document.write( "Question 633231: Prove nn^2 for n>=4 and nn^3 for n>=6 by using Principle of Mathematical Induction. \n" ); document.write( "
Algebra.Com's Answer #398832 by Edwin McCravy(20056)\"\" \"About 
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Prove n n² for n>=4
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document.write( "We begin by showing that it is true for the smallest possible \r\n" );
document.write( "value of n, which is 4:\r\n" );
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document.write( "4 4²\r\n" );
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document.write( "24 > 16 which is true.\r\n" );
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document.write( "Assume for n <= k\r\n" );
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document.write( "(1)  k k²                     <--- we are assuming this.\r\n" );
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document.write( "We need to show that assumption (1) leads to \r\n" );
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document.write( "(2)  (k + 1) (k + 1)²         <--- we do not know this yet!\r\n" );
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document.write( "Since (k + 1)² = k² + 2k + 1, the 2k + 1 tells us that we \r\n" );
document.write( "should add 2k + 1 to both sides of assumed inequality (1):\r\n" );
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document.write( "(3)   k! + 2k + 1 > k² + 2k + 1  = (k + 1)²   <--- we know this\r\n" );
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document.write( "Now we just need to show that the left side of (3) is less\r\n" );
document.write( "than the left side of (2)\r\n" );
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document.write( "That is, we need to show that  \r\n" );
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document.write( "(4)   k! + 2k + 1 < (k + 1)!         <--- we do not know this yet\r\n" );
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document.write( "which is equivalent to\r\n" );
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document.write( "(5)   k! + 2k + 1 < (k + 1)k!         <--- we do not know this yet\r\n" );
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document.write( "which is equivalent to\r\n" );
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document.write( "(6)   k! + 2k + 1 < k·k! + k!         <--- we do not know this yet\r\n" );
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document.write( "which is equivalent to\r\n" );
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document.write( "(7)   2k + 1 < k·k!                   <--- we do not know this yet\r\n" );
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document.write( "Which is equivalent to the following  (upon dividing through by k): \r\n" );
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document.write( "(9)   2 + \"1%2Fk\" < k!             <--- we DO know this.  \r\n" );
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document.write( "We DO know (9) is true.  Here's why:\r\n" );
document.write( "The left side is always less than 3 and since k >= 4, the right side k! is\r\n" );
document.write( "always 4! = 24 or greater so we do know (9) is true.\r\n" );
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document.write( "Now since all those inequalities are equivalent we can reverse the\r\n" );
document.write( "steps\r\n" );
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document.write( "(9)  2 + \"1%2Fk\" < k!          is true, therefore, multiplying thru by k,\r\n" );
document.write( "(8)         2k + 1 < k·k!        is true, therefore adding k! to both sides:\r\n" );
document.write( "(7)    k! + 2k + 1 < k·k! + k!   is true, therefore factoring the right:\r\n" );
document.write( "(6)    k! + 2k + 1 < k!(k + 1)   is true, and the right side is (k + 1)!, so\r\n" );
document.write( "(5)    k! + 2k + 1 < (k + 1)k!   is true.  So:\r\n" );
document.write( "(4)    k! + 2k + 1 < (k + 1)!    is true, which is the same as\r\n" );
document.write( "         (k + 1) k! + 2k + 1 \r\n" );
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document.write( "and since (3) is true, which is \r\n" );
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document.write( "(3)    k! + 2k + 1 > k² + 2k + 1  = (k + 1)², we have \r\n" );
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document.write( "(k + 1) k! + 2k + 1 > k² + 2k + 1 > (k + 1)²\r\n" );
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document.write( "which is what we needed to prove.\r\n" );
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document.write( "The proof of the other one is similar.  It amounts to taking\r\n" );
document.write( "the inequality you have to prove, simplifying it using equivalent\r\n" );
document.write( "inequalities to get something that you know is true, then you \r\n" );
document.write( "reverse the steps.\r\n" );
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document.write( "Edwin

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