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Mathematics LibreTexts

7.2: Equivalent Statements

  • Page ID
    24836
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    In other courses you will sometimes encounter a certain kind of theorem that is neither a conditional nor a biconditional statement. Instead, it asserts that a list of statements is "equivalent." You saw this (or will see it) in your linear algebra textbook, which featured the following theorem:

    Theorem Suppose A is an \(n \times n\) matrix. The following statements are equivalent:

    1. The matrix A is invertible.
    2. The equation Ax = b has a unique solution for every b \(\in \mathbb{R}\).
    3. The equation Ax = 0 has only the trivial solution.
    4. The reduced row echelon form of A is In.
    5. \(det(A) \ne 0\).
    6. The matrix A does not have 0 as an eigenvalue.

    When a theorem asserts that a list of statements is "equivalent," it is asserting that either the statements are all true, or they are all false. Thus the above theorem tells us that whenever we are dealing with a particular \(n \times n\) matrix A, then either the statements (a) through (f) are all true for A, or statements (a) through (f) are all false for A. For example, if we happen to know that \(det(A) \ne 0\), the theorem assures us that in addition to statement (e) being true, all the statements (a) through (f) are true. On the other hand, if it happens that \(det(A) = 0\), the theorem tells us that all statements (a) through (f) are false. In this way, the theorem multiplies our knowledge of A by a factor of six. Obviously that can be very useful.

    What method would we use to prove such a theorem? In a certain sense, the above theorem is like an if-and-only-if theorem. An if-and-only-if theorem of form \(P \Leftrightarrow Q\) asserts that P and Q are either both true or both false, that is, that P and Q are equivalent. To prove \(P \Leftrightarrow Q\) we prove \(P \Rightarrow Q\) followed by \(P \Rightarrow Q\), essentially making a "cycle" of implications from P to Q and back to P. Similarly, one approach to proving the theorem about the n × n matrix would be to prove the conditional statement \((a) \Rightarrow (b)\), then \((b) \Rightarrow (c)\), then \((c) \Rightarrow (d)\), then \((d) \Rightarrow (e)\), then \((e) \Rightarrow (f)\) and finally \((f) \Rightarrow (a)\). This pattern is illustrated below.

    Screen Shot 2020-02-02 at 9.20.16 PM.png

    Notice that if these six implications have been proved, then it really does follow that the statements (a) through (f) are either all true or all false. If one of them is true, then the circular chain of implications forces them all to be true. On the other hand, if one of them (say (c)) is false, the fact that \((b) \Rightarrow (c)\) is true forces (b) to be false. This combined with the truth of \((a) \Rightarrow (b)\) makes (a) false, and so on counterclockwise around the circle.

    Thus to prove that n statements are equivalent, it suffices to prove n conditional statements showing each statement implies another, in a circular pattern. But it is not necessary that the pattern be circular. The following schemes would also do the job:

    Screen Shot 2020-02-02 at 9.21.09 PM.png

    But a circular pattern yields the fewest conditional statements that must be proved. Whatever the pattern, each conditional statement can be proved with either direct, contrapositive or contradiction proof.

    Though we shall not do any of these proofs in this text, you are sure to encounter them in subsequent courses.