Two Directions, One Proof
In Working Backward From a Conclusion, we asked what would be enough to establish a goal. In Working Forward From Hypotheses, we began with the assumptions and derived consequences. These are complementary ways to plan a proof. The conclusion can suggest a useful intermediate target, while the hypotheses show which facts are actually available to reach it.
The two directions have different jobs. Backward analysis is a search strategy: inspect the desired conclusion and ask what sufficient condition would yield it. Forward reasoning provides justification: begin with the assumptions and establish the needed facts in a valid order. A finished proof ordinarily reads forward, even if its plan was discovered by looking backward from the goal.
For example, if the goal is an inequality, rearranging it or factoring a difference may reveal a condition that would be sufficient. The hypotheses may then supply the signs needed to prove that condition. The proof connects these pieces without treating the goal itself as an assumption.
Find a Bridge Between the Endpoints
A useful intermediate statement acts as a bridge: it follows from the hypotheses and is strong enough to give the conclusion. The bridge may be an inequality, a factorization, an integer representation, or a value that satisfies a definition. Its form depends on the problem.
Suppose a theorem has the form “if \(H\), then \(C\).” During planning, we can ask two questions. Forward: what consequences follow from \(H\)? Backward: what condition \(B\) would let us conclude \(C\)? The proof is successful when we can establish \(H\Longrightarrow B\) and \(B\Longrightarrow C\). We do not need \(B\) to be the only possible route; it only needs to be both reachable and sufficient.
Identify a condition that would be enough to prove the conclusion, using an equivalent rearrangement, a definition, or a known result.
Translate the assumptions into usable inequalities, representations, or other established facts.
Look for a consequence of the hypotheses that matches the sufficient condition suggested by the goal.
Verify signs, domains, and nonzero assumptions before applying an operation or a result.
Start with arbitrary objects and the hypotheses, prove the bridge, and then state explicitly how it yields the conclusion.
When the Goal Suggests a Factorization
An inequality can be difficult to approach if we only manipulate the hypothesis. Instead, compare the two sides of the conclusion. The difference between them may factor into expressions whose signs are directly controlled by the assumptions. The backward step helps reveal the right product; the forward steps establish its sign.
Worked Example: Factoring the Difference in an Inequality
Theorem. If \(x>2\), then \(x^2>2x\).
Backward analysis. The desired inequality is equivalent to \(x^2-2x>0\). Factoring gives \(x(x-2)>0\). This suggests establishing that both \(x\) and \(x-2\) are positive.
Forward analysis. From \(x>2\), we obtain \(x>0\) and \(x-2>0\). Thus the product in the backward analysis is positive.
Proof. Let \(x\in\mathbb R\) and suppose \(x>2\). Since \(2>0\), we have \(x>0\). Subtracting \(2\) from the hypothesis gives \(x-2>0\). Therefore the product of the positive numbers \(x\) and \(x-2\) is positive: $$ x(x-2)>0. $$ Expanding yields \(x^2-2x>0\). Adding \(2x\) to both sides gives \(x^2>2x\), as required.
The backward calculation was useful because it identified a condition that would imply the goal. It was not itself the proof: in particular, the proof did not assume \(x^2>2x\). It established the positivity of the factors from \(x>2\), then used the factorization to reach the conclusion.
When working backward through an algebraic statement, it is important to distinguish equivalence from sufficiency. In this example, rearranging the inequality and factoring preserve equivalence: \(x^2>2x\) holds exactly when \(x(x-2)>0\). In other problems, a chosen condition may only be sufficient, not necessary. Either kind of condition can help with a proof, provided the direction of implication is clear.
When a Definition Supplies the Bridge
Backward analysis is also useful when the conclusion asks for an object or a property defined by a particular form. If the goal says that an integer is odd, for instance, the definition tells us the form we must eventually produce: \(2m+1\) for some integer \(m\). Forward reasoning can then translate the hypothesis into a representation and calculate until that required form appears.
Worked Example: Using an Odd Representation to Prove a Square Is Odd
Theorem. If an integer \(n\) is odd, then \(n^2\) is odd.
Backward analysis. To prove that \(n^2\) is odd, the definition requires an integer \(m\) such that \(n^2=2m+1\). Thus we want to rewrite the square in that form and verify that the coefficient \(m\) is an integer.
Forward analysis. Since \(n\) is odd, there is an integer \(k\) such that \(n=2k+1\). Squaring that representation gives an expression with an odd final term.
Proof. Let \(n\in\mathbb Z\) be odd. By the definition of oddness, there is an integer \(k\) such that \(n=2k+1\). Therefore $$ n^2=(2k+1)^2=4k^2+4k+1 =2(2k^2+2k)+1. $$ Set \(m=2k^2+2k\). Since \(k\) is an integer, \(k^2\), \(2k^2\), and \(2k\) are integers, so their sum \(m\) is an integer. We have expressed \(n^2\) as \(2m+1\) for an integer \(m\). By the definition of oddness, \(n^2\) is odd.
The goal-directed part of the plan specified the exact form to aim for. The forward part supplied \(n=2k+1\), and the calculation met that target. Verifying \(m\in\mathbb Z\) is essential: a formula of the shape \(2m+1\) establishes oddness only when its \(m\) belongs to the required domain.
One Conclusion Can Need Several Hypotheses
Sometimes the conclusion combines information from more than one assumption. In that case, backward analysis may suggest an intermediate quantity that lies between the desired endpoints. Forward reasoning can then use each hypothesis to establish one part of the chain. The standard order properties of the real numbers, which are developed formally later in the course, allow us to add the same quantity to both sides of an inequality and to use transitivity.
Worked Example: Combining Two Strict Inequalities
Theorem. Let \(a,b,c,d\in\mathbb R\). If \(a<b\) and \(c<d\), then \(a+c<b+d\).
Backward analysis. The target compares \(a+c\) with \(b+d\). An intermediate expression using one term from each side is \(b+c\). If we can show \(a+c<b+c\) and \(b+c<b+d\), transitivity gives the target.
Forward analysis. From \(a<b\), adding \(c\) gives the first comparison. From \(c<d\), adding \(b\) gives the second.
Proof. Let \(a,b,c,d\in\mathbb R\) and suppose \(a<b\) and \(c<d\). Adding \(c\) to both sides of \(a<b\) gives $$ a+c<b+c. $$ Adding \(b\) to both sides of \(c<d\) gives $$ b+c<b+d. $$ By transitivity of the strict order, these inequalities imply \(a+c<b+d\). This is the required conclusion.
The intermediate expression \(b+c\) is not an extra hypothesis. It is a bridge created by applying the two given inequalities. The conclusion would not follow from either hypothesis alone in general; the proof uses both, and makes their separate contributions visible.
Planning Is Not the Same as Proving
Backward analysis can include rearrangements that are helpful for discovery. In the final proof, however, each line should follow from facts already established or from an earlier result. A chain written backward is not automatically a proof: if one starts at the conclusion and lists statements that would imply earlier statements, the reader may not know which of those statements has been established.
A safe practice is to use backward steps to propose a bridge, then turn the plan into a forward argument. Check that the assumptions really imply the bridge, and check that the bridge really implies the conclusion. If the plan uses an equivalence, preserve both directions where needed; if it uses a sufficient condition, prove that specific direction.
| Reasoning direction | Useful question | Role in the finished proof |
|---|---|---|
| Backward planning | What condition would be enough for this conclusion? | Suggests a useful target or intermediate fact. |
| Forward planning | What follows from the given hypotheses? | Shows whether the suggested target can be established. |
| Forward proof | Why is this next line true? | Justifies the argument from assumptions to conclusion. |
A failed route can still be informative. Perhaps the hypothesis does not give enough information to establish the first proposed bridge, or perhaps the bridge does not quite imply the conclusion. In that case, revise the target rather than forcing an unjustified step. A proof is complete only when every connection is valid, every domain condition is met, and the final statement is the original conclusion.
Check Your Understanding
For each question, consider both how the conclusion guides a plan and how the hypotheses support a proof.
- In the proof that \(x>2\) implies \(x^2>2x\), what factorization does the goal suggest, and which facts from the hypothesis establish the signs of its factors?
- Why does backward analysis of the claim that \(n^2\) is odd lead us to seek an integer \(m\) with \(n^2=2m+1\)?
- In the proof combining \(a<b\) and \(c<d\), what role does the intermediate expression \(b+c\) play?
- What two implications must be checked for an intermediate statement \(B\) to serve as a bridge between hypotheses and conclusion?
- Why is a backward calculation that begins with the desired conclusion not, by itself, a finished proof?
- If a proof writes an expression in the form \(2m+1\), what additional fact must it verify before concluding that the expression is odd?