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# 数学代写|离散数学代写Discrete Mathematics代考|MATH271 Relation between Proof by Contradiction and Proof by Contraposition

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## 数学代写|离散数学代写Discrete Mathematics代考|Relation between Proof by Contradiction and Proof by Contraposition

Observe that any proof by contraposition can be recast in the language of proof by contradiction. In a proof by contraposition, the statement
$\forall x$ in $D$, if $P(x)$ then $Q(x)$
is proved by giving a direct proof of the equivalent statement
$\forall x$ in $D$, if $\sim Q(x)$ then $\sim P(x)$.
To do this, you suppose you are given an arbitrary element $x$ of $D$ such that $\sim Q(x)$. You then show that $\sim P(x)$. This is illustrated in Figure 4.7.1.

Exactly the same sequence of steps can be used as the heart of a proof by contradiction for the given statement. The only thing that changes is the context in which the steps are written down.

To rewrite the proof as a proof by contradiction, you suppose there is an $x$ in $D$ such that $P(x)$ and $\sim Q(x)$. You then follow the steps of the proof by contraposition to deduce the statement $\sim P(x)$. But $\sim P(x)$ is a contradiction to the supposition that $P(x)$ and $\sim Q(x)$. (Because to contradict a conjunction of two statements, it is only necessary to contradict one of them.) This process is illustrated in Figure 4.7.2.

## 数学代写|离散数学代写Discrete Mathematics代考|Proof as a Problem-Solving Tool

Direct proof, disproof by counterexample, proof by contradiction, and proof by contraposition are all tools that may be used to help determine whether statements are true or false. Working with examples might have given you a sense that a statement of the form
For all elements in a domain, if (hypothesis) then (conclusion),
might be true. To explore further, imagine elements in the domain that satisfy the hypothesis. Ask yourself: Must they satisfy the conclusion? If you can see that the answer is “yes” in all cases, then the statement is true and your insight will form the basis for a direct proof. If after some thought it is not clear that the answer is “yes,” ask yourself whether there are elements of the domain that satisfy the hypothesis and not the conclusion. If you are successful in finding some, then the statement is false and you have a counterexample. On the other hand, if you are not successful in finding such elements, perhaps none exist. Perhaps you can show that assuming the existence of elements in the domain that satisfy the hypothesis and not the conclusion leads logically to a contradiction. If so, then the given statement is true and you have the basis for a proof by contradiction. Alternatively, you could imagine elements of the domain for which the conclusion is false and ask whether such elements also fail to satisfy the hypothesis. If the answer in all cases is “yes,” then you have a basis for a proof by contraposition.

Solving problems, especially difficult problems, is rarely a straightforward process. At any stage of following the guidelines above, you might want to try the method of a previous stage again. If, for example, you fail to find a counterexample for a certain statement, your experience in trying to find it might help you decide to reattempt a direct argument rather than trying an indirect one. Psychologists who have studied problem solving have found that the most successful problem solvers are those who are flexible and willing to use a variety of approaches without getting stuck in any one of them for very long. Mathematicians sometimes work for months (or longer) on difficult problems. Don’t be discouraged if some problems in this book take you quite a while to solve.

## 数学代写|离散数学代写Discrete Mathematics代考|Relation between Proof by Contradiction and Proof by Contraposition

$\forall x$ 在 $D$ ，如果 $\sim Q(x)$ 然后 $\sim P(x)$.

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