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{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 54 "High School Modul
es > Precalculus by Gregory A. Moore\n" }}{PARA 3 "" 0 "" {TEXT -1 4 "
    " }{TEXT 256 22 "Mathematical Induction" }}{PARA 0 "" 0 "" {TEXT 
-1 75 "\nAn exploration of the method of mathematical induction to pro
ve formulae.\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directions : Execute
 the Code Resource section first. Although there will be no output imm
ediately, these definitions are used later in this worksheet.]" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " \+
" }{TEXT 260 7 "0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "r
estart; with(plots): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "StandardPlot  :=  proc( ex
pr )\n   \n end proc:" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }
{TEXT 261 40 "1. The Concept - Sum of Integers Formula" }}{PARA 0 "" 
0 "" {TEXT -1 61 "\nHere is a formula the sum of the first n positive \+
integers.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "`1 + 2 + 3 + \+
.... + n` = n*(n+1)/2;" }}}{PARA 0 "" 0 "" {TEXT -1 120 "\nIn a sense \+
this formula is a function. Given an integer n, the sum of the first n
 integers is the value of this formula" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "S := n -> n*(n+1)/2;" }}}{PARA 0 "" 0 "" {TEXT -1 79 
"\nIt appears to work for n = 1, n = 2, in fact all n values through 1
5 at least." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "S(1); 1;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "S(2); 1 + 2;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "for k from 1 to 15 do \n       `wh
en k = `, k,`, the sum is `,\n        sum( j, j = 1..k),` and the form
ula gives `, S(k); od;" }}}{PARA 0 "" 0 "" {TEXT -1 7 "\nSo it " }
{TEXT 265 7 "appears" }{TEXT -1 536 " that this formula works - howeve
r, how do we know it doesn't fail for some larger value of n? We can't
 check an infinite number of possible values. \n\nHere is a way. Since
 we know it works for some values of n, that is a good start. But now \+
if we can show that whenever it works for one value, it also works for
 the next value, then we'll be certain that it always works. What we a
re saying is that if it works for n, then if we the next integer, n+1,
 we should get the value of the formula for S(n+1). Lets see if these \+
two are equal.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "S(n+1) =
 S(n) + (n+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "expand(lh
s(%)) = expand(rhs(%)) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 
"verify(lhs(%), rhs(%), testeq);" }}}{PARA 0 "" 0 "" {TEXT -1 177 "\nT
hey are! We have proven the formula is true for all integers n \263 1.
 This is, in essence, the method of mathematical induction. There are \+
two indispensable steps :\n            " }{TEXT 266 223 "\n           \+
 Mathematical Induction - to Verify a Formula\n\n            1. Show t
hat it works for at least one value of n\n            2. Show that whe
never it works for a general value of n, it also works for the next va
lue." }{TEXT -1 191 "\n\nThe first step is usually very easy - simply \+
plug in a value and show it works for that one value. The second step \+
involves expressing the same value in two different ways, as we did ab
ove." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" 
{TEXT -1 1 " " }{TEXT 262 30 "2. More Examples of Equalities" }}{PARA 
0 "" 0 "" {TEXT -1 97 "\nHere is another example. Lets do this in a mo
re streamlined way following the two steps above.\n\n" }{TEXT 267 19 "
        Example 2.1" }{TEXT -1 14 " : Prove that " }{XPPEDIT 18 0 "1/(
1*2);" "6#*&\"\"\"F$*&F$F$\"\"#F$!\"\"" }{TEXT -1 2 " +" }{XPPEDIT 18 
0 "1/(2*3);" "6#*&\"\"\"F$*&\"\"#F$\"\"$F$!\"\"" }{TEXT -1 2 " +" }
{XPPEDIT 18 0 "1/(3*4)" "6#*&\"\"\"F$*&\"\"$F$\"\"%F$!\"\"" }{TEXT -1 
7 " +...+ " }{XPPEDIT 18 0 "1/(n*(n+1))" "6#*&\"\"\"F$*&%\"nGF$,&F&F$F
$F$F$!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "n/(n+1)" "6#*&%\"nG\"\"
\",&F$F%F%F%!\"\"" }{TEXT -1 68 "  \n                                 \+
is true for all integers n > 0.\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "S := n -> n/(n+1);" }}}{PARA 0 "" 0 "" {TEXT -1 18 "
\n                 " }{TEXT 268 6 "Step 1" }{TEXT -1 19 " : Verify for
 n = 1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "S(1) = 1/(1*2);" }
}}{PARA 0 "" 0 "" {TEXT -1 15 "               " }}{PARA 0 "" 0 "" 
{TEXT 269 23 "                Step 2 " }{TEXT -1 63 ": Show that the f
ormula for S(n+1) holds if it holds for S(n). " }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 41 "S(n+1) = S(n) + 1/( (n+1)*( (n+1) + 1) );" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "simplify(lhs(%)) = simplify(
rhs(%)) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "testeq( lhs(%)
 = rhs(%) );" }}}{PARA 0 "" 0 "" {TEXT -1 10 "\nDone!\n\n\n\n" }{TEXT 
270 19 "        Example 2.2" }{TEXT -1 4 " :  " }{XPPEDIT 18 0 "1^3 + \+
2^3 + 3^3 +" "6#,**$\"\"\"\"\"$F%*$\"\"#F&F%*$F&F&F%%#%?GF%" }{TEXT 
-1 8 " .... + " }{XPPEDIT 18 0 "n^3 " "6#*$%\"nG\"\"$" }{TEXT -1 2 "= \+
" }{XPPEDIT 18 0 "(1 + 2 + 3 +     + n)^2" "6#*$,,\"\"\"F%\"\"#F%\"\"$
F%%#%?GF%%\"nGF%F&" }}{PARA 0 "" 0 "" {TEXT -1 251 "\nThis one is a li
ttle trickier, because we can't simply add another term to find S(n+1)
 because S(n) is a square. To get around this, we can take the sqarero
ot of S(n), add (n+1)^3, then square that result to demonstrate the ve
racity of this formula.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 
"restart:\nS := n -> sum( k, k = 1..n)^2 ;" }}}{PARA 0 "" 0 "" {TEXT 
-1 18 "\n                 " }{TEXT 271 6 "Step 1" }{TEXT -1 19 " : Ver
ify for n = 1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "S(1) = 1^3;
" }}}{PARA 0 "" 0 "" {TEXT -1 15 "               " }}{PARA 0 "" 0 "" 
{TEXT 272 23 "                Step 2 " }{TEXT -1 63 ": Show that the f
ormula for S(n+1) holds if it holds for S(n). " }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 34 "S(n+1) = sqrt( S(n) + (n+1)^3) ^2;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "simplify(lhs(%)) = simplify(rhs(%))
 ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "testeq( lhs(%) = rhs(
%) );" }}}{PARA 0 "" 0 "" {TEXT -1 7 "\nDone!\n" }}{PARA 0 "" 0 "" 
{TEXT -1 2 "\n\n" }{TEXT 273 19 "        Example 2.3" }{TEXT -1 14 " :
 Prove that " }{XPPEDIT 18 0 "1;" "6#\"\"\"" }{TEXT -1 2 " +" }
{XPPEDIT 18 0 "2;" "6#\"\"#" }{TEXT -1 2 " +" }{XPPEDIT 18 0 "2^2;" "6
#*$\"\"#F$" }{TEXT -1 7 " +...+ " }{XPPEDIT 18 0 "2^n;" "6#)\"\"#%\"nG
" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2^(n+1)-1;" "6#,&)\"\"#,&%\"nG\"\"
\"F(F(F(F(!\"\"" }{TEXT -1 68 "  \n                                 is
 true for all integers n > 0.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 22 "S := n -> 2^(n+1) - 1;" }}}{PARA 0 "" 0 "" {TEXT -1 18 "\n    \+
             " }{TEXT 274 6 "Step 1" }{TEXT -1 19 " : Verify for n = 1
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "S(1) = 1;" }}}{PARA 0 "" 
0 "" {TEXT -1 15 "               " }}{PARA 0 "" 0 "" {TEXT 275 23 "   \+
             Step 2 " }{TEXT -1 63 ": Show that the formula for S(n+1)
 holds if it holds for S(n). " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 24 "S(n+1) = S(n) + 2^(n+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 37 "simplify(lhs(%)) = simplify(rhs(%)) ;" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 26 "testeq( lhs(%) = rhs(%) );" }}}{PARA 0 "" 0 "" 
{TEXT -1 7 "\nDone!\n" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }
{TEXT 263 27 "3. Induction & Inequalities" }}{PARA 0 "" 0 "" {TEXT -1 
158 "\nWhen we are trying to prove an inequality rather than an equati
on, we go through the same two general steps, but we can't express the
 result as a function.\n\n" }{TEXT 279 18 "       Example 3.1" }{TEXT 
-1 16 " : Prove that   " }{XPPEDIT 18 0 "2*n+1 <= n^2;" "6#1,&*&\"\"#
\"\"\"%\"nGF'F'F'F'*$)F(F&F'" }{TEXT -1 28 "   is true for all integer
s " }{XPPEDIT 18 0 "n>=3" "6#1\"\"$%\"nG" }{TEXT -1 2 ".\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Inequal :=  2*n + 1 <= n^2;" }}}
{PARA 0 "" 0 "" {TEXT -1 18 "\n                 " }{TEXT 280 6 "Step 1
" }{TEXT -1 69 " : Verify for n = 3 (the smaller number we can start a
t in this case)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "subs( n =
 3, Inequal);\nverify(lhs(%), rhs(%) ,\{'less_than','equal'\});" }}}
{PARA 0 "" 0 "" {TEXT -1 15 "               " }}{PARA 0 "" 0 "" {TEXT 
281 23 "                Step 2 " }{TEXT -1 200 ": Show that the formul
a for S(n+1) holds if it holds for S(n).  Let's substitute n+1 for n, \+
in the left side of the inequality, \n               with the hope of \+
showing it is less than the right side." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "LHS := subs( n = n+1, lhs(Inequal) );" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "% = '(2*n + 1),` + `,2';" }}}{PARA 
0 "" 0 "" {TEXT -1 95 "\n                Since we know the the inequal
ity is true for n, we can make this statement : \n" }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 18 "LHS < ''n^2 + 2'';" }}}{PARA 0 "" 0 "" 
{TEXT -1 35 "              Also, by hypothesis :" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 6 "2 < n;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 12 "2 < 2*n + 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "'LHS'
 < 'n^2 + (2*n+1)';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "n^2 \+
+ (2*n+1): % = factor(%) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
14 "LHS < (n+1)^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "subs(
 n = n+1, Inequal);" }}}{PARA 0 "" 0 "" {TEXT -1 63 "So we have demons
trated the inequality is also true for n+1!\n\n\n" }{TEXT 276 19 "    \+
    Example 3.2" }{TEXT -1 16 " : Prove that   " }{XPPEDIT 18 0 "n^2 <
= 2^n;" "6#1*$)%\"nG\"\"#\"\"\")F'F&" }{TEXT -1 26 " is true for all i
ntegers " }{XPPEDIT 18 0 "n>=4" "6#1\"\"%%\"nG" }{TEXT -1 2 ".\n" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Inequal := n^2 <= 2^n;" }}}
{PARA 0 "" 0 "" {TEXT -1 18 "\n                 " }{TEXT 277 6 "Step 1
" }{TEXT -1 19 " : Verify for n = 4" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 69 "subs( n = 4, Inequal);\nverify(lhs(%), rhs(%) ,\{'les
s_than','equal'\});" }}}{PARA 0 "" 0 "" {TEXT -1 15 "               " 
}}{PARA 0 "" 0 "" {TEXT 278 23 "                Step 2 " }{TEXT -1 63 
": Show that S(n+1) follows if S(n) is true. We can assume that " }
{XPPEDIT 18 0 "n^2 <= 2^n" "6#1*$%\"nG\"\"#)F&F%" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "LHS := subs( n = n+1, lhs(In
equal) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "expand(%);" }}
}{PARA 0 "" 0 "" {TEXT -1 65 "\n                   Using the inequalit
y above in example 3.1...." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
27 "LHS := %; LHS <= n^2 + n^2;" }}}{PARA 0 "" 0 "" {TEXT -1 39 "     \+
              Using the fact that " }{XPPEDIT 18 0 "n^2 <= 2^n" "6#1*$
%\"nG\"\"#)F&F%" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "2*n^2 <= \+
2*(2^n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "2*n^2 <= 2^(n+1
);" }}}{PARA 0 "" 0 "" {TEXT -1 7 "\nDone!\n" }}{PARA 0 "" 0 "" {TEXT 
-1 3 "\n\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 264 15 "
4. Divisibility" }}{PARA 0 "" 0 "" {TEXT -1 134 "\nHere is another var
iation - demonstrating that an expression is divisible by some number \+
for all integers plugged into the formula.\n\n" }{TEXT 282 18 "       \+
Example 4.1" }{TEXT -1 15 " : Prove that  " }{XPPEDIT 18 0 "n^3 + 14*n
 + 3" "6#,(*$%\"nG\"\"$\"\"\"*&\"#9F'F%F'F'F&F'" }{TEXT -1 43 " is div
isible by 3 for all integers n > 0.\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "S := n -> n^3 + 14*n + 3;" }}}{PARA 0 "" 0 "" {TEXT 
-1 18 "\n                 " }{TEXT 283 6 "Step 1" }{TEXT -1 19 " : Ver
ify for n = 1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "S(1);\n%/3;
" }}}{PARA 0 "" 0 "" {TEXT -1 15 "               " }}{PARA 0 "" 0 "" 
{TEXT 284 23 "                Step 2 " }{TEXT -1 63 ": Show that the f
ormula for S(n+1) holds if it holds for S(n). " }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 7 "S(n+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 10 "expand(%);" }}}{PARA 0 "" 0 "" {TEXT -1 85 "\n              Lets
 take the difference of S(n+1) and S(n) - this will eliminate the " }
{XPPEDIT 18 0 "n^3" "6#*$%\"nG\"\"$" }{TEXT -1 1 "." }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 28 "difference := S(n+1) - S(n);" }}}{PARA 0 "
" 0 "" {TEXT -1 85 "             We expand the difference and find tha
t each term is divisible by three! " }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 10 "expand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
4 "%/3;" }}}{PARA 0 "" 0 "" {TEXT -1 211 "             This proves it.
 Do you see how? S(n+1) = S(n) + a polynomial divisible by three. Sinc
e both polynomials \n             on the right are divisible by 3, the
n so is the left side, since they are equal." }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 24 "S(n+1) = 'S(n) + 3*(%)';" }}}{EXCHG }}{PARA 0 "" 
0 "" {TEXT 259 36 "\n         \251 2002 Waterloo Maple Inc " }}}{MARK 
"0 1" 9 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }