int and double have different semantics. Consider division. 1/2 is 0, 1.0/2.0 is 0.5. In any given situation, one of those answers will be right and the other wrong.

That said, there are programming languages, such as JavaScript, in which 64-bit float is the only numeric data type. You have to explicitly truncate some division results to get the same semantics as Java int. Languages such as Java that support integer types make truncation automatic for integer variables.

In addition to having different semantics from double, int arithmetic is generally faster, and the smaller size (32 bits vs. 64 bits) leads to more efficient use of caches and data transfer bandwidth.

Doubles can be imprecise due to floating point rounding errors. For example, you would never use doubles if you were writing banking software that handles money. For example, try computing 1.89 * 792 in Java. Mathematically, you should get 1496.88 as the result. However, because of rounding errors, the value that Java (and other languages) produce is 1496.8799999999999. It's very close to 1496.88, but it's not exactly that amount. If you keep performing computations, the error will only increase in range, and then suddenly you're left wondering why your answers are off by 1 or 2, even when there is nothing wrong with the operations mathematically.

when you divide a by b i.e a/b

if both a & b are int then result will be int
else any or both a & b are double then result will be double

Edit:

Also see my answer to this Question Simple Divide Problem

int times = 5; int[ ] primes = new int [times]; int primeIndex = 0; int primesFound = 0; boolean isPrime = true; for (int i = 2; ; i++) { for (int j = 2; j < i; j++) { if (i % j == 0) { isPrime = false; break; } } if (isPrime) { primes [primeIndex++] = i; primesFound++; } else { isPrime = true; } if (primesFound == times) { break; } }