In finite precision floating point arithmetic the associative property of addition is not satisfied. This is, it is not always the case that $$(a+b)+c=a+(b+c)$$ Even $a=(a+b)-b$ is not always true.

To prove that $x+y<z$ is equivalent to $x<z-y$ with real numbers we can add $-y$ on both sides of $x+y<z$ to get $(x+y)-y<z-y$ and then from this $x=x+(y-y)<z-y$. But I can't repeat the last step for floating point.

Question 1: Are the inequalities $x+y<z$ and $x<z-y$ equivalent in finite precision floating point arithmetic?

Question 2: Are the inequalities $x+y\leq z$ and $x\leq z-y$ equivalent in finite precision floating point arithmetic?

I was too hasty and updated without checking properly. What I really indented to ask is if the breaking of the equivalence could be as drastic as:

Question 3: Could we have $x+y<z$ and $x>z-y$?

Possibly there are already answers below for this. If so, I will check them when I get a chance and select the answer.

Motivation: There is that interview problem that given an ordered array of numbers and a threshold it asks for the number of pairs of numbers of the array which sum is not larger than the threshold. Possible solutions aside, I wanted to understand the particular point of the dangers of switching between testing $x+y<z$ (and also $\leq$ depending on which variant of the problem is presented) and testing $x<z-y$ (or $\leq$).


Consider the case that x and y hold the largest representable value for the current type, and z is positive infinity. By the rules of floating point arithmetics, x + y is now infinity, which isn't less than infinity. On the other hand, z - y is also infinity, which the largest representable value is less than.

Another example would be when x is very small negative, and y = z and much larger in magnitude than x. Then x + y = y = z, but z - y = 0, which is larger than x.

The examples work for your second question as well if you flip the signs of all numbers - !(l > r) is nearly equivalent to l <= r. NaNs break the equivalence, but those aren't useful as examples to your question anyways.

  • $\begingroup$ In both of these examples the failing inequality is an $=$. Do you know if there are examples in which the failing inequality has the reverse sign? A failure for the equivalence with $\leq$. $\endgroup$ – plop Jul 5 '20 at 13:14
  • $\begingroup$ I don't believe that's possible. The right hand side is closer than the result of an operation than any value below/above the right hand side value whichever is away from the accurate result. $\endgroup$ – John Dvorak Jul 5 '20 at 13:36

Take x=1+u, y=1+2u, z=2+4u, where u is the value of the lowest mantissa bit. x+y = 2+3u gets rounded up to 2+4u, so x+y<z is false. z-y = 2+4u - (1+2u) = 1+2u, so x < z-y is true. That’s a counter example for the first case.

With x=1+3u you get a counter example for the second case.

  • $\begingroup$ In the second example, do we have $x+y=\operatorname{round}(2+5u)=2+6u>2+4u=z$ and $x=1+3u>1+2u=z-y$? $\endgroup$ – plop Jul 6 '20 at 20:24

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