document.write( "Question 1194395: I starting to learn about mathematical induction, and I'm not sure if my answers are correct. Please help.\r
\n" ); document.write( "\n" ); document.write( "Here is the question, my answers are inside brackets, the rest is provided by the exercise. Thank you.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Complete the following proof by mathematical induction that, for all n in W such that n >= 2, 3 ^ n > 2 ^ (n + 1)\r
\n" ); document.write( "\n" ); document.write( "We proceed by mathematical induction and begin by establishing the base of the induction.\r
\n" ); document.write( "\n" ); document.write( "3^ 2 = [9] > [8] =2^ 2+1\r
\n" ); document.write( "\n" ); document.write( "We see that 3 ^ n > 2 ^ (n + 1) when n=2.\r
\n" ); document.write( "\n" ); document.write( "Moving on to the induction step, we suppose that m in W is such that m >= 2
\n" ); document.write( "and as our induction hypothesis, we take the assumption that 3^m > 2 ^ (m + 1)
\n" ); document.write( "Noting that 3m+is the product of one more 3 than 3 ^m+1 and that 2 ^ (m + 2) is the product of one more 2 than 2 ^ (m + 1)\r
\n" ); document.write( "\n" ); document.write( "We calculate as follows:
\n" ); document.write( "2 ^ (m + 1) + 1 = [ 2 * 2 ^ (m + 1) ]
\n" ); document.write( "(choose from; 2+2^(m+1), 2.2^(m+1), 2^(m+1) +1)\r
\n" ); document.write( "\n" ); document.write( "= [ 2^ m+1 +2^ m+1 ]
\n" ); document.write( "(choose from; 2^m*2^m, 2^m+2^m, 2^(m+1)+2^(m+1))\r
\n" ); document.write( "\n" ); document.write( "< 3^m +2^m+1 by the induction hypothesis and additive monotonicity\r
\n" ); document.write( "\n" ); document.write( "< [ 3^m + 2 ^ m + 2^m ] by the induction hypothesis and additive monotonicity
\n" ); document.write( "(choose from; 3^m + 2^m + 2^m, 3^m +3^m, 3^m + 3^(m+1) )\r
\n" ); document.write( "\n" ); document.write( "< [ 3^m + 3^m + 3^m ] by order in W since 3^ m is in N
\n" ); document.write( "(choose from; 3^m *3^m, 3m +3^(m+1), 3^m + 3^m + 3^m )\r
\n" ); document.write( "\n" ); document.write( "= [3*3^m ]
\n" ); document.write( "(choose from, 3*3^m, 3^m*3^m, 3+3^m )
\n" ); document.write( "
\n" ); document.write( "=3^ m+1\r
\n" ); document.write( "\n" ); document.write( "We have shown that 3 ^ 2 > 2 ^ (2 + 1) and that, for all m in W such that m >= 2, if * 3 ^ m > 2 ^ (m + 1)
\n" ); document.write( "then 3 ^ (m + 1) > 2 ^ (m + 1) + 1
\n" ); document.write( "We may therefore conclude by mathematical induction that, for all n in W such that n >= 2; 3 ^ n > 2 ^ (n + 1) \n" ); document.write( "

Algebra.Com's Answer #826686 by math_helper(2461) $\"\"$ $\"About$

You can put this solution on YOUR website!

\r
\n" );
document.write( "\n" );
document.write( "I think your base case and hypothesis are fine.  You could have made the step case (m+1) clearer, as follows:\r
\n" );
document.write( "\n" );
document.write( "base:  n=2,  >    base case n=2 holds
\n" );
document.write( "hypothesis:  Assume  for some m, where m>2.\r
\n" );
document.write( "\n" );
document.write( "step case:  Let n=m+1:\r
\n" );
document.write( "\n" );
document.write( "      Need to show \r
\n" );
document.write( "\n" );
document.write( "  LHS:  
\n" );
document.write( "  RHS:  \r
\n" );
document.write( "\n" );
document.write( "We now apply the hypothesis \r
\n" );
document.write( "\n" );
document.write( "and can conclude LHS > RHS, i.e.    
\n" );
document.write( "(for A > 0 and B > 0, if A > B than certainly 3A > 2B).\r
\n" );
document.write( "\n" );
document.write( "or, re-writing,  \r
\n" );
document.write( "\n" );
document.write( "Done\r
\n" );
document.write( "\n" );
document.write( "Usually, in inductive proofs, you should be looking to reduce the 'n+1' case to something involving 'n'  so that you can directly apply the hypothesis, then include that new information into the 'n+1' scenario and draw the final conclusion.
\n" );
document.write( "     \n" );
document.write( "

\n" );