Questions tagged [absolute-value]

For questions about or involving the absolute value function also known as modulus function.

The absolute value function, usually denoted by $|x|$, is a function $\mathbb{R} \to [0, \infty)$ which can be defined in three equivalent ways:

  1. $|x| = \begin{cases}x &\ \text{if $x \ge 0$, and} \\ -x &\ \text{if $x < 0$.} \end{cases}$

  2. $|x| = \sqrt{x^2}$, and

  3. $|x| = \max \{x, -x\}$.

This definition extends to complex numbers as the square root of the norm: $|x+iy|=\sqrt{x^2+y^2}$. In both cases, the function may be interpreted geometrically as the distance of the input number from the origin.


More generally, an absolute value may be defined on an field (or integral domain) $k$ as a function $|\cdot | : k \to \mathbb{R}$ which satisfies the axioms

  1. (nonnegativity) $|x| \ge 0$ for all $x \in k$,

  2. (definiteness) $|x| = 0 \iff x = 0 \in k$,

  3. (multiplicativity) $|x y| = |x||y| $ for all $x,y\in k$ ), and

  4. (triangle inequality) $|x+y| \le |x| + |y|$ for all $x,y\in k$.

For example, if $p$ is a fixed prime number and $x \in \mathbb{Q}$, then there exists a unique $n \in \mathbb{Z}$ such that $x$ may be written as $$ x = p^n \frac{a}{b}, $$ where $\gcd(p, a) = \gcd(p, b) = 1$. The function which maps $x$ to $p^{-n}$ is an absolute value on $\mathbb{Q}$, called the $p$-adic absolute value.

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Solve $|1 + x| < 1$

I'm trying to solve $|1 + x| < 1$. The answer should be $ -2 < x < 0$ which wolframalpha.com agrees with. My approach is to devide the equation to: $1+x < 1$ and $1-x < 1$ and then solve those two: $ 1+x < 1 $ $ x < 0 $ $ 1 - x < 1$ $ -x < 0$ $x >…
iveqy
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$|x| - |y| \leq |x-y|?$

Is there a clever way to show that $$|x| - |y| \leq |x-y|$$ I believe I can think of a way to solve this using cases, but the book I'm working out of said that "A very short proof is possible if you write things in the right way" which has motivated…
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For which $a$ does $\lvert x+1\rvert+\lvert 2-x\rvert=a^2 -1$ have exactly two solutions?

If it is not a problem, I would really appreciate if someone could explain to me how to solve and graph the following equation: For which real numbers $a$ does the equation $$\lvert x+1\rvert +\lvert 2-x\rvert=a^2 -1$$ Have exactly two solutions?
vedran
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Solving $|\frac{x+1}{x}|< 1$

I need some help/suggestions solving the following math problem. I don't know how to continue from step 2. Find x. 1.) $\displaystyle\left|\frac{x+1}{x}\right|< 1$ 2.) $\displaystyle\frac{|x+1|}{|x|} < 1$
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no. of positive integral solutions of ||x - 1| - 2| + x = 3

What are the no. of positive integral solutions of ||x - 1| - 2| + x = 3 ? My effort ||x - 1| - 2| = 3 - x |x - 1| - 2 = 3 - x OR |x - 1| - 2 = x - 3 |x - 1| = 5 - x OR |x - 1| = x - 1 x - 1 = 5 - x OR x - 1 = x - 5 OR x - 1 $\geq$ 0 2x = 6 OR…
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Modulus Inequality

Solve the inequality $$2|x-3| > |3x+1|$$ Is sketching the only way I can solve ALL modulus equations and inequalities? Does an algebraic technique work for all modulus equations and inequalities?
user140161
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Real parameter equation

I'm having a problem with this question: For which values of the real parameter a the equation: $$||x|-1|=a$$ has exactly 4 solutions? The solution is this: $$0 < a < 1$$ What I tried was this: 1) Make the first condition, $a>0$, Split the equation…
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Need help on proofs

I'm not familiar with proofs and I dunno if I am doing them right. Here's one from Spivak's Calculus: 14. (a) Prove that |a| = |-a| (The trick is not to become confused by too many cases.) So I did: $$ |x| = \left\{\begin{aligned} &x &&: x \ge…
Kat
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Finding x-intercept of absolute values

What is the x-intercept of $|x-12|+8$? I don't know how to solve it with the absolute values and whether there will be two answers.
Jess
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Find values of the parameter m for absolute value equation to have two positive solutions

I solved an equation with absolute value and parameter and I got two different results when I applied two methods for this question. Could you please help me to spot and explain what am I doing wrong? Question: Find values of the parameter $m$, for…
Dracks
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Absolute value manipulation

Is it always true to say that if $|a|=b$ then $a=\pm b$ or am I missing something?
Salar
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$\left|x-1\right|=1-x$ solve the equation.

$\left|x-1\right|=1-x$ $\left|x-1\right| \Rightarrow \{ x-1, x\geq1; -x+1, x\lt1 \}$ When $x\geq1, x-1=1-x \Rightarrow 2x=2 \Rightarrow x=1$ When $x\lt1, -(x-1)=1-x \Rightarrow -x+1=1-x \Rightarrow 0=0$ so, true for all x All I can do is just…
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Can I Multiply Absolute Value Expressions

I know that for any two real numbers $a$ and $b$, that $$|a||b| = |ab|$$ And so I'd imagine that, assuming $x\in\mathbb{R}$, $$|x-a||x-b|=|(x-a)(x-b)|$$ since what is inside the absolute values are still real numbers. Still, I have not been able to…
Vpie649
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Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$

I'm solving the equation, $$|x-1| + |x-2| = 1$$ I'm making cases, $C-1, \, x \in [2, \infty) $ So, $ x-1 + x-2 = 1 \Rightarrow x= 2$ $C-2, \, x \in [1, 2) $ $x-1 - x + 2 = 1 \Rightarrow 1 =1 \Rightarrow x\in [1,2) $ $C-3, \, x \in (- \infty, 1)$ $…
William
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