Let's assume the following:
- x is a rational number.
- √y is an irrational number.


We can prove that x√y is an irrational number by contradiction.

Let's assume that x√y is a rational number. This means that we can express it in the form of a fraction where the numerator and denominator are integers and the denominator is not equal to zero.

Therefore, we can say that:

x√y = a/b

where a and b are integers, and b ≠ 0.

We can now square both sides of the equation to get:

x²y = a²/b²

Since x² is a rational number and y is an irrational number, we know that x²y is an irrational number.

On the other hand, a²/b² is a rational number because a and b are integers.

This creates a contradiction because we have shown that x²y is irrational and a²/b² is rational. Therefore, our assumption that x√y is rational must be false.


From the above proof, we can conclude that if x is rational and √y is irrational, then x√y is irrational.

let assume X +√y be rational no. then x+√y=p/q (Where p and q are integers and co-prime) => √y= p/q- x => √y is rational no. but, it contradict the fact √y is. irrational (I. e. given in question) hence, contradiction ,so x+√y is irrational. Thanks