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Chapter 1: Problem 54

Show that a nonzero polynomial in \(\mathbb{R}[x]\) of degree \(n\) has at most \(n\) roots in \(\mathbb{R}\).

Short Answer

Expert verified
Using induction, we showed that a polynomial of degree 1 has at most 1 root. We then assumed the statement holds for all polynomials of degree n-1 and showed that it also holds for a polynomial of degree n. Thus, a nonzero polynomial in \(\mathbb{R}[x]\) of degree n has at most n roots in \(\mathbb{R}\).

Step by step solution

01

Prove the base case (polynomial of degree 1)

Let's consider a polynomial of degree 1: \(P(x) = ax+b\), where \(a \neq 0\). To find the roots of this polynomial, we need to solve the equation \(ax+b = 0\). We find the only root by solving for x: \[x = -\frac{b}{a}\] Since there is only one root, the statement holds for a polynomial of degree 1.
02

Inductive assumption (polynomial of degree n-1)

Assume the statement is true for all polynomials of degree n-1: A nonzero polynomial in \(\mathbb{R}[x]\) of degree n-1 has at most n-1 roots in \(\mathbb{R}\).
03

Prove for a polynomial of degree n

Now, consider a polynomial of degree n: \(P(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0\), with \(a_n \neq 0\). Suppose P(x) has a root r. Then, by the factor theorem: \(P(x) = (x-r)Q(x)\), where Q(x) is a polynomial of degree n-1. Now, let's find the remaining roots of the polynomial P(x). Since we know r is a root, we can focus on the polynomial Q(x), which has a degree of n-1. By the inductive assumption, we know that Q(x) has at most n-1 roots. This means the polynomial P(x) has at most n-1 roots in addition to the root r, giving a total of at most n roots in \(\mathbb{R}\).
04

Conclusion

Using induction, we have shown that a nonzero polynomial in \(\mathbb{R}[x]\) of degree n has at most n roots in \(\mathbb{R}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Real Analysis
Real analysis is a fundamental part of mathematics that deals with real numbers and real-valued functions. It lays the groundwork for understanding continuity, limits, and calculus in general. In the context of our exercise, real analysis provides us with the necessary tools to discuss polynomials and their roots within the real number system.

When dealing with polynomials, a key concept in real analysis is the notion that a polynomial of degree n can produce a graph with at most n distinct crossings with the x-axis, which represent its roots. The degree of a polynomial signals the highest power of the variable x, and it determines the maximum number of solutions that can exist for the equation P(x) = 0, where P(x) is the polynomial in question.
Inductive Proof
Inductive proof is a powerful mathematical technique used to show that a statement holds true for all natural numbers. It consists of a base case and an induction step. The base case verifies that the statement holds for the initial value, often n=1. The induction step involves assuming the statement for n=k, where k is any natural number, and then proving that the statement holds for n=k+1.

In our exercise, the inductive proof starts with verifying a polynomial of degree 1 has exactly one root, which is the base case. Then, the induction hypothesis assumes any polynomial of degree n-1 has at most n-1 roots. Using this assumption, the exercise then demonstrates that a polynomial of degree n must have at most n roots. This type of logical progression ensures the validity of the statement for all degrees of polynomials.
Factor Theorem
The factor theorem is a vital concept from algebra that connects the roots of a polynomial to its factors. The theorem states that if a number r is a root of the polynomial P(x), then \(x-r\) is a factor of P(x). This means P(x) can be expressed as \(P(x) = (x-r)Q(x)\), where Q(x) is another polynomial with one degree less than P(x).

In the step-by-step solution, the factor theorem plays a crucial role in reducing our problem to a polynomial of lower degree. After identifying a root r of the polynomial P(x), the problem is simplified to finding roots of Q(x), a polynomial with a degree n-1, which by the induction assumption, has at most n-1 roots. Consequently, P(x) has at most n roots, including r, which completes the proof.

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Most popular questions from this chapter

Show that if \(n \in \mathbb{N}\) and \(a_{1}, \ldots, a_{n}\) are nonnegative real numbers, then \(\left(a_{1}+\cdots+a_{n}\right)^{2} \leq n\left(a_{1}^{2}+\cdots+a_{n}^{2}\right) .\) (Hint: Write \(\left(a_{1}+\cdots+a_{n}\right)^{2}\) as \(t_{1}+\cdots+t_{n}\) where \(\left.t_{k}:=a_{1} a_{k}+a_{2} a_{k+1}+\cdots+a_{n-k+1} a_{n}+a_{n-k+2} a_{1}+\cdots+a_{n} a_{k-1} .\right)\)

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 all \(p(x), q(x) \in \mathbb{R}[x]\), 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 every 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 \(a_{1}, \ldots, a_{n}\) be positive real numbers. Prove that $$ \sqrt[n]{a_{1} \cdots a_{n}} \geq \frac{n}{r}, \quad \text { where } \quad r:=\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}, $$ and that equality holds if and only if \(a_{1}=\cdots=a_{n}\). [Note: The above result is sometimes called the G.M.-H.M. Inequality and \(n / r\) is called the harmonic mean of \(\left.a_{1}, \ldots, a_{n} .\right]\)

Let \(I_{n}=\left[a_{n}, b_{n}\right], n \in \mathbb{N}\), be closed and bounded intervals in \(\mathbb{R}\) such that \(I_{n} \supseteq I_{n+1}\) for each \(n \in \mathbb{N}\). If \(x=\sup \left\\{a_{n}: n \in \mathbb{N}\right\\}\) and \(y=\inf \left\\{b_{n}: n \in \mathbb{N}\right\\}\) then show that \(x \in I_{n}\) and \(y \in I_{n}\) for every \(n \in \mathbb{N}\).

Using only the algebraic properties A1-A5 on page 2, prove the following. (i) 0 is the unique real number such that \(a+0=a\) for all \(a \in \mathbb{R}\). In other words, if some \(z \in \mathbb{R}\) is such that \(a+z=a\) for all \(a \in \mathbb{R}\), then \(z=0\). (ii) 1 is the unique real number such that \(a \cdot 1=a\) for all \(a \in \mathbb{R}\). (iii) Given any \(a \in \mathbb{R}\), an element \(a^{\prime} \in \mathbb{R}\) such that \(a+a^{\prime}=0\) is unique. (As noted before, this unique real number \(a^{\prime}\) is denoted by \(-a .\).) (iv) Given any \(a \in \mathbb{R}\) with \(a \neq 0\), an element \(a^{*} \in \mathbb{R}\) such that \(a \cdot a^{*}=1\) is unique. (This unique real number \(a^{*}\) is denoted by \(a^{-1}\) or by \(1 / a .\) ) (v) Let \(a \in \mathbb{R}\). Then \(-(-a)=a .\) Further, if \(a \neq 0\), then \(\left(a^{-1}\right)^{-1}=a\). (vi) Let \(a, b \in \mathbb{R}\). Then \(a(-b)=-(a b)\) and \((-a)(-b)=a b\).
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