Hi,

Correct me, If I am wrong,

**Quote:**

If d is a positive integer is \sqrt{d}an integer?

The Integer set contains, all the whole number and the negatives of all the natural number ie -1,-2,0,1,2

As per your statement, If Suppose 3 is a positive integer, \sqrt{3} will also be positive integer, which is not true as sqrt3 is approx 1.73, which by definition doesn't fall in integer set.

If & Only If d is a perfect square, the \sqrt{d} will a integer as in 121 will give +11, -11.

Do let me know, If I am missing something or mis-stated something.

Regards,

Tushar