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Chapter 1: Problem 11

If \(a, b \in \mathbb{R}\) with \(a

Short Answer

Expert verified
In this exercise, we have proven that there are infinitely many rational and irrational numbers between any two real numbers \(a\) and \(b\) with \(a < b\). For rational numbers, we showed that taking the midpoint between the numbers repeatedly generates infinitely many rational numbers between \(a\) and \(b\). For irrational numbers, we considered the product of irrational weights and the sum of the numbers, which generates infinitely many irrational numbers between \(a\) and \(b\).

Step by step solution

01

Show that there exist infinitely many rational numbers between \(a\) and \(b\)

To show that there are infinitely many rational numbers between \(a\) and \(b\), let's consider the midpoint between \(a\) and \(b\). This midpoint can be found by taking the average of \(a\) and \(b\), which is \(\frac{a+b}{2}\). Now, we have a rational number between \(a\) and \(b\), as the midpoint will always be between \(a\) and \(b\). We can repeat this process indefinitely by taking a new midpoint between the old midpoint and either \(a\) or \(b\). In each step, we will find a new rational number between the old midpoint and one of the original numbers \(a\) or \(b\). Therefore, there are infinitely many rational numbers between \(a\) and \(b\).
02

Show that there exist infinitely many irrational numbers between \(a\) and \(b\)

In order to show that there are infinitely many irrational numbers between \(a\) and \(b\), we can use a similar approach as for rational numbers. Let's consider the number \(\frac{\sqrt{2}(a+b)}{2}\). We know that \(\sqrt{2}\) is an irrational number, and, since the product of an irrational and a rational number is irrational, we also know that \(\frac{\sqrt{2}(a+b)}{2}\) is irrational. Now, we have an irrational number between \(a\) and \(b\). We can repeat this process indefinitely by considering a new irrational number between the old irrational number and either \(a\) or \(b\) (for example, by taking the weighted average of the old irrational number and one of the original numbers \(a\) or \(b\) using irrational weights, such as \(\sqrt{2}\)). In each step, we will find a new irrational number between the old irrational number and one of the original numbers \(a\) or \(b\). Therefore, there are infinitely many irrational numbers between \(a\) and \(b\). So, we have shown that there exist infinitely many rational numbers as well as infinitely many irrational numbers between \(a\) and \(b\).

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Most popular questions from this chapter

Use the algebraic properties and the order properties of \(\mathbb{R}\) to prove that (i) \(a^{2}>0\) for any \(a \in \mathbb{R}, a \neq 0\) (ii) Given \(a, b \in \mathbb{R}\) with \(0

Given any \(p(x), q(x) \in \mathbb{R}[x]\), not both zero, a polynomial \(d(x)\) in \(\mathbb{R}[x]\) satisfying (i) \(d(x) \mid p(x)\) and \(d(x) \mid q(x)\), and (ii) \(e(x) \in \mathbb{R}[x], e(x) \mid p(x)\) and \(e(x)|q(x) \Longrightarrow e(x)| d(x)\) is called a greatest common divisor, or simply a GCD, of \(p(x)\) and \(q(x)\). In case \(p(x)=q(x)=0\), we set the GCD of \(p(x)\) and \(q(x)\) to be 0 . Prove that for any \(p(x), q(x) \in \mathbb{R}[x]\), a GCD of \(p(x)\) and \(q(x)\) exists, and is unique up to multiplication by a nonzero constant, that is, if \(d_{1}(x)\) as well as \(d_{2}(x)\) is a GCD of \(p(x)\) and \(q(x)\), then \(d_{2}(x)=c d_{1}(x)\) for some \(c \in \mathbb{R}\) with \(c \neq 0\). Further, show that any GCD of \(p(x)\) and \(q(x)\) can be expressed as \(u(x) p(x)+v(x) q(x)\) for some \(u(x), v(x) \in \mathbb{R}[x]\). (Hint: Consider a polynomial of least degree in the set \(\\{u(x) p(x)+v(x) q(x)\) : \(u(x), v(x) \in \mathbb{R}[x]\) with \(u(x) p(x)+v(x) q(x) \neq 0\\} .)\)

Let \(I\) be an interval and \(f: I \rightarrow \mathbb{R}\) be any function. (i) If \(f\) is monotonically increasing as well as monotonically decreasing on \(I\), then show that \(f\) is constant on \(I\). (ii) If \(f\) is convex as well as concave on \(I\), then show that \(f\) is given by a linear polynomial (that is, there are \(a, b \in \mathbb{R}\) such that \(f(x)=a x+b\) for all \(x \in I)\).

Show that \(n ! \leq 2^{-n}(n+1)^{n}\) for every \(n \in \mathbb{N}\), and that equality holds if and only if \(n=1\).

Let \(S\) be a nonempty subset of \(\mathbb{R}\). If \(S\) is bounded above, then show that the set \(U_{S}=\\{\alpha \in \mathbb{R}: \alpha\) is an upper bound of \(S\\}\) is bounded below, \(\min U_{S}\) exists, and \(\sup S=\min U_{S} .\) Likewise, if \(S\) is bounded below, then show that the set \(L_{S}=\\{\beta \in \mathbb{R}: \beta\) is a lower bound of \(S\\}\) is bounded above, \(\max L_{S}\) exists, and inf \(S=\max L_{S}\)
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