Show that the sum of rational no with an irrational no is always irrational?

Answer 1

Suppose x is a rational number and y is an irrational number.
Let x + y = z, and assume that z is a rational number.

The set of rational number is a group.
This implies that since x is rational, -x is rational [invertibility].
Then, since z and -x are rational, z - x must be rational [closure].

But z - x = y which implies that y is rational.

That contradicts the fact that y is an irrational number. The contradiction implies that the assumption [that z is rational] is incorrect.

Thus, the sum of a rational number x and an irrational number y cannot be rational.

Answer 2

Suppose x is a rational number and y is an irrational number.

Let x + y = z, and assume that z is a rational number.

The set of rational number is a group.

This implies that since x is rational, -x is rational [invertibility].

Then, since z and -x are rational, z - x must be rational [closure].

But z - x = y which implies that y is rational.

That contradicts the fact that y is an irrational number. The contradiction implies that the assumption [that z is rational] is incorrect.

Thus, the sum of a rational number x and an irrational number y cannot be rational.