Say that `x` is very close to `n0.5`. What do you think will be the value of `n/(x*x)`?

### A good answer might be:

It should be very close to `1.0`

# Ending Test

But, due to the lack of precision with single precision, the best we can hope for is to calculate an `x` such that:

    0.999999  <  n/x*x  <  1.00001

The left and the right bound in the above have 6 decimal digits of precision. If 1.00000 is subtracted from this value the result is:

    -0.000001  <  n/x*x - 1  <  0.00001

To be safe, terminate the loop when:

    | n/x*x - 1 |  <  0.00001

... where `|x|` means the absolute value of `x`.

### QUESTION 12:

Is there a floating point absolute value instruction?