
Why is the Archimedean Property called a theorem rather than an axiom of the real numbers? Sketch the argument that proves it from completeness.
Think about your answer, then reveal below.
Model answer: It is a theorem because it can be derived from the completeness axiom (least upper bound property) without being assumed independently. Proof sketch: Suppose, for contradiction, that ℕ is bounded above in ℝ. By completeness, ℕ has a supremum M. Since M − 1 < M is not an upper bound, there exists n ∈ ℕ with n > M − 1, so n + 1 > M. But n + 1 ∈ ℕ, contradicting M being an upper bound. Therefore ℕ has no upper bound in ℝ — which is exactly the Archimedean Property (taking a = 1).
Distinguishing axioms from theorems matters because it reveals the logical structure of the real numbers. Only one completeness axiom is needed; the Archimedean Property comes for free. Ordered fields that lack completeness (like certain non-Archimedean fields) show that completeness is genuinely doing the work — remove it and infinitely large elements become possible. The proof is also a template for other arguments: assume a supremum exists, then construct an element that exceeds it, deriving a contradiction.