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Is a - b + c - d = a + c - b - d mathematically correct?

I believe this statement can be correct but only sometimes if the order of evaluation does not matter so if I were to do {(a - b) + c} - d and choose numbers that would evaluate to {(a + c) - b} - d where b and c are both the same numbers then this may be correct.

Is there a more mathematical and logical explanation for this?

I also think it has to do with associativity but that would prove that this statement is never correct since addition and multiplication are associative (separately) but not addition and subtraction together

C2K
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  • If the values are all 0 then it is always correct. The same problem with dibision - it's only correct when the denominator is not 0. So you can't tell if it is correct without knowing the values. – Jerry Jeremiah Feb 03 '19 at 20:31
  • The folks at [Math SE](https://math.stackexchange.com/) might have something to say about this. – Him Feb 03 '19 at 20:33
  • It is mathematically correct, at least by convention of what + 'means' (i.e. that it refer to a commutative associative binary operation etc). But it's not true in programming, for example, (-9999999999999999 + 1 + 1 === -9999999999999999 + (1 + 1)) === false (in javascript). – Countingstuff Feb 03 '19 at 20:44

1 Answers1

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This highly depends on the definition of + and -. So far as you have written, they are but free untyped infix symbols so it's hard to tell.

A simple example. Suppose values are of fixed-width floating point type (like one of those defined in IEEE-754, for instance). Next, if we have

a = 10e100
b = -10e-100
c = -10e100
d = -10e-100

and the expressions are evaluated greedily left-to-right, then

a - b + c - d = ((a - b) + c) - d

When the type has enough order bits to contain decimal orders of -100 and 100, but its mantissa is not wide enough to correctly represent 10e100 + 10e-100, specifically, the RHS argument is simply lost in this expression, then the value of the whole large expression is

((10e100 - -10e-100) + -10e100) - -10e-100 =
= (10e100 + -10e100) - -10e-100 = 0 - -10e-100 = 10e-100

But the second expression would evaluate to

((a + c) - b) - d = ((10e100 + -10e100) - -10e-100) - -10e-100 = 20e-100

So you see, the result can differ by 100% depending on the order of evaluation.

bipll
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