Are x<=y and x-y<=0 equivalent in floating point arithmetic?

Question

Assume that x,y are two floating point numbers. Then is that true:

x<=y <==> x-y<=0

in floating point arithmetic?

Thank for your ideas.

[Edit] Let us assume, additionally, that neither x nor y are NaN.

Is that possible that x<=y holds but not x-y<=0 or x-y<=0 holds but not x<=y.

Answer 1

[Note: I'm ignoring infinities and NaNs in this answer, as either trivially leads to non-equivalence.]

If you've disabled subnormal numbers (or flush-to-zero behaviour), then it's possible to produce an underflow with the subtraction, resulting in non-equivalence between your two expressions.

For example:

#include <stdio.h>
#define CSR_FLUSH_TO_ZERO (1 << 15)

// Note: GCC-specific
void disable_ftz(void) {
    unsigned csr = __builtin_ia32_stmxcsr();
    csr |= CSR_FLUSH_TO_ZERO;
    __builtin_ia32_ldmxcsr(csr);
}

int main(void) {
    disable_ftz();

    float x = 2.8e-45;
    float y = 1.4e-45;

    printf("%e\n", x);           // 2.802597e-45
    printf("%e\n", y);           // 1.401298e-45

    printf("%d\n", x <= y);      // 0
    printf("%d\n", (x-y) <= 0);  // 1
    return 0;
}

Note that this requires some compiler-specific magic on an x86. However, it's permissible to have a floating-point implementation that doesn't have subnormals at all, and on such systems no magic would be required to achieve the same non-equivalence.