Subtraction on Numbers is Not Associative

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Theorem

The operation of subtraction on the numbers is not associative.

That is, in general:

$a - \paren {b - c} \ne \paren {a - b} - c$


Proof

By definition of subtraction:

\(\ds a - \paren {b - c}\) \(=\) \(\ds a + \paren {-\paren {b + \paren {-c} } }\)
\(\ds \) \(=\) \(\ds a + \paren {-b} + c\)


\(\ds \paren {a - b} - c\) \(=\) \(\ds \paren {a + \paren {-b} } + \paren {-c}\)
\(\ds \) \(=\) \(\ds a + \paren {-b} + \paren {-c}\)

So we see that:

$a - \paren {b - c} = \paren {a - b} - c \iff c = 0$

and so in general:

$a - \paren {b - c} \ne \paren {a - b} - c$

$\blacksquare$


Examples

$5$ minus $3$ minus $2$

\(\ds 5 - \paren {3 - 2}\) \(=\) \(\ds 5 - 1\) \(\ds = 4\)
\(\ds \paren {5 - 3} - 2\) \(=\) \(\ds 2 - 2\) \(\ds = 0\)


Sources