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According to my textbook, the statement $n^2-n-2=0 \Leftarrow (n=2 \text{ and } n=-1)$ is true (full solution was not provided).

I am not sure why the statement must be true. My reasoning is as follows:

$$n^2-n-2=0 \Leftarrow (n=2 \text{ and } n=-1)$$

is the same as

$$(n=2 \text{ and } n=-1) \Rightarrow n^2-n-2=0$$

The hypothesis $(n=2 \text{ and } n=-1)$ is false, since $n$ can only take on 1 value at a time. Since the hypothesis is always false, the implication will always be true regardless of the truth value of the conclusion.

Is that how I am suppose to deduce the answer?

EDIT:

The textbook I am referring to is "An Introduction to Mathematical Reasoning: numbers, sets and functions" by Peter J. Eccles.

mauna
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1 Answers1

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The above expression is only a bad "formalization" of the fact that the equation :

$n^2−n−2 = 0$

has two solutions : $2$ and $-1$.

Thus, if $n^2−n−2 = 0$, i.e. $(n-2)(n+1)=0$, then $n = 2 \lor n = -1$.

Comment

See Peter J. Eccles, An Introduction to Mathematical Reasoning: numbers, sets and functions (1997), page 19 :

$n^2−n−2 = 0 \Rightarrow (n = 2 \text{ or } n = -1)$.

Your formula is on page 20; but be careful ... all are parts of on Exercise asking :

Which of the following universal statements are true and which are false for integers $n$ ?

Mauro ALLEGRANZA
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    That is indeed a grievous error if this occurred in a mathematical logic text. I suspect (or hope!) that the comment of @fuglede to the OP's question is really the explanation (see above). – MPW Jun 05 '14 at 11:27
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    @MauroALLEGRANZA Actually, the question is on page 20, (viii) and is still written in the same manner – mauna Jun 05 '14 at 12:41