From a Hypothesis to Its Consequences
In Introduction to Mathematical Proof, we proved that \(x>1\) implies \(x^2>1\) by starting with the hypothesis, establishing positivity of two factors, and reaching the conclusion. That argument illustrates the method we now study systematically: direct proof.
For a universally quantified claim
we begin by fixing an arbitrary \(x\in D\) and supposing \(P(x)\) holds. We then establish \(Q(x)\) without imposing any additional restriction on \(x\). Because the input was arbitrary, the argument proves the implication throughout the domain.
As explained in the previous tutorial, inputs for which \(P(x)\) is false require no further argument: the implication is true there by its logical meaning. In ordinary proof writing, this observation need not be repeated at the end of every direct proof.
The word “direct” describes the logical direction of the argument, not its length or difficulty. A direct proof may involve several intermediate deductions, a carefully chosen expression, or an application of another theorem.
Use Definitions at Both Ends
Definitions often tell us both what the hypotheses provide and what the conclusion requires. For elementary examples, recall the standard descriptions of even and odd integers:
Here \(\mathbb Z\) denotes the integers. These definitions include negative integers; for example, \(-3=2(-2)+1\) is odd. Also, \(0=2\cdot0\) is even.
| Information available | What the definition permits or requires |
|---|---|
| Hypothesis: \(n\) is even. | Introduce an integer \(k\) with \(n=2k\). |
| Hypothesis: \(n\) is odd. | Introduce an integer \(k\) with \(n=2k+1\). |
| Goal: an integer expression \(E\) is even. | Find an integer \(r\) and establish \(E=2r\). |
| Goal: an integer expression \(E\) is odd. | Find an integer \(r\) and establish \(E=2r+1\). |
An existential statement in a hypothesis supplies a witness we may name. An existential statement in the conclusion requires a witness we must produce and verify. The distinction determines how a definition is used in the argument.
Worked Example: The Sum of Two Even Integers
Claim. If \(m\) and \(n\) are even integers, then \(m+n\) is even.
Proof. Let \(m,n\in\mathbb Z\) be arbitrary even integers. By the definition of evenness, there exist \(a,b\in\mathbb Z\) such that \(m=2a\) and \(n=2b\). Therefore
Since integers are closed under addition, \(a+b\in\mathbb Z\). Thus \(m+n\) is twice an integer, so it is even by definition.
The proof begins and ends with the same definition, used in different directions. It first translates the hypotheses into equations and then translates the final equation into the desired property.
A Complete Direct Proof with a Constructed Witness
Before writing the proof, identify its endpoint. To show \(mn\) is odd, we need an integer \(r\) for which \(mn=2r+1\). The hypotheses provide representations of \(m\) and \(n\), so multiplying those representations is a natural first calculation.
Proof. Let \(m,n\in\mathbb Z\) be arbitrary odd integers. By the definition of oddness, there exist integers \(a,b\) such that \(m=2a+1\) and \(n=2b+1\). Using the distributive law,
Set \(r=2ab+a+b\). Closure of the integers under multiplication and addition gives \(r\in\mathbb Z\). Hence \(mn=2r+1\) for an integer \(r\), so \(mn\) is odd. This proves the claim for every pair of odd integers.
There is no need to treat positive and negative odd integers separately. Every step remains valid for arbitrary integer witnesses \(a\) and \(b\). The proof also does not require the two inputs to be distinct.
The statement “\(r\in\mathbb Z\)” is essential. Merely expressing a number as \(2r+1\) with a real \(r\) would not establish oddness. For instance, \(4=2(3/2)+1\), but \(3/2\) is not an integer.
Direct Proofs of Inequalities
For inequalities, a useful direct approach is to show that the difference between the proposed larger and smaller quantities is positive. This is the same idea used in the previous tutorial’s factorization of \(x^2-1\).
Worked Example: Comparing Reciprocals
Claim. For real numbers \(a,b\), if \(0<a<b\), then \(1/b<1/a\).
Proof. Let \(a,b\in\mathbb R\) satisfy \(0<a<b\). Both \(a\) and \(b\) are positive, so the reciprocals are defined and \(ab>0\). Also, \(b-a>0\). Consequently,
because a quotient of positive real numbers is positive. Adding \(1/b\) to the inequality gives \(1/a>1/b\), which is the required conclusion.
The positivity hypothesis does two jobs: it excludes zero denominators and determines the sign of their product. A direct proof should expose these roles rather than hide them in an unexplained instruction to “take reciprocals.”
The weaker hypothesis \(a<b\) would not support this argument. With \(a=-1\) and \(b=1\), both reciprocals exist, but the asserted conclusion would be \(1<-1\), which is false. If \(a=0\), one of the reciprocals is not even defined.
Planning Backward, Writing Forward
A direct proof need not be discovered in the order in which it is written. Studying the desired conclusion can reveal a useful identity or construction. The final argument must nevertheless justify that conclusion from information already available.
Worked Example: A Number Strictly Between Two Inputs
Claim. For all \(a,b\in\mathbb R\), if \(a<b\), then there exists \(c\in\mathbb R\) such that \(a<c<b\).
Planning. The goal asks for a witness between the inputs. The midpoint \(c=(a+b)/2\) is a candidate. To verify it, we will need both \(c-a>0\) and \(b-c>0\); each difference simplifies to \((b-a)/2\).
Proof. Let \(a,b\in\mathbb R\) satisfy \(a<b\). Set \(c=(a+b)/2\), which is real by closure of real arithmetic and because \(2\neq0\). Since \(b-a>0\),
The first inequality gives \(a<c\), and the second gives \(c<b\). Thus this real \(c\) is a witness satisfying both required inequalities.
The planning paragraph proposes a witness; the proof verifies it. No positivity assumption on \(a\) or \(b\) was needed. Only their order mattered, so the argument applies to negative inputs and inputs on opposite sides of zero as well.
By contrast, starting with the unproved assertion \(a<(a+b)/2<b\) and manipulating it until one reaches \(a<b\) does not, by itself, prove the claim. Deducing a known hypothesis from a desired conclusion establishes the wrong direction unless the reverse steps are also justified.
Using an Earlier Result Directly
Direct proofs can use theorems as well as definitions. The essential obligation is to check that the theorem applies to the particular object at hand.
Worked Example: Substituting into an Established Theorem
Claim. For every real \(x\), if \(x>2\), then \((x-1)^2>1\).
Proof. Let \(x\in\mathbb R\) satisfy \(x>2\), and set \(u=x-1\). Then \(u\in\mathbb R\) and \(u>1\). By the theorem proved in Introduction to Mathematical Proof, a real number greater than \(1\) has square greater than \(1\). Applying that theorem to \(u\) gives \(u^2>1\). Since \(u=x-1\), we obtain \((x-1)^2>1\).
There is no need to repeat the earlier factorization proof. Naming the input \(u\), checking its domain and inequality, and invoking the established result supplies a complete justification.
A Procedure for Writing a Direct Proof
Introduce arbitrary inputs in their domains and assume exactly the hypotheses of the implication.
Unpack definitions to obtain usable equations or inequalities, and identify what would establish the conclusion.
Calculate, construct witnesses, or apply earlier results while checking domains, signs, and other required conditions.
Explain why the final expression or inequality satisfies the original goal for every permitted input.
A useful final check is to read each sentence and ask whether it is an assumption allowed by the claim, a definition, an established fact, or a consequence of earlier lines. Any sentence that merely restates the desired conclusion without justification marks a gap.
The strength of direct proof is that the dependence of the conclusion on the hypotheses remains visible. Each assumption provides information, and each deduction moves that information toward the precise claim to be established.
Check Your Understanding
Write complete direct arguments, introducing arbitrary inputs and checking the conditions needed for each calculation.
- Prove that the sum of two odd integers is even. Identify the integer witness that establishes the conclusion.
- Prove that the product of an even integer and any integer is even. Explain why your proof includes zero and negative inputs.
- An attempted proof begins, “Let \(m\) and \(n\) be even, so \(m=2k\) and \(n=2k\) for some integer \(k\).” What extra restriction has been introduced? Rewrite the opening correctly.
- Prove that if \(0<a<b\), then \(a^2<b^2\), by considering \(b^2-a^2\). Identify exactly where positivity is used.
- For arbitrary real \(a<b\), set \(c=(2a+b)/3\). Prove directly that \(a<c<b\), including verification that \(c\) lies in the required domain.
- Using the theorem from the previous tutorial that \(u>1\) implies \(u^2>1\), prove that \(x>0\) implies \((x+1)^2>1\). State the substituted input and check the theorem’s hypothesis.