int x = 1; int y = 9; System.out.println( Math.sqrt( x/y ) );
0.0
The "trick" in this question is that the division of
x by y is the first thing done
in evaluating the statement.
Since both x and y are integers,
integer division is done and the result is integer 0.
Next, the integer 0 is converted to double precision 0.0,
and it is that value that is sent to sqrt(),
which computes 0.0 as a result.
Although the question was a "trick" this situation often occurs naturally in programs and you should watch out for it.
Does the following correct the problem?
int x = 1; int y = 9; System.out.println( Math.sqrt( (double)x/y ) );