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Chapter 1: Problem 20

Prove that the absolute value function, that is, \(f: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x)=|x|\), is not a rational function.

Short Answer

Expert verified
We assumed that the absolute value function \(f(x) = |x|\) is a rational function and found that it must be written as \(f(x) = \frac{P(x)}{Q(x)}\), where P(x) and Q(x) are polynomials and Q(x) is not the zero polynomial. We then observed that P(x) must be equal to x in the positive domain and -x in the negative domain, as the function is continuous and undefined nowhere. However, this is not a polynomial, as a polynomial is required to be continuous and differentiable everywhere. This contradiction implies that the absolute value function is not a rational function.

Step by step solution

01

The absolute value function is defined as \(f(x) = |x|\), which can be written as \(f(x) = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}\) #Step 2: Assume that the function is rational#

Assume that the absolute value function is a rational function, which means it can be written as the quotient of two polynomials: \(f(x) = \frac{P(x)}{Q(x)}\), where P(x) and Q(x) are polynomials and Q(x) is not the zero polynomial. #Step 3: Observe properties of the polynomials #
02

Notice that if f(x) is a rational function, then both P(x) and Q(x) must have the same degree on both the positive and negative domains. This is because f(x) has the same behavior on both domains, i.e., it is continuous and undefined nowhere. #Step 4: Consider polynomial behavior for x >= 0 #

For \(x \ge 0\), we have \(f(x) = x\). Therefore, the rational function must reduce to a polynomial of degree 1 for nonnegative x-values. In other words, we must have \(Q(x)=1\) for \(x\ge0\), since the degree of Q(x) cannot change. This implies that Q(x) is a constant polynomial. #Step 5: Consider polynomial behavior for x < 0 #
03

For \(x < 0\), we have \(f(x) = -x\). Now the rational function must reduce to a polynomial of degree 1 for negative x-values. Since Q(x) is a constant polynomial, we must have: \(-x = -P(x)\). #Step 6: Show contradiction #

We found that for \(x\ge0\), we have \(x = P(x)\) and for \(x <0\), we have \(-x = - P(x)\). Therefore, P(x) must be equal to x in the positive domain and -x in the negative domain. However, this is not a polynomial, because a polynomial is required to be continuous and differentiable everywhere. Thus, we have reached a contradiction, which means that our assumption that f(x) = |x| is a rational function is incorrect. So, we conclude that the absolute value function is not a rational function.

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Most popular questions from this chapter

Let \(p(x), q(x) \in \mathbb{R}[x]\) be relatively prime polynomials of positive degree. Show by an example that the polynomials \(u(x), v(x) \in \mathbb{R}[x]\) such that \(u(x) p(x)+v(x) q(x)=1\) need not be unique. Show, however, that there are unique \(u(x), v(x) \in \mathbb{R}[x]\) such that \(u(x) p(x)+v(x) q(x)=1\) and either \(u(x)=0\) or deg \(u(x)<\operatorname{deg} q(x)\). In this case show that either \(v(x)=0\) or \(\operatorname{deg} v(x)<\operatorname{deg} p(x) .\) (Hint: Exercise 58.)

Let \(p(x), q_{1}(x)\) and \(q_{2}(x)\) be nonzero polynomials in \(\mathbb{R}[x]\) of degrees \(m, d_{1}\), and \(d_{2}\), respectively. Assume that \(q_{1}(x)\) and \(q_{2}(x)\) are relatively prime and that \(d_{1}\) and \(d_{2}\) are positive. If \(m

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 \(n \in \mathbb{N}\) and let \(a_{1}, \ldots, a_{n}\) and \(b_{1}, \ldots, b_{n}\) be any real numbers. Assume that not all \(a_{1}, \ldots, a_{n}\) are zero. Consider the quadratic polynomial $$ q(x)=\sum_{i=1}^{n}\left(x a_{i}+b_{i}\right)^{2} $$ Show that the discriminant \(\Delta\) of \(q(x)\) is nonnegative, and \(\Delta=0\) if and only if there is \(c \in \mathbb{R}\) such that \(b_{i}=c a_{i}\) for all \(i=1, \ldots, n\). Use this to give an alternative proof of Proposition \(1.12 .\)

Let \(m, n \in \mathbb{Z}\). Show that \(m\) and \(n\) are relatively prime if and only if \(\operatorname{GCD}(m, n)=1\). Also show that \(m\) and \(n\) are relatively prime if and only if \(u m+v n=1\) for some \(u, v \in \mathbb{Z} .\) Is it true that if a positive integer \(d\) satisfies \(u m+v n=d\) for some \(u, v \in \mathbb{Z}\), then \(d=\operatorname{GCD}(m, n) ?\)
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