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Am I right in saying that the absolute value symbols act like a function such that if $x$, for example, is $x<0$ then $x=-x$

In other words $x$ will be positive regardless of what value you give to $x$, right?

In which case, isn't $|z|$ always positive? Considering $|z|=\sqrt{a^2+b^2}$, where $z=a+bi$.

Jean-Claude Arbaut
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Thomas Winkworth
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    Actually $|z|$ is always non-negative. It's not clear to me what your objection is. – Michael Albanese Apr 12 '15 at 18:03
  • $|z|$ is always nonnegative; $|0|=0$. – vadim123 Apr 12 '15 at 18:03
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    I can't tell if this is relevant, but notice that $\sqrt{z^2}$ is not necessarily a real number (e.g. if $z=i$ or any non-real number) and does not equal $|z|$. – Milo Brandt Apr 12 '15 at 18:10
  • Now, Thomas, the way your question is worded makes me think that you're complaining that the absolute value symbol is somehow redundant, unnecessary or pointless. Am I reading you wrong? – Robert Soupe Apr 13 '15 at 01:19
  • @Robert Soupe This question is half a rant and half a way of trying to understand why they chose the absolute value symbol. For real numbers they mean one thing, for complex numbers they mean another. I understand that they both find the distance from the origin. But in terms of them being a function it's very easy to get confused. I personally think that they should have used a different symbol for each, perhaps a symbol that looks similar but not identical. – Thomas Winkworth Apr 13 '15 at 09:09
  • Unlike 20th century composers, mathematicians have never liked inventing new symbols. If a symbol already exists that means something slightly similar to what they want, they prefer to overload that symbol. And if there's no existing symbol with a similar meaning, then maybe just overload parentheses with yet another meaning. –  Apr 13 '15 at 16:39

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Yes, the absolute value of a number is always non-negative (Zero if the input is zero, positive otherwise).

This is the point of the absolute value function; it is in some sense a measure of the size of the number. To be more precise, the absolute value of a number is the distance from the number to the origin.

Eff
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  • What do you mean? I'm not sure if $|z|$ can be equal to zero, however if it can all $|z|$ is telling you is that it is not less than zero, right? Since it'll always be nonnegative. – Thomas Winkworth Apr 12 '15 at 18:10
  • $|z| = 0$ if and only if $z=0$. Can you clarify a bit on what you are confused about? – Eff Apr 12 '15 at 18:11
  • Is this just with complex numbers? I thought that the absolute value symbols are essentially just used to "force" a positive result. Whereas with complex numbers it's seems they're used to represent a function to find the distance from the origin – Thomas Winkworth Apr 12 '15 at 18:17
  • Sorry "nonnegative" rather than positive – Thomas Winkworth Apr 12 '15 at 18:18
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    What I have written is true for both real and complex numbers. Consider the real number line, then the number $-5$ is of distance $5$ away from $0$ (the origin), hence $|-5| = 5$. For real numbers, however, this turns out to be the same as just "forcing the number to be non-negative." – Eff Apr 12 '15 at 18:21
  • Awesome, this is why I love math – Thomas Winkworth Apr 12 '15 at 18:30
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Am I right in saying that the absolute value symbols act like a function such that if $x$, for example, is $x<0$ then $x=-x$

Yes, if you would add bars: $\lvert x \rvert = -x$ for $x < 0$.

The definition is $$ \lvert x \rvert = \left\{ \begin{array}{cccl} x & \mbox{ for } & x > 0 & \Rightarrow \mbox{positive sign} \\ 0 & \mbox{ for } & x = 0 & \Rightarrow \mbox{no sign} \\ -x & \mbox{ for } & x < 0 & \Rightarrow \mbox{positive sign} \\ \end{array} \right. $$ Therefore $\lvert x \rvert$ is always positive or zero, thus non-negative.

In other words $x$ will be positive regardless of what value you give to $x$, right?

No, $\lvert 0 \rvert = 0$, which is not positive and not negative.

In which case, isn't $|z|$ always positive? Considering $|z|=\sqrt{a^2+b^2}$, where $z=a+bi$.

No, if $r = 0$, thus $z=0$, then $\lVert z \rVert = 0$, which is not positive.

Now we are talking about the complex absolute value, which can be viewed as 2-norm on $\mathbb{R}^2$: $$ \lVert z \rVert^2 = z \bar{z} = (\mbox{Re } z)^2+ (\mbox{Im } z)^2 $$ Among the properties it shares with the real version are: $$ \lVert z \rVert \ge 0 \\ \lVert z \rVert = 0 \iff z = 0 $$

mvw
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