$6 \leq 2x \leq 4$ for all $3 \leq x \leq 2$ is true but $6 \leq 2x \leq 4$ for some $3 \leq x \leq 2$ is false
Could someone help me out why the second one is false.
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3There is no $x$ such that $3\le x\le 2$. – David C. Ullrich Jun 17 '19 at 15:15
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The second one is $\exists x [(3 \le x \le 2) \land (6 \le 2x \le 4)]$ and is FALSE (1st conjunct FALSE). – Mauro ALLEGRANZA Jun 17 '19 at 15:35
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The first one is $\forall x [(3 \le x \le 2) \to (6 \le 2x \le 4)]$ and is TRUE (antecedent is FALSE). – Mauro ALLEGRANZA Jun 17 '19 at 15:35
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See the post When can $p \to q$ be true while $p \land q$ be false – Mauro ALLEGRANZA Jun 17 '19 at 15:37
3 Answers
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There is no $x$ satisfying $3\leq x\leq 2$. So, any statement is true for all such $x$ (since there are none).
However, there does not exist a value of $x$ which has this property (see above), which makes the second statement false.
log_math
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The second one says that there is at least one value of $x$ such that $3\leq x\leq 2$ for which $6\leq 2x\leq 4$. There are no values of $x$ satisfying $3\leq x\leq 2$, so stating that "there is at least one $x$ satisfying this" is incorrect.
If you were to suppose there was such an $x$ then you would have that $3\leq 2$ which is false... $3$ is not less than $2$ in this context.
Compare to the first which says that for every $x$ satisfying $3\leq x\leq 2$ (if any exist in the first place) you will have $6\leq 2x\leq 4$. This is vacuously true since there are no $x$ satisfying it.
JMoravitz
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Note the direction of the inequality signs. Neither inequality is ever true, for any $x$.
If you say "for any $x$ such that the second inequality is true, the first inequality is true", then that's vacuously true. That's what the first statement says. However, the second statement says that there is some $x$ which fulfills the first inequality, and we know there are none. This makes the second statement false.
Arthur
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