Why is the following true? $$ |x|= \begin{cases} x,&x\ge 0\\ -x,&x<0 \end{cases} $$ Isn't the modulus of a number always positive? According to the above formula $|-4|=-4$ because $4<0$, which is incorrect. Please explain this to me. Thank you.
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Ok. I got it. Thank you guys – Tamás Oct 25 '13 at 10:39
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Remember that $-x$ is positive when $x$ is negative. – ShreevatsaR Oct 25 '13 at 10:42
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It's a common misunderstanding that $-x$ is negative; read it “the (additive) opposite of $x$”, rather than “the negative of $x$” and you'll be OK. – egreg Oct 25 '13 at 10:45
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1Indeed, if $4<0$ then $|-4|=-4$. :( SCNR – Carsten S Oct 25 '13 at 17:38
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No, the formula says that $|-4|=-(-4)=4$. The $x$ in this case is $-4$, and $-(-4)=4$, not $-4$. When you get the value $-4$, you’re not applying the definition correctly: you’re taking the negative of $4$, not of $-4$, but the number inside the absolute value is not $4$.
Brian M. Scott
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According to your sentence : "Isn`t the modulus of a number always positive? According to the above formula |-4|=-4 because 4<0, which is incorrect." . You have said , "4<0" ? Is it right ? infact -4 < 0 . Infact , $ |-4| = -(-4) $ which is perfectly fine.
Way to infinity
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$\vert x\rvert\ge0$ for any $x \in \mathbb{R}$. The piecewise-defined function for $\lvert x\rvert$ is
$$ \vert x\rvert= \begin{cases} x& x>0\\ 0& x=0\\ -x& x<0 \end{cases} $$
We read it literally as
$$ \vert x\rvert= \begin{cases} x& \text{if $x$ is a positive real number}\\ 0& x=0\\ -x& \text{if $x$ is a negative real number} \end{cases} $$
If $x$ is a negative real number then $-x$ is a positive real number. I think it is obvious!
kiss my armpit
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The prefix unary operand - changes the sign of its operand. So a negative number's sign is changed to positive by this operation.
ncmathsadist
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