Absolute Value and Order - GRE Quantitative Reasoning

$|a|<|b|\iff a^2<b^2$. Squaring both sides preserves the inequality since squares equal the squares of absolutes and are monotonic for non-negative values.

$x\le -6$ or $x\ge 2$. Solving |x+2|>=4 splits into x+2 <= -4 or x+2 >=4, subtracting 2.

$-3\le x\le 2$. Solving |2x+1|<=5 gives -5 <= 2x+1 <=5, subtract 1 then divide by 2.

$1<x<5$. Solving |x-3|<2 expands to -2 < x-3 < 2, adding 3 to all parts.

$|0-4|$ is larger. At x=0, distance to 4 is 4, greater than distance to 1 which is 1.

$|2-9|=7$ is larger. Distance from 2 to 9 is 7, exceeding distance to 6 which is 4.

$\left| -\frac{5}{4} \right|=\frac{5}{4}$ is larger. Absolute value turns -5/4 positive to 1.25, which is greater than -1.5.

$|-7|$ is larger. Larger magnitude negative input to absolute value yields larger output, as 7>5.

$|-4|=4$ is larger. Absolute value makes |-4| positive 4, which exceeds -3 on the number line.

$-2<1<|-3|$. Ordering -2 (negative), then 1 (positive), then 3 from | -3 |.

$|x|=\sqrt{x^2}$. The square root of x squared yields the non-negative root, matching the absolute value.

$-\frac{3}{2}<|-1|<\frac{4}{3}$. Ordering starts with negative -1.5, then 1 from | -1 |, then positive 1.333.

$|x|=\begin{cases}x,&x\ge 0\\-x,&x<0\end{cases}$. The absolute value function returns the non-negative value by taking x when x is non-negative and its negation when x is negative.

$|a-b|$ is the distance between $a$ and $b$. On the number line, the absolute difference |a-b| measures the positive distance separating points a and b.

No largest value; $|x|$ is unbounded above. Since x can be arbitrarily large positive or negative, |x| can exceed any bound with no maximum.

Minimum is $0$, occurring at $x=0$. The absolute value |x| is always non-negative, reaching its minimum of 0 only when x=0.

$-a\le x\le a$. For positive a, |x|<=a includes all x within or at distance a from 0.

$x<-a$ or $x>a$. For positive a, |x|>a means x is more than distance a from 0, either left of -a or right of a.

$x\le -a$ or $x\ge a$. For positive a, |x|>=a includes all x at or beyond distance a from 0.

No solution; empty set. Since |x| is always >=0, it cannot be less than 0, yielding no solutions.

$x=0$. Since |x|>=0 and equals 0 only at x=0, |x|<=0 holds solely there.

$x\ne 0$. |x|>0 excludes only x=0 where |x|=0, including all other reals.

No solution; empty set. For negative a, |x|>=0 cannot be less than a negative number, so no solutions.