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"# Algebra Revisited 2018\n",
"## Modern Algebra: A Logical Approach, Book Two, \n",
"## circa 1966 [Allen, Pearson]\n",
"## Chapter 6: Functions and Other Relations\n",
"\n",
"# Bounded Functions \n",
"\n",
"p.325 (exercises 3, 5, 6) \n",
"p.326 (exercise 8 proof)"
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"source": [
"### (3) \n",
"### M = {$(x, y)\\mid y = \\sqrt{x}\\; \\land\\; x\\in[0, \\infty)$}"
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"%display typeset\n",
"x = var('x')\n",
"f(x) = sqrt(x)\n",
"domain = (x, 0, 1000)\n",
"p = plot(f(x), domain)\n",
"p.show(xmin=-10, xmax=50, ymin=-2, ymax=10)\n"
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"Since $x \\ge 0,\\; f(x) \\ge 0$ \n",
"Thus M is bounded below. \n",
"However, M is not bounded above."
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""
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"source": [
"limit(f(x), x=0, dir='+')"
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"limit(f(x), x=oo)"
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"$\\lim\\limits_{x \\to 0^+} \\sqrt{x} = 0$ \n",
"$\\lim\\limits_{x \\to \\infty} \\sqrt{x} = \\infty$ \n",
"### (5)\n",
"### S = {$(x, y)\\mid y = \\frac{1}{x-4}\\; \\land\\; x\\gt 4$}"
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"%display typeset\n",
"x = var('x')\n",
"f(x) = 1/(x - 4)\n",
"domain = (x, 4, 1000)\n",
"p = plot(f(x), domain)\n",
"p.show(xmin=0, xmax=10, ymin=-1, ymax=10)"
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"source": [
"Since $x \\gt 4,\\; f(x) \\gt 0$ \n",
"Thus S is bounded below. \n",
"However, S is not bounded above."
]
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"limit(f(x), x=4, dir='+')"
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"limit(f(x), x=oo)"
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"$\\lim\\limits_{x \\to 4^+} \\frac{1}{x-4} = +\\infty$ \n",
"$\\lim\\limits_{x \\to \\infty} \\frac{1}{x-4} = 0$"
]
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"source": [
"### (6)\n",
"### T = {$(x, y)\\mid y = \\frac{1}{x^2+2}$} "
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"x = var('x')\n",
"f(x) = 1/(x^2+2)\n",
"p = plot(f(x), (x, -10, 10))\n",
"p.show(xmin=-10, xmax=10, ymin=0, ymax=1)"
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"\n",
"Since $x \\in \\mathbb{R} \\land x^2 \\ge 0$, the Absolute Maximum is $f(0) = \\frac{1}{2}$ \n",
"$\\lim\\limits_{x \\to \\infty} \\frac{1}{x^2+2} = 0$ \n",
"$\\lim\\limits_{x \\to -\\infty} \\frac{1}{x^2+2} = 0$ \n",
"## (8) \n",
"#### Given that F = {$(x,y) \\mid y = \\frac{2x}{x^2+1}$}, show that 1 is the absolute maximum of F. \n",
"\n",
"We will need to use Theorem 8-3 and it's corollaries.\n",
"\n",
"If $a,b \\in \\mathbb{R}$, \n",
"then ($ab \\gt 0 \\longleftrightarrow$ a and b have like signs) \n",
"and ($ab \\lt 0 \\longleftrightarrow$ a and b have opposite signs).\n",
"\n",
"Corollary 1: \n",
"(1) If $a \\in \\mathbb{R}$, then $a^2 \\ge 0$ \n",
"(2) If $a \\in \\mathbb{R} \\land a \\ne 0$, then $a^2 \\gt 0$ \n",
"\n",
"Proof: \n",
"If a = 0, then 0*0 = 0 \n",
"If a $\\ne 0$, then the two factors of a*a are the same and have the same sign. It follows from Theorem 8-3 that $a^2 \\gt 0$. \n",
"\n",
"Corollary 2: 1 > 0 and -1 < 0 \n",
"Proof that 1 > 0 follows from the fact that 1 = 1*1. \n",
"Then the proof that -1 < 0 follows from the Corollary to Theorem 5-3. \n",
"Theorem 5-3: If $a, b \\in \\mathbb{R}$, then $a \\gt b \\longleftrightarrow a - b \\lt 0$ \n",
"\n",
"Corollary to Theorem 5-3: A real number is negative if and only if its opposite is positive, and a real number is negative if and only if its opposite is positive. $\\\\$\n",
"\n",
"Corollary 3: If $a \\in \\mathbb{R}$ and $a \\ne 0$, then a and $\\frac{1}{a}$ have like signs. $\\\\$\n",
"\n",
"To prove that 1 is the absolute maximum of F = {$(x,y) \\mid y = \\frac{2x}{x^2 + 1}$}, we need to prove \n",
"(i) $\\frac{2x}{x^2 + 1} = 1$ is true when $x = 1$, and \n",
"(ii) $\\frac{2x}{x^2 + 1} \\le 1$ is true. \n",
"\n",
"We know that $\\frac{2x}{x^2 + 1} \\le 1$ if $2x \\le x^2 + 1$, and that $2x \\le x^2 + 1$ is true if $0 \\le x^2 - 2x + 1 = (x-1)^2$. This suggests a proof of the statement, \"If $x \\in \\mathbb{R}$, then $\\frac{2x}{x^2+1} \\le 1$\" \n",
"\n",
"(1) $\\;$ $x \\in \\mathbb{R} \\longrightarrow (2)\\; (x-1) \\in \\mathbb{R} \\longrightarrow (3)\\; (x-1)^2 \\ge 0 \\longrightarrow (4)\\; x^2 - 2x + 1 \\ge 0 \\longrightarrow (5)\\; x^2 + 1 \\ge 2x \\longrightarrow (6)\\; 1 \\ge \\frac{2x}{x^2+1}$.\n",
"$\\\\$ \n",
"REASONS: \n",
"(1)$\\;$ Given, (2)$\\;$ $\\mathbb{R}$ is closed under addition, (3)$\\;$ Theorem 8-3 Corollary 1, (4)$\\;$ Distributive property, (5)$\\;$ Addition property of inequality, (6)$\\;$ Multiplication property of inequality. \n",
"\n",
"\n",
"NOTE: In step 6 we assumed that $\\frac{1}{x^2+1}$ was positive. This can be proved as follows: \n",
"$\\\\$ \n",
"(1)$\\;$ $x \\in \\mathbb{R} \\longrightarrow (2)\\; x^2 \\ge 0$, (3)$\\;$ $1 \\gt 0 \\longrightarrow (4)\\; x^2 + 1 \\gt 0 \\longrightarrow (5)\\; \\frac{1}{x^2+1} \\gt 0$. \n",
"\n",
"REASONS: (1)$\\;$ Given, (2)$\\;$ Theorem 8-3 Corollary 1, (3)$\\;$ Theorem 8-3 Corollary 2, (4)$\\;$ Addition propert of inequality, (5)$\\;$ Theorem 8-3 Corollary 3."
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"x = var('x')\n",
"f(x) = 2*x/(x^2+1)\n",
"p = plot(f(x), (x, -10, 10))\n",
"p.show(xmin=-10, xmax=10, ymin=-2, ymax=2)"
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