For questions about or involving the absolute value function.

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9
votes
1answer
24k views

Reverse Triangle Inequality Proof

I've seen the full proof of the Triangle Inequality \begin{equation*} |x+y|\le|x|+|y|. \end{equation*} However, I haven't seen the proof of the reverse triangle inequality: \begin{equation*} ...
4
votes
4answers
832 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 ...
8
votes
4answers
18k 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 :)
16
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
4k 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 ...
2
votes
2answers
83 views

Proving that $x_n\to L$ implies $|x_n|\to |L|$, and what about the converse?

Problem 3. Show that for a sequence $(x_n)$ the following are true: (i) $\lim x_n=0$ if and only if $\lim |x_n|=0$. (ii) $\lim x_n=L$ implies $\lim |x_n|=|L|$. Is the converse true? Prove or ...
14
votes
4answers
666 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 $$A=\dfrac{2}{\sqrt{10}}$$ ...
3
votes
3answers
4k views

Equality holds in triangle inequality iff both numbers are positive, both are negative or one is zero

How do we show that equality holds in the triangle inequality $|a+b|=|a|+|b|$ iff both numbers are positive, both are negative or one is zero? I already showed that equality holds when one of the ...
1
vote
6answers
526 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,
3
votes
4answers
4k 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
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
174 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.
3
votes
2answers
600 views

When does the equality hold in the triangle inequality? [duplicate]

Hi guys could you please help me on this question I'm confused. question: when does the equality hold in the triangle inequality: my attempt : $|x + y| \leq |x| + |y|$ this implies $(|x+y|)^2 = ...
1
vote
4answers
230 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
534 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
143 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
574 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
364 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
337 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
182 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 $$ ...
3
votes
4answers
7k 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 ...
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 ...
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
17k 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
370 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
350 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
138 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
252 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
7k 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
62 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
2answers
2k views

Prove that the absolute value of a product is the product of the absolute values of factors.

Theorem. $|a||b|=|ab|$ Proof. Applying the definition of absolute value, the left hand side of the equation could be either $a\times(-b)$ or $(-a)\times(b)$ or $a\times b$ or $(-a)\times(-b)$. For ...
1
vote
1answer
182 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
40 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
66 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
600 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
327 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
390 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 $n$ ...
6
votes
2answers
728 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
344 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
144 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
122 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 ...
7
votes
2answers
1k 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 ...
6
votes
3answers
222 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
359 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
630 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
844 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
75 views

Is there a number whose absolute value is negative?

I've recently started to think about this, and I'm sure a couple of you out there have, too. In Algebra, we learned that $|x|\geq0$, no matter what number you plug in for $x$. For example: ...