For questions about or involving the absolute value function.

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9
votes
1answer
21k 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 ...
4
votes
4answers
598 views

The median minimizes the sum of absolute deviations

Suppose we have a set $S$ of real numbers. Show that $$\sum_{s\in S}|s-x| $$ is minimal if $x$ is equal to the median. This is a sample exam question of one of the exams that I need to take and I ...
8
votes
4answers
14k 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 :)
15
votes
6answers
2k 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.
7
votes
2answers
3k 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 ...
14
votes
4answers
641 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 $$A=\dfrac{2}{\sqrt{10}}$$ ...
1
vote
6answers
483 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,
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
166 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.
1
vote
4answers
211 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 ...
6
votes
3answers
533 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
135 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
569 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
329 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 $y < 0$ $x\geq0$ ...
2
votes
3answers
330 views

Question about solving absolute values.

I solved the following problem from by book, but the answer of this problem at the end of book is $x \leq 3$. Please tell me how I can get this answer.
1
vote
2answers
175 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
1k 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
3k 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 ...
2
votes
4answers
3k 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 ...
7
votes
2answers
167 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
4answers
16k views

Is the absolute value function a linear function?

I'm inclined to say yes, as it doesn't involve exponentiation, roots, logarithmic or trigonometric functions, but I watched a video where the teacher said that the absolute value function is "clearly ...
3
votes
2answers
341 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
346 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
2answers
63 views

Prove triangle inequality using the properties of absolute value

So I was given the task of proving the following variant of the triangle inequality using only the properties of the absolute value: $\vert\lvert x\rvert -\lvert y \rvert \rvert \leq \lvert ...
2
votes
2answers
248 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
1answer
6k 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
1answer
57 views

Continuity of absolute value

Let $f(x)$ be a continuous function. Prove that $\left|f(x)\right|$ is also continuous. Is it correct to say that, by the reverse triangle inequality, $\left|f(x)-f(c)\right| \geq ...
1
vote
1answer
178 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
0answers
35 views

integrate an absolute value periodic function

$$ \int_{-\frac{\pi}{2}}^{t} |\cos{t}|dt = \sin(t-\pi\lfloor(\frac{t}{\pi}+\frac{1}{2})\rfloor)+ 2\lfloor(\frac{t}{\pi}+\frac{1}{2})\rfloor $$ In know that this integral holds. It can be obtained by ...
0
votes
1answer
33 views

Finding $\lim_{t\to 0}\frac{|t-2|}{t}$ and $\lim_{t\to \infty}\frac{|t-2|}{t}$

Find $$\lim_{t\to 0}\frac{|t-2|}{t}$$ and $$\lim_{t\to\infty}\frac{|t-2|}{t}$$ Usually I would simply the top and bottom but I'm not sure what to do for absolute values. Any help would be ...
0
votes
2answers
57 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 ...
0
votes
3answers
584 views

Integrating absolute value function

I'm working on a problem drawing phase plane diagrams in my applied mathematics course. I'm supposed to draw the phase line diagram of $x''+\vert x\vert=0.$ In the process, I get to the differential ...
-2
votes
1answer
292 views

Example of a function $f$ which is nowhere continuous but $|f|$ should be continuous at all points [duplicate]

So I had an exam today and one of the questions were: Give an example of a function $f$ which is nowhere continuous but $|f|$ should be continuous at all points. At first I had no idea how to do it ...
7
votes
4answers
374 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 $n$ ...
3
votes
4answers
5k views

How to solve inequalities with absolute values on both sides?

If you have an inequality that has two absolute value bars like $|4x+1|<|3x|$, how do you go about doing this? I know that if $4x+1<3x$, then those $x$'s will work but what else do I do? I think ...
6
votes
2answers
651 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 ...
5
votes
5answers
258 views

How to write an expression in an equivalent form without absolute values?

The question I have in front of me is the very first problem in Trench's Introduction to Real Analysis: Write the following expression in equivalent form not involving absolute values: $a+b+|a-b|$ ...
4
votes
4answers
131 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 ...
3
votes
3answers
119 views

absolute value inequalities

When answer this kind of inequality $|2x^2-5x+2| < |x+1|$ I am testing the four combinations when both side are +, one is + and the other is - and the opposite and when they are both -. When I ...
6
votes
3answers
165 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 $x,y < 0$, $x \ge 0$ ...
4
votes
2answers
348 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 ...
2
votes
3answers
541 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
2answers
2k 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
800 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
127 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 $x<a, |x-a|=-(x-a)$ ...
0
votes
2answers
128 views

solving absolute value equation 2

My question is : Solve simultaneously- $$\left\{\begin{align*}&|x-1|+|y-2|=1\\&y = 3-|x-1|\end{align*}\right.$$ I tried to solve this question by the method told by Marvis as I had ...
7
votes
2answers
893 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
981 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
171 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
154 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 ...