Absolute Value


For a real number, the absolute value of \(x\), denoted \( \lvert x \rvert \), is defined as

x={x,if x0x,if x<0. \left|x \right| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0. \end{cases}


Evaluate 3(1755) \left| 3\left( 17-55 \right) \right| .

3(1755)=3(38)=114=114 \begin{aligned} \left| 3\left( 17-55 \right) \right| &= \left| 3\left( -38 \right) \right| \\ &= \left| -114 \right| \\ &= 114 _\square \end{aligned}


If x=5+11i2 x=\frac{5+\sqrt{11}i}{2} and y=511i2 y=\frac{5 -\sqrt{11}i}{2} , what is the value of x+3y2 \left| x + 3y \right|^2 ?

x+3y2=5+11i2+3(511i2)2=5+11i+15311i22=20211i22=1011i2=(102+112)2=1112=111 \begin{aligned} \left| x + 3y \right|^2 &= \left| \frac{5+\sqrt{11}i}{2} + 3\left( \frac{5 -\sqrt{11}i}{2} \right) \right|^2 \\ &= \left| \frac{5+\sqrt{11}i + 15 -3\sqrt{11}i}{2} \right|^2 \\ &= \left| \frac{20-2\sqrt{11}i}{2} \right|^2 \\ &= \left| 10 - \sqrt{11}i \right|^2 \\ &= \left( \sqrt{10^2 + \sqrt{11}^2} \right)^2 \\ &= \sqrt{111}^2 \\ &= 111 _\square \end{aligned}

Application and Extensions

What is the absolute value of 210a 2 - 10a ?

The answer depends on the value of a a . From the definition above, we have

210a={210a,if 210a0(210a),if 210a<0 \left|2-10a \right| = \begin{cases} 2-10a, & \text{if } 2-10a \geq 0 \\ -(2-10a), & \text{if } 2-10a < 0 \end{cases}

Solving the inequality on the right, we obtain

210a0210a15a. \begin{aligned} 2-10a &\geq 0 \\ 2 & \geq 10a \\ \frac{1}{5} &\geq a. \end{aligned}

It follows that 210a0a15. 2-10a \geq 0 \Longleftrightarrow a \leq \frac{1}{5}.


210a={210a,if a1510a2,if a>15 \left|2-10a \right| = \begin{cases} 2-10a, & \text{if } a \leq \frac{1}{5} \\ 10a - 2, & \text{if } a > \frac{1}{5} \end{cases} _\square

Note by Arron Kau
7 years, 4 months ago

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thank you very much :-)

Shovik Nandy - 7 years, 3 months ago

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I like this format of note, I think I'll Start Share So !

Explanation More Exercises Solved !

Gabriel Merces - 7 years, 4 months ago

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I didn't quite understand the problem having i i . At one stage the i i disappears. Can you please explain?

Shabarish Ch - 7 years, 4 months ago

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It is because the absolute value of a complex number (a+bia + bi) is defined as a2+b2\sqrt{a^2 + b^2} , i.e., a+bi|a + bi| = a2+b2\sqrt{a^2 + b^2}. In the above question, 1011|10 - \sqrt11| = 102+112\sqrt{10^2 + \sqrt{11}^2}

Aman Bansal - 7 years, 4 months ago

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Didn't know that, thanks!!

Shabarish Ch - 7 years, 4 months ago

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I am not sure but as far as my mind think I can say this is because when the sign of absolute value was removed.Then sign was changed to positive & i(iota) has nothing to do when there is is postive root.So,It would have been removed...

Archiet Dev - 7 years, 4 months ago

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ii is defined as 1\sqrt{-1} i.e. imaginary unit. So, those were complex numbers (real number + imaginary numbers).

Vishnuram Leonardodavinci - 7 years, 4 months ago

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Archiet Dev - 7 years, 4 months ago

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In ending step ..note that to take modulus of imag eniry number is different than simple modulas

Puneet Mehra - 6 years, 9 months ago

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