A real number r is rational if and only if there exists two integer a and b such that
\(r = a/b\) ( b is not equal to zero)
Let us assume that x is an irrational number and y is a rational number .
Suppose for the sake of contradiction that \(x + y\) is not irrational
\(y - a/b\)
since s is irrational , x cannot be written as the ratio of two integers
Since \(x + y\) in not irrational,
\(x + y\) is rational .
By the definition of rational, there exists integers c and d not equal to zero such that
\(x + y = c/d\)

\(\text{since} y = a/b\)

\(x + a/b = c/d\)

\(x = c/d -a/b\)

\(x = (bc - ad)/bd\) Since a,b,c,d are integers bc - ad and bd are also integers. Moreover bd is zero as b and d are both non zero. However , this then implies that x is a rational number (as it is written as the ratio of tow integers). which is in contradiction with the fact that with the fact that x is an irrational number. Thus our supposition that \(x + y\) is not irrational is false. Which means that \(x + y\) is irrational .

\(\text{since} y = a/b\)

\(x + a/b = c/d\)

\(x = c/d -a/b\)

\(x = (bc - ad)/bd\) Since a,b,c,d are integers bc - ad and bd are also integers. Moreover bd is zero as b and d are both non zero. However , this then implies that x is a rational number (as it is written as the ratio of tow integers). which is in contradiction with the fact that with the fact that x is an irrational number. Thus our supposition that \(x + y\) is not irrational is false. Which means that \(x + y\) is irrational .