Existence: produce one
The claim: there is a whole number n with . A claim that something exists is proved by producing one such thing and checking it.
Take n = 2: . That proves the claim. Nothing about any other value of n needs to be said.
Building the object
Usually the object is found by building it, and the construction is itself the proof. The claim: 2x + 3 = 11 has a solution.
Work back from the equation: subtract 3 to get 2x = 8, then halve to get x = 4. Then check forwards: 2 × 4 + 3 = 8 + 3 = 11. The working finds the candidate, and the check proves that it works.
Uniqueness: any two coincide
A uniqueness claim says there is at most one such object. Producing one says nothing about whether a second exists, so uniqueness needs its own argument: take any two objects with the property, and show that they are equal.
Suppose a and b both solve 2x + 3 = 11, so that 2a + 3 = 11 and 2b + 3 = 11. Both left sides equal 11, so 2a + 3 = 2b + 3. Subtract 3 from each side: 2a = 2b. Halve each side: a = b. Any two solutions are the same number, so there was only ever one.
Together the two arguments prove that 2x + 3 = 11 has exactly one solution, x = 4.
The line y = 2x + 3 against the level y = 11. The line rises all the way across, so it reaches 11 at one x only, x = 4.
Two separate claims
“There is exactly one” bundles two claims: at least one exists, and no more than one does. A proof has to establish each of them, and neither does the other’s work.
Existence without uniqueness: has the solution n = 2, and also n = 3, since . Factorized, , which is 0 exactly when n = 2 or n = 3. So “there is a whole number n with ” is true, and “there is exactly one” is false.
Uniqueness without existence: no whole number squares to 2, since and . So “there is at most one whole number n with ” is true, because there are none, and “there is exactly one” is false.
meets the x-axis at x = 2 and at x = 3. Producing the root 2 proved existence; the second root shows that existence alone says nothing about uniqueness.
A quantity that only moves one way
A common way to prove uniqueness: show that the quantity only increases. If it only increases as x increases, two different values of x give two different values of the quantity, so it takes any one value at most once. 2x + 3 only increases, so it equals 11 at most once, which is the uniqueness proof again in other words.
Existence then often comes from a change of sign. A quantity that changes without jumps from a negative value to a positive one passes through 0 somewhere in between. In the applications below, the gap between two hikers is −12 km at the start and 10 km four hours later and only increases, so it is 0 exactly once; and the shortage of strawberries at a market is 300 kg at $2 per kg and −150 kg at $12 per kg and only falls as the price rises, so exactly one price clears the market.
The usual mistakes
Stopping after existence. One object found leaves open whether there is a second, as n = 3 shows for .
Checking several values for uniqueness. Checking shows which values fail, and it cannot rule out the values not tried.
Starting the uniqueness argument from the object already found. It must take any two objects with the property, a and b, and force them to be equal.
Keeping a root that means nothing. In the applications, the quadratic has a second root outside the hours or prices the problem describes, so it is not a second answer.
Worked example: Two Hikers on One Mountain Trail, One Going Up and One Coming Down
Question A trail runs 12 km from the foot of a mountain to a hut at the top. At 6:00 am Ana starts up from the foot and slows as the trail steepens: h hours after 6:00 am she is 3.5h − h24 km from the foot, and she never stops or turns back until she reaches the hut at noon. At the same moment Ben starts down from the hut at a steady 3 km/h, and he reaches the foot at 10:00 am. (a) How far apart along the trail are they at 6:00 am and at 10:00 am, and who is higher up each time? Use this to prove that they pass each other exactly once between 6:00 am and 10:00 am. (b) At what time do they pass, and how far from the foot of the trail?
1.Measure both hikers from the foot of the trail. Ben is 12 − 3h km from the foot, so the gap, Ana's distance minus Ben's, is g(h) = 3.5h − h24 − (12 − 3h) = 6.5h − h24 − 12. They are at the same place exactly when g(h) = 0.
The gap is Ana's distance from the foot minus Ben's: 6.5h − h24 − 12. 2.At 6:00 am, g(0) = −12: Ben is 12 km higher, at the hut. At 10:00 am, g(4) = 26 − 4 − 12 = 10: Ana is 10 km from the foot and Ben is at the foot, so Ana is 10 km higher. Neither can jump along the trail, so the gap changes from −12 to 10 without a break and must pass through 0: they meet at least once.
At 6:00 am the gap is −12 and at 10:00 am it is 10, so it passes through 0. 3.Ana only moves up the trail and Ben only moves down it, so Ana's distance from the foot only increases and Ben's only decreases. The gap therefore only increases and can be 0 at most once. (a) They are 12 km apart at 6:00 am with Ben higher and 10 km apart at 10:00 am with Ana higher, and they pass exactly once.
(a) The gap only grows, so they pass exactly once. 4.Solve g(h) = 0. Multiply by −4: h2 − 26h + 48 = 0, which factorizes as (h − 2)(h − 24) = 0. The root h = 24 is long after Ben's 4 hours of walking, so it is rejected, which agrees with the single meeting found in (a).
h2 − 26h + 48 = 0 gives h = 2 or h = 24, and 24 is rejected. 5.(b) They pass at h = 2, which is 8:00 am. Ana is then 3.5 × 2 − 224 = 7 − 1 = 6 km from the foot. Check: Ben has walked 3 × 2 = 6 km down from the hut, so he is 12 − 6 = 6 km from the foot as well.
(b) At h = 2, 8:00 am, both are 6 km from the foot.
Answer: (a) 12 km apart at 6:00 am with Ben higher, and 10 km apart at 10:00 am with Ana higher, so they pass exactly once; (b) at 8:00 am, 6 km from the foot
Common mistakes
- Treating Ana as walking at a steady 2 km/h, her average for the climb, which puts the meeting at 12 ÷ (2 + 3) = 2.4 hours after 6:00 am, at 8:24 am. Ana walks faster early on and slower near the top, so she reaches the meeting point sooner.
- Keeping h = 24 as a second meeting. The formula for Ana describes her walk only until noon and Ben's walk ends at 10:00 am, so a root outside those hours describes nothing that happens on the trail.
Worked example: The Price of Strawberries at a Farmers' Market, Set by Supply and Demand
Question At a farmers' market, when strawberries sell at $p per kg, growers bring 15p + 30 kg and shoppers want to buy 720p kg. The market price settles where the amount brought equals the amount wanted. (a) Find the shortage or surplus at $2 per kg and at $12 per kg, and use them to prove that there is exactly one positive price at which the two amounts are equal. (b) Find that price, and the amount of strawberries sold at it.
1.Let the shortage be the amount wanted minus the amount brought: E(p) = 720p − (15p + 30). The two amounts are equal exactly when E(p) = 0.
The shortage is the amount wanted minus the amount brought: 720p − (15p + 30). 2.At $2 per kg, E(2) = 360 − 60 = 300: shoppers want 300 kg more than growers bring. At $12 per kg, E(12) = 60 − 210 = −150: growers bring 150 kg more than shoppers want. E changes without jumps for positive prices, so it passes through 0 between $2 and $12, and at least one such price exists.
A shortage of 300 kg at $2 and a surplus of 150 kg at $12: the curves cross in between. 3.Suppose two prices p1 < p2 both made the amounts equal. The higher price brings more, 15p2 + 30 > 15p1 + 30, and is wanted less, 720p2 < 720p1. Then 720p2 < 720p1 = 15p1 + 30 < 15p2 + 30, so the amounts are not equal at p2, a contradiction. (a) There is a shortage of 300 kg at $2 and a surplus of 150 kg at $12, and exactly one such price.
(a) Two crossing prices would contradict each other, so there is exactly one. 4.Solve 720p = 15p + 30. Multiply both sides by p: 15p2 + 30p − 720 = 0. Divide by 15: p2 + 2p − 48 = 0, so (p + 8)(p − 6) = 0. A price cannot be negative, so p = −8 is rejected and p = 6.
p2 + 2p − 48 = 0 gives p = 6 or p = −8, and −8 is rejected. 5.(b) The price is $6 per kg, and the amount sold is 15 × 6 + 30 = 120 kg. Check: shoppers want 7206 = 120 kg, the same amount.
(b) $6 per kg, where 120 kg is both brought and wanted.
Answer: (a) A shortage of 300 kg at $2 and a surplus of 150 kg at $12; exactly one price; (b) $6 per kg, with 120 kg sold
Common mistakes
- Stopping at the change of sign. A shortage at $2 and a surplus at $12 show that at least one price works, but not that only one does; that needs the argument that a higher price always brings more and is always wanted less.
- Keeping p = −8 as a second equilibrium. A price of −$8 per kg means nothing at a market, and the amount wanted, 720p, describes the shoppers only at positive prices.