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The problem I am working on is:

Determine the truth value of each of these statements if the domain consists of all integers.

a) $∀n(n+1>n)$

b) $∃n(2n=3n)$

c) $∃n(n=−n)$

d) $∀n(3n≤4n)$

The only part I am having difficulty with is part (d). The answer key declares that this statement is true. But isn't it really a false statement? Wouldn't any negative number render this statement false?

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I think the domain is positive integers for the d part or maybe you should think of the negative integers as -1.n, $n\in Z^+$, so when you multiply the inequality with -1, the direction of the inequality changes. – ciceksiz kakarot Feb 9 '13 at 19:52
The answer key is wrong -- since (c) only makes sense if the domain is positive and negative integers, and presumably the same domain is to be assumed in (c) and (d). – Peter Smith Feb 9 '13 at 19:56
@PeterSmith Well, c) is true for the value 0. – Mack Feb 9 '13 at 20:01
The answer key is clearly wrong for (d). – Brian M. Scott Feb 9 '13 at 20:07
@PeterSmith $-1 \cdot 0 = 0$ – Mack Feb 9 '13 at 22:03
up vote 3 down vote accepted

Given that the domain of $n$, as stated, is all* $n\in \mathbb{Z}$, then your reasoning is correct and $d$ is indeed false. Negative integers would serve as your counterexample showing the statement is false. So the answer key must be wrong, or there was a typo in the problem set!

If the domain of $n$ were $\mathbb{N}$, and depending on how one defines the natural numbers $\mathbb{N}$: would is any integer $n \geq 0$ (or an integer $n\geq 1$).

Hence, in either case, negative numbers are excluded from the domain of $n\in \mathbb{N}$.

Hence, $(d)$ would be true, if the domain were in fact $n \geq 0$: given ANY $n\in \mathbb{N},\;3n\leq 4n$, since $3\leq 4$ is clearly true.

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+1 for ruling out that bad $d$. :D – Babak S. Feb 9 '13 at 20:46
@amWhy Yes, and thank you very much! – Mack Feb 9 '13 at 21:53

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