How does the closure property prove that the sum of a rational and irrational number is irrational?

Answer 1

The closure property itself does not directly prove that the sum of a rational and an irrational number is irrational. However, the concept of closure is closely related to this proof.

The closure property states that if you perform an operation (such as addition) on two elements of a certain set (such as the set of rational numbers), the result will also be an element of that set. In other words, when you add two rational numbers, the result is always a rational number.

To prove that the sum of a rational and an irrational number is irrational, we can use proof by contradiction.

Let's suppose that the sum of a rational and an irrational number is rational. So, we have a rational number "a" and an irrational number "b", and their sum is a rational number "c". Mathematically, we can express this as:

a + b = c

We know that the rational numbers are a closed set under addition. Hence, the sum of two rational numbers is always rational. Therefore, the sum of the rational number "a" and the irrational number "b" (which is assumed to be rational) should also be rational.

Now, we can rewrite the equation as follows:

b = c - a

Since "c" is rational (as we assumed), and "a" is rational, the difference c - a must also be rational (since rational numbers are closed under subtraction). However, we initially assumed that "b" was irrational, which contradicts our assumption.

Hence, the assumption that the sum of a rational and an irrational number is rational must be incorrect. Therefore, the sum of a rational and an irrational number is always irrational.

This proof relies not only on the concept of closure (which ensures that the difference c - a remains within the set of rational numbers), but also on the properties of rational and irrational numbers.