What is Absolute value: Definition and 367 Discussions
In mathematics, the absolute value or modulus of a real number x, denoted |x|, is the non-negative value of x without regard to its sign. Namely, |x| = x if x is positive, and |x| = −x if x is negative (in which case −x is positive), and |0| = 0. For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3. The absolute value of a number may be thought of as its distance from zero.
Generalisations of the absolute value for real numbers occur in a wide variety of mathematical settings. For example, an absolute value is also defined for the complex numbers, the quaternions, ordered rings, fields and vector spaces. The absolute value is closely related to the notions of magnitude, distance, and norm in various mathematical and physical contexts.
First lets focus on ##|x|## which is defined as distance between ##x##and ##0##. But if we look into it closely
$$13=|-11-2|$$ which is distance between -11 and 2 but $$13=|11-(-2)|$$ which means this is distance between 11 and -2. Which is it?
In the same way $$x=|x-0|$$ is distance between 0...
Im a having trouble understanding how this exactly works.
$$ |x^2 - 4| < |x^2+2| $$ So I know the usual thing to do when you have absolute values,here it is even simpler since the right part of the inequality is always positive so I just have these 2 cases.
1. ## x^2-4 >= 0 ## and 2. ## x^2-4...
I was given a problem to solve that goes like this ##\frac{3}{|x+3|-1}\geq |x+2|## . I got the correct solution for all possible cases and here they are; for ##|x+3|\geq0## and ##|x+2|\geq## i got ##x\epsilon <-2, -2\sqrt{3} ]## and for ##|x+3|\leq0## , ##|x+2|\leq0## I got ##x\epsilon [-5...
I am a bit confused, so if anyone can explain to me which way is right I would be very thankful.
I think that the way in pic 1 is right because of the properties written next to the procedure but the professor who posts videos on youtube solved it the way as written in pic 2 where he didn't...
Problem statement : Let me copy and paste the problem to the right as it appears in the text.
Solution attempt (mine) : There are mainly three cases to consider.
(1) ##\boldsymbol{x\ge 3\; :}## Using the relevant equations given above, the problem statement reduces to $$x-3+x-2 = 1\Rightarrow...
Problem Statement : I copy and paste the problem from the text to the right.
Attempt (mine) : Given the inequality ##\dfrac{x}{x+2}\le \dfrac{1}{|x|}##. We see immediately that ##x\ne 0, -2##. At the same time, since ##|x|\ge 0\Rightarrow \frac{x}{x+2}\ge 0##.
Now if ##\frac{x}{x+2}\le...
(I could solve the problem but could not make sense of the solution given in the text. Let me put my own solutions below first).
1. Problem Statement : I copy and paste the problem to the right as it appears in the text.
2. My attempt : There are three "regions" where ##x## can lie.
(1)...
How does one manipulate rational absolute inequalities?
For example, I want to transform the absolute value inequality ##|x-3|<1## to ##\frac{|x+3|}{5x^2}<A \ ##, for some number ##\text{A}##, to find an upper and lower bound on the latter term using the constraint in the first term, and not...
See attachment.
I don't understand the solution given by David Cohen.
1. Note: x^2 is nonnegative for any real number x. This is because any value for x when squared is positive. Yes?
2. If x is greater than or equal to 0, then I can say that -2 - x^2 is negative in value, right?
3. What...
Absolute Value (algebraic version)
Rule:
| x | = x when x ≥ 0
| x | = -x when x > 0
Rewrite each expression without using absolute value notation.
Question 1
| x^4 + 1 |
I say x^4 + 1 is a positive value.
My answer is x^4 + 1.
Question 2
|-sqrt{3} - sqrt{5} |
The value of -sqrt{3} -...
Absolute Value (algebraic version)
Rule:
| x | = x when x ≥ 0
| x | = -x when x > 0
Rewrite each expression without using absolute value notation.
Question 1
|1 - sqrt{2} | + 1
The value 1 - sqrt{2} = a negative value.
So, -(1 - sqrt{2}) = - 1 + sqrt{2}.
When I put it all together, I get...
see attached image, it asks to repesent it in x-graph
constant "a" isn't conditioned.
Do I need to separate it into a few cases of the constant a and represent each in one x-graph?
Reals $x,\,y$ and $z$ satisfies $3x+2y+z=1$. For relatively prime positive integers $p$ and $q$, let the maximum of $\dfrac{1}{1+|x|}+\dfrac{1}{1+|y|}+\dfrac{1}{1+|z|}$ be $\dfrac{q}{p}$. Find $p+q$.
First, I try to define the function in the figure above: ##V(t)=100\left[sin(120\{pi}\right]##.
Then, I use the fact that absolute value function is an even function, so only Fourier series only contain cosine terms. In other words, ##b_n = 0##
Next, I want to determine Fourier coefficient...
Solve for y: $\quad |y+3|\le 4$
a.$\quad y \le 1$
b.$\quad y\ge 7$
c.$\quad -7\le y\le1$
d. $\quad -1\le y\le7$
e. $\quad -7\ge y \ge 1$
Ok I think this could be solved by observation but is risky to do so...
Hello everyone,
I've been struggling quite a bit with this problem, since I'm not sure how to approach it correctly. The inequality form reminds me of the equation of a circle (x^2 + y^2 = r^2), but I have no idea how to be sure about it. Would it help just to simplify the inequality in terms...
So far I've got the real part and imaginary part of this complex number. Assume: ##z=\sin (x+iy)##, then
1. Real part: ##\sin x \cosh y##
2. Imaginary part: ##\cos x \sinh y##
If I use the absolute value formula, I got ##|z|=\sqrt{\sin^2 {x}.\cosh^2 {y}+\cos^2 {x}.\sinh^2 {y} }##
How to...
How to solve these two absolute value problems?
1.
##|3x - 5| > |x + 2|##
My attempt:
From what I read in my textbook, the closest properties of absolute value is the one that uses "equal" sign
##|3x - 5| = |x + 2|##
##3x - 5 = x + 2##
##3x -x = 5 + 2##
##2x = 7##
##x = \frac{7}{2}##
##|3x -...
given
$|y+3|\le 4$
we don't know if y is plus or negative so
$y+3\le 4 \Rightarrow y\le 1$
and
$-(y+3)\le 4$
reverse the inequality
$ y+3 \ge -4$
then isolate y
$y \ge -7$
the interval is
$-7 \le y \le 1$
D={(x,y)∈ℝ2: 2|y|-2≤|x|≤½|y|+1}
I am struggling on finding the domain of such function
my attempt :
first system
\begin{cases}
x≥2y-2\\
-x≥2y-2\\
x≥-2y-2\\
-x≥-2y-2
\end{cases}
second system
\begin{cases}
x≤y/2+1\\
x≤-y/2+1\\
-x≤y/2+1\\
-x≤-y/2+1\\
\end{cases}
i draw the graph and get the...
Reading The Theoretical Minimum by Susskind and Friedman. They state the following...
$$\left|X\right|=\sqrt {\langle X|X \rangle}\\
\left|Y\right|=\sqrt {\langle Y|Y \rangle}\\
\left|X+Y\right|=\sqrt {\left({\left<X\right|+\left<Y\right|}\right)\left({\left|X\right>+\left|Y\right>}\right)}$$...
Problem Statement: Prove that |a|=|-a|
Relevant Equations: ##|a|= a, ## if ## a \geq 0 ## and
-a, if ## a \leq 0 ##
Somewhat stumped on where to start...
i know that we need to use cases. If we consider ##a\geq 0##, then are we allowed to use the fact that ##|-a|=|-1|\cdot|a| = |a| ##?
This...
In transforming an integral to new coordinates, we multiply the “volume” element by the absolute value of the Jacobian determinant.
But in the one dimensional case (where “change of variables” is just “substitution”) we do not take the absolute value of the derivative, we just take the...
Solve the absolute value equation.
|3x - 2|/|2x - 3| = 2
Solution:
|3x - 2| = 2|2x - 3|
3x - 2 = 2(2x - 3)
3x - 2 = 4x - 6
Solving for x, I get x = 4.
However, the textbook has two answers for this problem.
The answer is also 8/7.
How do I find 8/7?
I have the expression ##|nr^n|^{1/n}##. A quick question is whether I can allow the exponent to go inside of the absolute value. I know that if it were an positive integral exponent then because of the multiplicativity of the absolute value function that would be allowed. But I'm not sure what...
Homework Statement
Find k so that y - 36x = k is a normal to the curve y = 1 / abs(x-2).
Homework EquationsThe Attempt at a Solution
My problem is regarding the absolute value. I know that the tangent to the curve must be (-1/36). In the solutions manual, it is said that by knowing the sign...
Homework Statement
Find solutions to the given equality
2. Relevant equation
$$ x^2 +3|x-1|=1 $$
The Attempt at a Solution
The above can be rewritten as:
$$ |x-1| = {1-x^2\over 3}$$
If my understanding of absolute values is correct, the above simply means that:
$$ x-1 = {1-x^2\over 3}...
Homework Statement
1. Show that for all real numbers x and y:
a) |x-y| ≤ |x| + |y|
Homework Equations
Possibly -|x| ≤ x ≤ |x|,
and -|y| ≤ y ≤ |y|?
The Attempt at a Solution
I tried using a direct proof here, but I keep getting stuck, especially since this is my first time ever coming...
Rule:
Suppose a>0, then |x|>a if and only if x>a OR x<-a
So |x|>|x-1| becomes:
x>x-1 which is false (edit: or more accurately doesn't give the whole picture, it implies true for all x)
OR
x<-x+1
2x<1
x<1/2 which is false
Explain, in your own words, why there are no real numbers that satisfy the absolute value equation | x^2 + 4x | = - 12.
Can we say there is no real number solution here? If so, is the answer then imaginary taught in some advanced math class?
Homework Statement
Consider a continuous random variable X with the probability density function fX(x) = |x|/5 , – 1 ≤ x ≤ 3, zero elsewhere.
I need to find the cumulative distribution function of X, FX (x).
2. Homework Equations
The equation to find the cdf.
The Attempt at a Solution
FX(x)...
This is rather basic, and may be a misconception of the notation, however, I can't make the following sum up:
The following is given:
x_n(t) = 1 -nt , (if 0 <= t <= 1/n) and 0, (if 1/n < t <= 1)
However, this part I can't grasp this part in the book:
\begin{equation}
||x_n||^2 = \int_0^1...
Homework Statement
Rewrite |x| < 1 and |x| > 1 by eliminating the absolute value sign
Homework Equations
|x| < 1 = -1 < x < 1
|x| > 1 = ?
The Attempt at a Solution
I know that |x| < 1 can be rewritten as -1 < x < 1 but I'm not sure about |x| > 1. Am I right to assume that |x| > 1 = -1 > x > 1?
Hi there,
I'm having trouble understanding this math problem:
|x| + |x-2| = 2
The answer says its: 0<=x<=2
I understand you need different "cases" in order to solve this. For example, cases for when x is less than 0, when x-2 is less than 0, etc.
Thanks,
blueblast
I am reading Manfred Stoll's Book: "Introduction to Real Analysis" ... and am currently focused on Chapteer 4: Limits and Continuity ...
I need some help with an inequality involving absolute values in Example 4.1.2 (a) ...
Example 4.1.2 (a) ... reads as follows:
In the above text we read ...