I teach three interchangeable definitions of the absolute value function, encouraging students to choose a favorite and be able to swap between them freely.

In each definition, the absolute value function is a [real function][(\(\mathbb{R}\to\mathbb{R},\) meaning that for every real input there is exactly one real output)] that I denote with \(\DeclareMathOperator{\absop}{abs}


This is the geometric definition, the intuition and the purpose.

\(\abs{x}\) represents the distance between \(x\) and the origin. This is equivalent to the definition that \(\abs{x-y}\) represents the distance between \(x\) and \(y\). (Why?)(why-equiv-distance)

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We will show equivalence by first assuming the distance definition for \(\abs{x}\) and deriving that of \(\abs{x-y}\) from it, and second by assuming the definition for \(\abs{x-y}\) and deriving \(\abs{x}\) from it.

Going from \(\abs{x}\) as the distance between \(x\) and the origin \(0\) to \(\abs{x-y}\) is a matter of shifting<!TODO> to the right by \(y.\) This means that \(\abs{x-y}\) as a function of \(x\) feels like \(\abs{x}\) if the origin were at \(y\) instead of \(0,\) and thus we have \(\abs{x-y}\) as the distance between \(x\) and \(y\) instead of \(0.\)

Going the other direction is easy: Simply let \(y=0,\) and then the distance \(\abs{x-y}\) between \(x\) and \(y=0\) becomes \(\abs{x},\) and so \(\abs{x}\) must be the distance between \(x\) and the origin.

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This is the definition most courses use, the most direct for computation.

When computing the value of \(\abs{x},\) if \(x\) isn't negative, leave it alone; if it is, swap it to positive. In precise notation, that means:

\abs{x} = \begin{cases}
x & \text{if $x\ge 0$} \\
-x & \text{if $x< 0$}

Square Root

This is a convenient trick.

For calculators that don't have an absolute value, you can use \(\abs{x}=\sqrt{x^2}\) to compute it. This is also surprisingly useful for proving a few properties, especially for understanding why the solution to \(x^2=4\) is \(x=\pm 2.\) (Why?)(why-square-root-solution)

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After taking the square root of both sides of \(x^2=4,\) we get \(\sqrt{x^2}=2.\) Many students try to simplify \(\sqrt{x^2}\) to \(x,\) but as mentioned above and proven below, \(\sqrt{x^2}=\abs{x}\) instead. So we have \(\abs{x}=2,\) which is easy to solve using one of the other definitions to get \(x=\pm2.\)

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How do we know these definitions are interchangeable?

The first two definitions are equivalent somewhat by fiat: we define the distance of a nonnegative number from the origin to be the number itself. It then makes sense that the distance for a negative number be the opposite of itself, since you've gone that far in the negative direction.

This square root definition works because the square root function is actually the principal square root function. (That is...)(that-is-principal-square-root)

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When looking for a square root of \(y,\) you want to find an \(x\) for which \(x^2=y,\) and when there is a solution, there's usually two, since \((-x)^2=(-1)^2x^2=x^2.\) When there's a positive and a negative choice for square roots, the principal square root function \(\sqrt{y}\) is conveniently defined to return the positive choice.

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So that means \(\abs{x}=\sqrt{x^2}\) is the same as saying \(\abs{x}^2=x^2\) together with \(\abs{x}\ge 0.\) When \(x\) is also not negative, this is the same as \(\abs{x}=x,\) and when \(x\) is negative, it is \(\abs{x}=-x,\) so this definition is equivalent to the piecewise definition.