My workbook have this question:

Let be the following expression. What is its logical value on $\mathbb{Z}$ and $\mathbb{R}$?

$$\forall \space a \space (6a^2 \geq6 \Rightarrow a \geq 1)$$

First, after some algebra I rewrite the expression as:

$$\forall \space a \space (a \leq -1 \space \vee a\geq1 \Rightarrow a \geq 1)$$

Then a set up two conditions, $P(a)=6a^2 \geq 6$ and $Q(a)=a \geq1$. So the expression satys like:

$$\forall \space a \space (P(a) \Rightarrow Q(a))$$

Then I thought. On $\mathbb{Z}$, $P(a)$ is possible because there is the $0$, so the condition is not universal. $Q(a)$ it's also possible, because it is only true when $a \geq 1$. So the implication is not universal on $\mathbb{Z}$, and then the expression is false.

On $\mathbb{R}$, the same happen because $\mathbb{Z} \subset \mathbb{R}$. Is this correct?Thanks.

  • 1
    $\begingroup$ Be careful: as dwarandae commented, it must be $\,\Bbb Z\subset \Bbb R\, $ , and *not* $\,\Bbb Z\in \Bbb R\,$ $\endgroup$ – DonAntonio Jan 11 '13 at 4:25

The correct one is what you wrote first: $\,a\leq-1\,\,\,\vee\,\,\, a\geq 1\,$ , but from this it does not follow that $\,a\geq 1\,$ , but rather $\,|a|\geq 1\,$ , which is equivalent with the first two inequalities above.

Thus, the logical value of $\,\forall\,a\in\Bbb Z\,\,\vee\,\,\Bbb R\;\;(6a^2\geq 6\Longrightarrow a\geq 1)\,$ is false, as $\,a=-1\,$ provides a contradiction in both sets.


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