Question

Use the well-ordering principle to show that if x and y are real numbers with x<y, then there is a rational number r with x<r<y. [Hint: Use the Archimedean property, given in Appendix 1, to find a positive integer A with .$\mathit{A}\mathbf{>}\mathbf{1}\mathbf{/}\mathbf{(}\mathit{y}\mathbf{-}\mathit{x}\mathbf{)}$ Then show that there is a rational number with denominator A between x and y by looking at the numbers $\mathbf{\lfloor }\mathbf{x}\mathbf{\rfloor }\mathbf{+}\mathbf{j}\mathbf{/}\mathbf{A}$, where is a positive integer.]

Short answer

The statement “there exists a rational number r with $x<r<y$” is true by assuming it is false with the help of well-ordering principle and the counterexamples.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The value of the first real number is x.
• The value of the second real number is y.
• The value of the rational number isr .
• The value of the first positive integer is A .
• The value of the second positive integer isj .

Significance of the principle of well ordering

The principle of well ordering mainly works with the division algorithm. The principle states that a particular set should contain a minimum of one element.

Proving the problem statement with the help of well ordering principle

Let the statement “there exists a rational number r with $x<r<y$” is not true because of the contradiction. Let the values set be $X\subseteq N$for which $n>0$and also $x<r<y$that has $A=1/(y-x)$and $r=\lfloor x\rfloor +\frac{n}{A}$. As the statement is not true X, then is a non-empty set.

Using the principle of well ordering X, has a minimum of one element. Let the least element be . It is to be noted that exists inside . X As $a\ne 0$, then$a>0$ and$a-1\in N$ .

As is the counterexample being the smallest element, then the counterexample if not$a-1$ as$a-1<x$ which satisfies$x<\lfloor x\rfloor +\frac{a}{A}<y$ .

Then $\lfloor x\rfloor +\frac{a}{A}$also satisfies the particular statement, that is clear contradiction which is “there exists a rational number r with $x<r<y$” is not true. Hence, by proving “the statement is false”, the statement becomes true.

Thus, the statement “there exists a rational number r with $x<r<y$” is true by assuming it is false with the help of well-ordering principle and the counterexamples.