f(x,y) = 333.75y6 + x2(11x2y2 - y6 -121y4 - 2) + 5.5y8 + x/2y
for x = 77617 and y = 33096. All numerical inputs in this calculation are exact machine numbers, so any errors we get in the result are due to the computation. Looking at the computed results from a Fortran program, which Rump did on a IBM S/370, and others have repeated on many other machines, we see that when using single precision the result is
f = 1.172603940053178...
The fact that the answer does not change with increasing precision is often taken as confirmation that the correct answer has been obtained. However, the correct answer is, in fact,
f = -0.827396059946...
So we didn't even get the sign right!
The problem here is due to rounding errors, combined with other difficulties, such as cancellation errors, that are inherent in the use of floating point arithmetic. A frequent reaction when people see this example is "so what, this will never happen to me" and "even if it does happen to me, it will be no big deal." So consider now a couple of real world examples.