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“There is exactly one” makes two claims: that at least one object exists, and that no second one does. This lesson proves each separately. Existence can be shown by constructing an object or, without construction, by an argument such as the intermediate value theorem; uniqueness is shown by assuming two such objects and proving they are equal. Keeping the two halves apart is what makes these proofs complete.
Existence and uniqueness are separate claims
“There exists a unique x with property P” makes two claims:
- Existence: at least one x has property P.
- Uniqueness: if x and y both have property P, then x = y.
Prove them separately. Existence can be shown by constructing an example and checking it, or by an argument that one must exist without naming it. Uniqueness is usually proved by assuming two objects with the property and showing they are equal, or by contradiction.
Existence without construction
Worked example 1
An irrational power that is rational
Worked example 1
An irrational power that is rational
Prove that there exist irrational numbers a and b such that a^b is rational.
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Consider one number and split into cases
Method
Let q = square root of 2^(square root of 2). Either q is rational or it is irrational.Reason
Two cases that cover every possibility give a proof even if we cannot tell which one holds.Working
Answer
If q is rational, take a = b = square root of 2.
If q is irrational, take a = q and b = square root of 2. Then
a^b = (square root of 2^(square root of 2))^(square root of 2) = square root of 2² = 2,which is rational. In both cases suitable a and b exist. blacksquare
Quick check
The proof never decides whether square root of 2^(square root of 2) is rational. A constructive proof names the numbers: take a = square root of 2 and b = log ₂9. Then b = 2 log ₂3 is irrational, because log ₂3 is irrational, and a^b = 2^(1/2 log ₂9) = 2^(log ₂3) = 3.
A unique root
You may use the intermediate value theorem: if f is continuous on [a,b] and f(a), f(b) have opposite signs, then f(c) = 0 for some c between a and b.
Exam-style question 1
Exactly one real root
Exam-style question 1
Exactly one real root
Prove that the equation x³ + x-1 = 0 has exactly one real root. [5]
Write your answer, then compare it with the solution
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Existence
Method
Let f(x) = x³ + x-1. Then f(0) = -1 and f(1) = 1.Reason
f is a polynomial, so it is continuous, and the intermediate value theorem applies.Working
Answer
There is a root c with 0 < c < 1.
Uniqueness
Method
f'(x) = 3x² + 1 > 0 for every real x, so f is strictly increasing.Reason
A strictly increasing function takes each value at most once.Working
Answer
Suppose f(c) = f(d) = 0 with c ≠ d, say c < d. Then f(c) < f(d), contradicting f(c) = f(d). So the root is unique. blacksquare
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Tick a point only if your written proof contains it. This self-check does not record a mark.
Common mistake
A calculator graph showing one crossing is evidence, not proof: it shows only a window of the graph. Proving existence alone answers “at least one”, not “exactly one”. The root is about 0.6823, which a calculator can confirm once the proof is complete.
Uniqueness by comparing two representations
The division algorithm says: for integers n and d with d > 0, there are unique integers q and r with n = qd + r and 0 ≤ r < d. Existence comes from taking q as the greatest integer with qd ≤ n.
Check your understanding 1
Why the remainder is unique
Check your understanding 1
Why the remainder is unique
Suppose n = qd + r = q'd + r' with 0 ≤ r < d and 0 ≤ r' < d. Prove that r = r' and q = q'.
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Subtracting, (q-q')d = r'-r, so d divides r'-r. Both remainders lie in [0,d), so -d < r'-r < d. The only multiple of d strictly between -d and d is 0, so r' = r. Then (q-q')d = 0 with d ≠ 0, so q = q'.
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Tick a point only if your written proof contains it. This self-check does not record a mark.
Proof by construction of more general objects, such as a function or a sequence with given properties, uses the same check: build the object explicitly, then verify every required property.
Existence by construction
A constructive proof names the object and then checks every required property. The check is part of the proof: an object that has not been verified has not been shown to exist.
Exam-style question 2
A long run with no primes
Exam-style question 2
A long run with no primes
Prove that, for every positive integer n, there exist n consecutive positive integers none of which is prime. [4]
Write your answer, then compare it with the solution
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Name the integers
Method
Take (n + 1)! + 2, (n + 1)! + 3, …, (n + 1)! + (n + 1).Reason
These are n consecutive integers, and (n + 1)! is divisible by every integer from 2 to n + 1, so each added number can be matched with a divisor.Working
Answer
There are (n + 1)-2 + 1 = n of them, and they are consecutive.
Verify each one is composite
Method
For each j with 2 ≤ j ≤ n + 1, show that j is a proper divisor of (n + 1)! + j.Reason
A positive integer with a divisor strictly between 1 and itself is not prime.Working
Answer
j divides (n + 1)! and divides j, so j divides their sum. Also 1 < j < (n + 1)! + j. So (n + 1)! + j is composite for every j, and the run contains no prime. blacksquare
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Tick a point only if your written proof contains it. This self-check does not record a mark.
Common mistake
Showing a run for n = 4, such as 122 to 125, is an example, not a proof for every n. A construction is complete only when the divisor is shown to be proper: j > 1 and j is less than the number it divides.
Common mistakes
- Proving that a solution exists and calling that uniqueness, or the reverse.
- Proving uniqueness by finding one solution and saying “this is the only one” without an argument.
- Using the intermediate value theorem without checking that the function is continuous on the whole interval.
- Assuming two objects are different at the start of a uniqueness proof and then deriving a contradiction from that unstated assumption.
- Constructing an object without checking that it satisfies every condition required.
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