Question

let i be the set of real numbers that are not rational; elements of i are called irrational numbers. prove if a < b, then there exists x ∈ i √ suchthata

Answer 1

To prove that if a < b, then there exists an irrational number x such that a < x < b, we can show that there are infinitely many irrational numbers between any two rational numbers. Let's consider the square root of 2 as an example to demonstrate this.

Step-by-step explanation:

In order to prove the statement, "if a < b, then there exists x ∈ i √ such that a < x < b," we need to show that there exists an irrational number between any two real numbers. Let's assume a and b are two real numbers with a < b. Since there are infinitely many irrational numbers between any two rational numbers, we can conclude that there exists an irrational number x such that a < x < b.

To prove this, let's consider the square root of 2 (√2) which is an irrational number. Since √2 is between two consecutive perfect squares, we can see that it lies between 1 and 2. So, if a < 1 and b > 2, then a < √2 < b.

Thus, we have shown that if a < b, then there exists an irrational number x such that a < x < b.