This exercise is from Apostol's Calculus:

$|1 + 3x| \le 1$ implies $x \ge -{\frac 23}$

I answered false, but the right answer is true. My understanding is — of course all $x$ are bigger than $-{\frac 23}$, but if $x$ is $1000$ the first inequality is obviously wrong.

• Your confusion is a logical one. Do you understand what implies means? Commented Jan 1, 2014 at 16:02
• Your thinking is right, but you have your logic reversed. You answered correctly if you misinterpreted the statement as: $x\ge -2/3 ~\text{ implies } |1+3x| \le 1$. Based on your explanation, this is your misinterpretation. Commented Jan 1, 2014 at 16:07
• @GitGud: "A implies B" means if A is true than B must be true. Commented Jan 1, 2014 at 16:16
• For $x=1000$, is $A$ true? Commented Jan 1, 2014 at 16:17
• @GitGud, no, but you're moving reverse. Commented Jan 1, 2014 at 16:17

It doesn't look like it's the arithmetic that confuses you. Hopefully we agree that depending on what $x$ is, we can be in one of three situtations:

• $|1 + 3x| \le 1$ is false and $x \ge -{\frac 23}$ is also false (for example when $x=-3$).
• $|1 + 3x| \le 1$ is true and $x \ge -{\frac 23}$ is also true (for example when $x=-\frac13$).
• $|1 + 3x| \le 1$ is false but $x \ge -{\frac 23}$ is true (for example when $x=3$).

In each of these cases it holds that $|1 + 3x| \le 1$ implies $x \ge -{\frac 23}$, because the only way for "$P$ implies $Q$" not to hold is if $P$ is true and $Q$ is false. And that is never the case with these formulas.

Note that strictly speaking a thing like "$|1 + 3x| \le 1$ implies $x \ge -{\frac 23}$" in its entirety is a claim that depends on the value of $x$. For example, "$x\ge0$ implies $x=10$" is true whenever $x$ happens to be $10$ or negative, but is false when $x$ is $0$ or positive and different from $10$.

When the exercise asks about "$|1 + 3x| \le 1$ implies $x \ge -{\frac 23}$" what it really means is

For every possible $x$ it holds that $|1 + 3x| \le 1$ implies $x \ge -{\frac 23}$

which happens to be true here because there is no $x$ that makes the implication false. The logical notion of "$P$ implies $Q$" is designed to be useful in a situation where $P$ and $Q$ depend on some parameters, and what we're really interested in is being able to say "for every choice of $x$, $y$, $z$, it will be that $P(x,y,z)$ implies $Q(x,y,z)$". That's just the mathematical way of saying "whenever you find some $x$, $y$, $z$ that make $P(x,y,z)$ true, those $x,y,z$ will also make $Q(x,y,z)$". By design that tells you nothing about what happens to $Q$ in choices where $P$ is false.

$$\begin{array}{l} |1+3x|\le 1 &\iff -1 \le 1 +3x \le 1\\ &\iff -2\le 3x \le 0\\ &\iff -\frac{2}{3}\le x \le 0\\ &\iff -\frac{2}{3}\le x \text{ and } x \le 0\\ &\implies -\frac{2}{3}\le x \end{array}$$

$$|1+3x|\leqslant 1\iff -1\leqslant 1+3x\leqslant 1$$ Solving for $x$: $$-2\leqslant 3x\leqslant 0\iff \dfrac{-2}3\leqslant x\leqslant0\implies\dfrac{-2}3\leqslant x$$ Your confusion regarding Apostol's statement is rather a logical one. Read in those two links what implies mean: $1$, $2$.

For real $b\ge0, y$ $$|y|\le b\iff -b\le y\le b$$

$$\implies|1+3x|\le1\iff -1\le 1+3x\le1$$