For questions about or involving the absolute value function.

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3
votes
1answer
12k views

Reverse Triangle Inequality Proof

I've seen the full proof of the Triangle Inequality $|x+y|\le|x|+|y| $ However, I haven't seen the proof of the reverse triangle inequality: $||x|-|y||\le|x-y|$ Could you please prove this using ...
13
votes
6answers
1k views

Show that the $\max{ \{ x,y \} }= \dfrac{x+y+|x-y|}{2}$.

Show that the $\max{ \{ x,y \} }= \dfrac{x+y+|x-y|}{2}$. I do not understand how to go about completing this problem or even where to start.
6
votes
2answers
2k views

How to use triangle inequality to establish Reverse triangle inequality

I need to use $|a+b| \leq |a|+|b|$ to show that $||a|-|b|| \leq |a-b|$. I have tried to represent $||a|-|b||$ as $||a|+(-|b|)|$, and then get $||a|+(-|b|)| \leq |a|+|-|b||$, but that isn't leading ...
2
votes
3answers
2k views

Proof of triangle inequality

I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)
1
vote
6answers
408 views

How could we solve $x$, in $|x+1|-|1-x|=2$?

How could we solve $x$, in $|x+1|-|1-x|=2$? Please suggest a analytical way that I could use in other problems too like this $ |x+1|+|1-x|=2$ and of this genre. Thank you,
14
votes
4answers
537 views

How find this inequality $\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$

let $a,b,c,d,e\in R$,and such $$a^2+b^2+c^2+d^2+e^2=1$$ find this value $$A=\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$$ I use computer have this $$A=\dfrac{2}{\sqrt{10}}$$ ...
5
votes
1answer
1k views

Prove variant of triangle inequality containing p-th power for 0 < p < 1

Sorry if this is a trivial question, but I am kind of stuck with proving the following inequality and have been searching for a while: $\rho \left( \sum\limits_i^n d_i \right) \leq \sum\limits_i^n ...
3
votes
4answers
151 views

Is $\sqrt{x^2}=|x|$ or $=x$? Isn't $(x^2)^\frac12=x?$ [duplicate]

$|x|=\sqrt{x^2}$ as Wolfram|Alpha shows. But, as $(x^2)^\frac12=x$, I can't understand where am I wrong interpreting Square-root.
6
votes
3answers
515 views

How prove this inequality: $\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|$? [duplicate]

Let $x_{1},x_{2},\cdots,x_{n}$ be real numbers. Show that $$\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|.$$ I think this problem may be solved using nice methods, but I can't find ...
4
votes
5answers
123 views

How is it, that $\sqrt{x^2}$ is not $ x$, but $|x|$?

As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is ...
4
votes
3answers
492 views

Proof for Integral Inequality $|\int f| \le \int |f|$ - is it sufficient enough?

Claim: If f is integrable, $\left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx$ Proof (attempt): We know $-|f|\le f \le|f|$, so $\int-|f| \le \int f \le \int|f|$.* Since, if $-b<a<b$, we say ...
2
votes
5answers
204 views

Prove the triangle inequality [duplicate]

I want to porve the triangle inequality: $x, y \in \mathbb{R} \text { Then } |x+y| \leq |x| + |y|$ I figured out that probably the cases: $x\geq0$ and $y \geq 0$ $x<0$ and $y < 0$ $x\geq0$ ...
1
vote
2answers
144 views

Solving $|x-2| + |x-5|=3$ [duplicate]

Possible Duplicate: How could we solve $x$, in $|x+1|-|1-x|=2$? How should I solve: $|x-2| + |x-5|=3$ Please suggest a way that I could use in other problems of this genre too Any help ...
10
votes
7answers
957 views

what does $|x-2| < 1$ mean?

I am studying some inequality properties of absolute values and I bumped into some expressions like $|x-2| < 1$ that I just can't get the meaning of them. Lets say I have this expression $$ ...
1
vote
3answers
2k views

Proving absolute value inequalities

I need help proving the last two cases for the following inequality: $\bigl|\lvert x\rvert-\lvert y\rvert\bigr| \le \lvert x-y\rvert$. Case 1: $x > 0$ and $y > 0$: the inequality simplifies ...
7
votes
2answers
166 views

Maximum of the difference

What is the maximum value of $f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{2012})$ where $x_{1}, x_{2}, … , x_{2012}$ are distinct integers in the set ${1, 2, 3, …, 2012}$ and $f$ is the absolute ...
3
votes
2answers
245 views

Prove the monotonicity of the expectation of a binary random variable function

Consider $R$ independent binary random variables $y^1, \ldots, y^R$ over the space $\{-1, +1\}$ such that $\Pr(y^j = 1) = p^j \geq 0.5$ and $\Pr(y^j = -1) = 1 - p^j$, $\forall j = 1,\ldots,R$. ...
3
votes
2answers
312 views

Why exactly can you take the absolute value of one side of this inequality and assume it is still true?

Exercise: Show that if $(b_n) \to b$, then the sequence of absolute values $\left| b_n \right|$ converges to $\left| b \right|$. Solution (partial): By the triangle inequality, ...
2
votes
1answer
5k views

Solving inequality with two absolute values

Hey, ! In my pre-calculus class the teacher showed the solution of the following example: \begin{align} \vert x-3 \vert \lt \vert x - 4 \vert + x \end{align} He started by stated the domains ...
1
vote
4answers
133 views

Proving the inequality $|a-b| \leq |a-c| + |c-b|$ for real $a,b,c$

Let $a,b,c$ real numbers. Prove the inequality $|a-b| \leq |a-c| + |c-b|$. Prove that equality holds if and only if $a \leq c \leq b$ or $b \leq c \leq a$. I've tried starting with just $a \leq ...
1
vote
1answer
173 views

Absolute value of a real number

My question is: Solve: $|x-4|< a$, where $a$ belongs to the real numbers. Solve this by considering various cases depending upon whether $a$ is negative, positive or zero. What I have tried ...
0
votes
2answers
46 views

Absolute value question false solution

|x| = 3x – 2 Why does this statement eventually give you a solution that isn't valid. So this equation comes out: x = 3x - 2 2 = 2x x = 1 OR x = -3x + 2 4x = 2 x = 1/2 However 1/2 doesn't ...
7
votes
4answers
307 views

Inequality for absolute values

How do you show either of the equivalent inequalities: $$2(|a|+|b|+|c|)\leq |a+b+c|+|a+b-c|+|a-b+c|+|a-b-c|$$ or $$|x+y|+|x+z|+|y+z|\leq |x|+|y|+|z|+|x+y+z|$$ Hold for complex numbers or in $n$ ...
4
votes
4answers
109 views

Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value.

Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value. I thought of it this way: $$f(x)=\begin{cases}2x-1-3(2x+4)+7 ...
1
vote
3answers
1k views

Proving square root of a square is the same as absolute value

Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't ...
5
votes
3answers
133 views

How would I prove $|x + y| \le |x| + |y|$?

How would I write a detailed structured proof for: for all real numbers $x$ and $y$, $|x + y| \le |x| + |y|$ I'm planning on breaking it up into four cases, where both $x,y < 0$, $x \ge 0$ ...
5
votes
2answers
357 views

Is there a lower-bound version of the triangle inequality for more than two terms?

The triangle inequality $|x+y|\leq|x|+|y|$ can be generalized by induction to $$|x_1+\ldots+ x_n|\leq|x_1|+\ldots+|x_n|.$$ Can we generalize the version $|x+y|\geq||x|-|y||$ to $n$ terms too? I need ...
3
votes
2answers
327 views

Why do definitions of distinct conic sections produce a single equation?

I understand how to get from the definitions of a hyperbola — as the set of all points on a plane such that the absolute value of the difference between the distances to two foci at $(-c,0)$ and ...
1
vote
2answers
949 views

Proof the maximum function $\max(x,y) = \frac {x +y +|x-y|} {2}$ [duplicate]

I want to prove the maximum function max: $\mathbb{R} \rightarrow \mathbb{R}$, which is defined by $$\max(x,y) = \begin{cases}x, \text { if } x \geq y , \\ y, \text { if } x < y \end{cases}$$ ...
1
vote
1answer
609 views

Prove by contradiction or contrapositive? If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$.

Prove: If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$ This looks as though it's true from the start. Take $x=-4, y=4$. $|-4+4|<|-4|+|4|$ $0<8$ is true. The question is asking for a ...
0
votes
1answer
105 views

Determine all the values of the parameter $a$ for which the inequality $3-|x-a|>x^2$ is satisfied by at least one negative $x$.

I wanted to know, how can I determine all the values of the parameter $a$ for which the inequality $3 - |x-a| > x^2$ is satisfied by at least one negative $x$. I tried for $x<a, |x-a|=-(x-a)$ ...
7
votes
2answers
433 views

How does one DERIVE the formula for the maximum of two numbers

I want to derive (not prove that this is true) the formula $\max (x,y) = \dfrac{x + y + |y-x|}{2}$ I was reading a proof (which they have the result ahead of time already) that we do cases and then ...
4
votes
3answers
774 views

Proving two integral inequalities

Can anyone help me to prove that these integral inequalities hold? Here $x$ is a real value: $$ \left| \int_a^b\ f(x) dx \right| \leq \int_a^b\ |f(x)| dx $$ Here $z$ is a complex value: $$ \left| ...
4
votes
3answers
167 views

Is there an alternate definition for $\{ z \in \mathbb{C} \colon \vert z \vert \leq 1 \} $.

Is there a method of constructing a subset of a reasonably arbitrary ring so that when the construction is applied the $\mathbb{C}$ the result is $B = \{ z \in \mathbb{C} \colon |z| \leq 1 \} $? My ...
3
votes
3answers
142 views

Absolute Value inequality help: $|x+1| \geq 3$

Find the solutions to the inequality: $$|x+1| \geq 3$$ I translate this as: which numbers are at least $3$ units from $1$? So, picturing a number line, I would place a filled in circle at the ...
3
votes
2answers
85 views

How prove this $|x_{p}-y_{q}|>0$

let $$x_{1}=\dfrac{1}{8},x_{n+1}=x_{n}+x^2_{n},y_{1}=\dfrac{1}{10},y_{n+1}=y_{n}+y^2_{n}$$ show that: for any $p,q\in N^{+}$ we have $$|x_{p}-y_{q}|>0$$
3
votes
4answers
216 views

the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$

what is the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$ ? I solved it by first assuming: $-1 < \frac{2x - 3}{2x + 3 } < 1$ ended with: $x > 0 > -3/2$ Is that a ...
2
votes
7answers
130 views

for $n$ an integer, why is $n^0=1$ ??

This is so going to cost me.... I was wondering why for any integer $n$: $n^0 =1$. Perhaps It's because $n$ is a round number and if $m$ is a non negative integer, also round then: $$n^m = 1 \cdot ...
2
votes
3answers
136 views

Inequalities - Absolute Value $|2x-1| \leq |x-3|$

$$|2x-1| \leq |x-3|$$ Answer is $$-2 \leq x \leq \frac43$$ My Question is HOW?
2
votes
3answers
192 views

Absolute ratios

I'm curious about the following idea: suppose we have two values $P$ and $Q$, and the magnitude of the ratio $\frac{P}{Q}$ is between $0$ and $\infty$. If $P$ is smaller, then it's between $0$ and ...
2
votes
2answers
233 views

Why is the derivative of $\frac{|x|}{x}$ equal to $\emptyset$ at $x=0$?

I got a bit of a confusion here. If $\varphi(x)=\frac{|x|}{x}$, then $$ \varphi(x) = \left.\Bigg\{ \begin{array}{cc} 1 &if \ x>0\\ \emptyset & if \ x=0\\ -1 & if \ x <0 \end{array} ...
2
votes
3answers
349 views

Solve an absolute value equation simultaneously

My question is : Solve simultaneously $$\left\{\begin{align*}&|x-1|-|y-2|=1\\&y = 3-|x-1|\end{align*}\right.$$ What I did : $y=3 - |x-1|$ is given. Thus $y = 3-(x-1)$ or $y = ...
1
vote
4answers
96 views

How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$?

I am able to solve equation of the type $ |5x+1|=|11-2x|$. I square both the side and my equation becomes $ (5x+1)^2=(11-2x)^2 $ further simplification gives me $ (5x+1)=\pm (11-2x)$. I get have ...
1
vote
3answers
72 views

Value and simplify

I want to find the value and simplify square root 36 ? Square root of 36 is 6 But I would know how to find the value and simplify it .
1
vote
2answers
59 views

Finding $x$ from inequality: $\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}$

Find $x$ in $\mathbb{Z}$ satisfying this inequality: $$\left | \frac{3^n + 2}{3^n + 1} - 1 \right | \le \frac{1}{28}.$$ I tried something, but I don't think it's correct. $$-\frac{1}{28} ...
1
vote
2answers
115 views

Solving $| ax + b | \gt c$

Does $| a x + b | > c$ always result in two solutions, $x \gt \dfrac{c - b}{a}$, and $x \lt\dfrac{-c - b}{a}$? If I understand correctly, the first solution, $x > \dfrac{c - b}{a}$, is only ...
1
vote
1answer
469 views

double integral of an absolute function

I'm just a little unsure of how to tackle this one. I understand that typically you would separate the integral into two for where x is positive or negative, I'm just unsure of how to separate it for ...
1
vote
0answers
88 views

Fields with their own absolute value

Let $F\hspace{.02 in}$ be a field. $\:$ Let $E\hspace{.02 in}$ be a non-zero subring of $F$. Let $\hspace{.03 in}\leq\hspace{.03 in}$ be a total order on $E\hspace{.02 in}$ that makes $E\hspace{.02 ...
0
votes
3answers
76 views

Absolute value quadratic inequalities not the usual?

$ | -x^2 + 6x | \gt 13 $,for example. I would start off solving $ -x^2 + 6x = \pm 13 $ and either get 4 solutions, 3 solutions or two simply do the the nature of the graph. Without knowing if the two ...
0
votes
1answer
16 views

Trying to differentte $\ln(|2+f(x)|)=2+e^{x*x}$

I am trying to solve this differential $\ln(|2+f(x)|)=2+e^{x*x}$ so far I did this much; $$ \ln(|2+f(x)|)=2+e^{x*x}\\ |2+f(x)|=e^{2+e^{x*x}}\\ \text{now I have two situations/solutions, because of ...