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2019 Paper 2 Q3
D: 1500.0 B: 1500.0
polynomials triangle inequality proof by induction roots of polynomials inequality geometric series bound on roots rational root theorem

For any two real numbers \(x_1\) and \(x_2\), show that $$|x_1 + x_2| \leq |x_1| + |x_2|.$$ Show further that, for any real numbers \(x_1, x_2, \ldots, x_n\), $$|x_1 + x_2 + \cdots + x_n| \leq |x_1| + |x_2| + \cdots + |x_n|.$$

  1. The polynomial f is defined by $$f(x) = 1 + a_1 x + a_2 x^2 + \cdots + a_{n-1} x^{n-1} + x^n$$ where the coefficients are real and satisfy \(|a_i| \leq A\) for \(i = 1, 2, \ldots, n-1\), where \(A \geq 1\).
    1. If \(|x| < 1\), show that $$|f(x) - 1| \leq \frac{A|x|}{1 - |x|}.$$
    2. Let \(\omega\) be a real root of f, so that \(f(\omega) = 0\). In the case \(|\omega| < 1\), show that $$\frac{1}{1 + A} \leq |\omega| \leq 1 + A. \quad (*)$$
    3. Show further that the inequalities \((*)\) also hold if \(|\omega| \geq 1\).
  2. Find the integer root or roots of the quintic equation $$135x^5 - 135x^4 - 100x^3 - 91x^2 - 126x + 135 = 0.$$

Solution: Claim: \(|x_1 + x_2| \leq |x_1| + |x_2|\) Proof: Case 1: \(x_1, x_2 \geq 0\). The inequality is equivalent to \(|x_1 + x_2| = x_1 + x_2 = |x_1|+|x_2|\) so it's an equality. Case 2: \(x_1, x_2 \leq 0\). The inequality is equivalent to \(|x_1+x_2| = -x_1-x_2 = |x_1|+|x_2\), so it's also an equality in this case. Case 3: (wlog) \(|x_1| \geq |x_2| > 0\) and \(x_1x_2 < 0\) then \(|x_1+x_2| = x_1-x_2 \leq x_1 \leq |x_1|+|x_2|\) We can prove this by induction, we've already proven the base case and: \(|x_1+x_2 + \cdots + x_n| \leq |x_1 + x_2 + \cdots x_{n-1}| + |x_n| \leq |x_1| + |x_2| + \cdots + |x_n|\)

  1. \(\,\) \begin{align*} && |f(x) - 1| &= |a_1 x + a_2x^2 + \cdots + a_{n-1}x^{n-1} + x^n| \\ &&&\leq |a_1x| + |a_2x^2| + \cdots + |a_{n-1}x^{n-1}| + |x^n| \\ &&&\leq |a_1||x| + |a_2||x|^2 + \cdots + |a_{n-1}||x|^{n-1} + |x|^n \\ &&&\leq A|x| + A|x|^2 + \cdots + A|x|^{n-1} + |x|^n \\ &&&=A|x| \frac{1-|x|^{n-1}}{1-|x|} + |x|^n \\ &&&= \frac{A|x|-A|x|^{n}+|x|^{n+1}-|x|^n}{1-|x|} \\ &&&= \frac{A|x|-|x|^n(\underbrace{A-|x|+1}_{\geq0})}{1-|x|} \\ &&&\leq \frac{A|x|}{1-|x|} \end{align*}
  2. If \(f(\omega) = 0\) then \begin{align*} && 1 & \leq \frac{A|\omega|}{1-|\omega|} \\ \Leftrightarrow && 1-|\omega| &\leq A |\omega| \\ \Leftrightarrow && 1 &\leq (1+A) |\omega| \\ \Leftrightarrow && \frac{1}{1+A} &\leq |\omega| \\ \end{align*} We also know \(\omega \leq 1 < 1 + A\)
  3. If \(\omega\) is a root of \(f(x)\) then \(1/\omega\) is a root of \(1 + a_{n-1}x + a_{n-2}x^2 + \cdots + a_1x^{n-1}+x^n\) and so \(1/\omega\) satisfies that inequality, ie \begin{align*} && \frac{1}{1+A} && \leq &&|1/\omega| && \leq &&1 + A \\ \Leftrightarrow &&1+A && \geq&& |\omega| && \geq&& \frac{1}{1 + A} \end{align*}
  4. First notice that it's equivalent to: \(0 = x^5 - 1x^4 - \frac{100}{135}x^3-\frac{91}{135}x^2-\frac{126}{135} + 1\) therefore all integer roots must be between \(-2,-1\) and \(1\) and \(2\). \(1\) doesn't work. \(-1\) works. Clearly \(2\) cannot work by parity argument, therefore the only integer root is \(-1\).

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2011 Paper 3 Q2
D: 1700.0 B: 1516.0
polynomials rational root theorem proof by contradiction irrationality integer roots divisibility monic polynomial

The polynomial \(\f(x)\) is defined by \[ \f(x) = x^n + a_{{n-1}}x^{n-1} + \cdots + a_{2} x^2+ a_{1} x + a_{0}\,, \] where \(n\ge2\) and the coefficients \(a_{0}\), \(\ldots,\) \(a_{{n-1}}\) are integers, with \(a_0\ne0\). Suppose that the equation \(\f(x)=0\) has a rational root \(p/q\), where \(p\) and \(q\) are integers with no common factor greater than \(1\), and \(q>0\). By considering \(q^{n-1}\f(p/q)\), find the value of \(q\) and deduce that any rational root of the equation \(\f(x)=0\) must be an integer.

  1. Show that the \(n\)th root of \(2\) is irrational for \(n\ge2\).
  2. Show that the cubic equation \[ x^3- x +1 =0 \] has no rational roots.
  3. Show that the polynomial equation \[ x^n- 5x +7 =0 \] has no rational roots for \(n\ge2\).

Solution: Let \(\f(x) = x^n + a_{{n-1}}x^{n-1}+ \cdots + a_{2} x^2+ a_{1} x + a_{0}\), and suppose \(f(p/q) = 0\) with \((p,q) = 1\), the consider \begin{align*} && 0 &= q^{n-1}f(p/q) \\ &&&= \frac{p^n}{q} + \underbrace{a_{n-1}p^{n-1} + a_{n-2}p^{n-2}q + \cdots + a_0q^{n-1}}_{\in \mathbb{Z}} \\ \end{align*} But \(p^n/q \not \in \mathbb{Z}\) unless \(q = 1\) therefore \(p/q\) must be an integer, ie all rational roots are integers.

  1. Note that \(\sqrt[n]2\) is a root of \(x^n - 2 =0\), but this has no integer solutions. (We can try all factors of \(2\)). Therefore all its roots must be irrational, ie \(\sqrt[n]2\) is irrational for \(n \geq 2\)
  2. If \(n\) is a root of \(x^3 - x+1\) then it must be \(1\) or \(-1\) by the rational root theorem, ie \(1-1+1 \neq 0\) and \(-1 + 1 +1 \neq 0\), therefore no integer roots, therefore no rational roots.
  3. Suppose \(m\) is an integer root of \(x^n - 5x + 7 = 0\) then by considering parity we must have \(m^n - 5m + 7 \equiv 1 \pmod{2}\) therefore we cannot have any rational roots.

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