Contradiction and existence proofs
Use an impossible consequence carefully and distinguish finding a witness from showing that a witness must exist.
On this page
The question for today
What you’ll understand
- State the exact temporary assumption in a contradiction proof.
- Locate the impossible consequence rather than merely an inconvenient one.
- Compare constructive witnesses with indirect existence arguments.
Sometimes a statement becomes easier after assuming it is false. That assumption can force an object to have incompatible properties, allowing us to reject the assumption in classical logic. The method is powerful precisely when the impossible consequence is unmistakable. Existence proofs require similar care: showing that a search is difficult is no evidence of nonexistence, while proving existence need not tell you an efficient way to locate the object.
A contradiction has to be logically impossible
To prove a claim P by contradiction, temporarily assume its negation and derive a statement incompatible with the assumptions already in force. You may derive that an integer is simultaneously even and odd, that a positive number is nonpositive, or that an explicitly nonzero quantity is zero. The contradiction must concern the same object under the same hypotheses. Two different integers can have opposite parity without difficulty, so keep witness names and dependencies visible.
Do not label a surprising result a contradiction merely because it disagrees with an unproved intuition. A divergent series or a nonintuitive geometric construction may be perfectly valid mathematics. Identify the exact earlier fact that the derived statement violates. Also distinguish contradiction from contraposition: the latter directly proves a logically equivalent implication, while contradiction assumes the entire desired claim fails. Both can use negated hypotheses, but their proof obligations are not identical.
Existence can be constructive or indirect
A constructive existence proof supplies an object and verifies every required property. To prove there is an integer between ten and twelve, give eleven and check both inequalities. In a parameterized claim, the witness may be a formula or an algorithm depending on the already chosen parameters. The proof must establish that the output belongs to the stated domain. A proposed rational witness for an integer existence claim cannot be rescued by satisfying only the equation.
An indirect proof may show that denying all possible witnesses is impossible, without singling out a witness by the proof’s end. For a finite set, one may establish that a maximum exists by comparing its elements, which is constructive; completeness arguments in analysis can establish existence without a finite search. “Nonconstructive” does not mean uncertain. It means the argument’s logical guarantee should not be confused with a procedure for computing the object at a desired cost.
Minimal representations reveal hidden assumptions
The irrationality of the square root of two illustrates how an assumption creates a useful representation. If the number were rational, it could be written as a quotient of integers in lowest terms with positive denominator. Squaring yields an even numerator square, hence an even numerator. Substitution then forces an even denominator as well. This contradicts lowest terms, rather than contradicting the mere existence of an even numerator or denominator.
The lowest-terms condition must be introduced before using it. Every rational number admits such a representation by dividing numerator and denominator by their greatest common divisor. Without that reduction, discovering that both are even is completely compatible with rationality: two over four is an ordinary rational representation. This example shows why a proof’s setup is substantive reasoning. The final contradiction is only as strong as the exact properties you were justified in assuming about the objects.
Work it through
There is no largest integer
Prove that the integers have no largest element by contradiction.
Assume that an integer M is largest, meaning every integer n satisfies n ≤ M. This is an existential assumption about one integer that dominates every other integer, not a claim about a large finite list.
Form the integer M + 1. Closure of integer addition guarantees that it belongs to the same domain over which the assumed maximality applies, so maximality implies M + 1 ≤ M.
Subtracting M gives 1 ≤ 0, contrary to the usual order of integers. Therefore the assumed largest integer cannot exist. Notice that constructing the successor also supplies a direct refutation of each candidate.
For every integer proposed as a largest element, its integer successor is larger, so no largest integer exists.
The living notebook
Explore the model
Interactive lab · function
A contradiction needs an exact invariant
Set a to 1 and inspect while changing position x. Suppose a negative target were the output: compare that supposition with the algebraic nonnegativity of squares. The picture illustrates the conflicting signs, but its resolution does not prove the sign property over all real inputs.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forName the exact property that the temporary assumption violates.
Pause & reason
Check your understanding
Your turn
Try it yourself
Prove constructively that between any real numbers a < b there is another real number.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use a formula that treats both endpoints symmetrically.
Compare with the solution
Choose c = (a + b)/2. It is a real number by closure of real addition and division by the nonzero number two. The differences c − a and b − c both equal (b − a)/2, which is positive because b > a. Therefore a < c < b. The formula supplies a witness as well as verifying its membership and both strict inequalities.
Keep the idea
Take it with you
- A contradiction must violate a stated definition, assumption, or established theorem.
- Existence and an efficient construction are different guarantees.
- A carefully chosen representation can carry the decisive hypothesis.
Read further
Original sources and deeper explanations to follow at your own pace.
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