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 October 4th, 2011, 12:35 PM #1 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 Mathematical Induction Proof I have the equation 2^(k+1) < (k + 3)! for any integer (k) greater or equal to 0. I am supposed to prove it by mathematical induction but I am lost. I already proved the base case, but how am I supposed to prove that the left side will always be less than the right side. So far this is what I have: 2^(k+1) = 2*2*2*2...*2 (k +1) times (k+3)! = 1*2*3*...*(k+1)*(k+2)*(k+3) does anybody know how to prove the top line will always be less than the bottom line. Any Help Would Bw Appreciated October 4th, 2011, 01:26 PM #2 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Mathematical Induction Proof You're using induction, remember? So write 2^(k+1) in terms of 2^k and (k+3)! in terms of (k + 2)! and see if you can prove that if the inequality held for the last pair it will hold for the next. October 4th, 2011, 01:33 PM #3 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 Re: Mathematical Induction Proof The form I actually wrote is the (k+1) step. So it is in that form, but I just can't figure out how to prove it logically, even though I know it is true October 4th, 2011, 01:51 PM   #4
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Re: Mathematical Induction Proof

Quote:
 Originally Posted by doleary22 The form I actually wrote is the (k+1) step. So it is in that form, but I just can't figure out how to prove it logically, even though I know it is true
You wrote down what (k+3)! is, but not what it is in terms of (k+2)!. Ditto for 2^(k+1) and 2^k.

Once you do this the problem is actually quite easy. October 4th, 2011, 02:26 PM #5 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 Re: Mathematical Induction Proof The original problem was 2^k < (k+2)!. So then for my inductive step I added k+1 to each side to try to prove that it would work for the next number. So then I got 2^k+1 < (k +1 + 2) which is 2^k+1 < (k+3). I do not understand what forms they are supposed to be in to make the proof. Thanks For The Help October 4th, 2011, 05:33 PM #6 Global Moderator   Joined: May 2007 Posts: 6,762 Thanks: 697 Re: Mathematical Induction Proof For k=1, 2 < 3!=6 Assume 2^k < (k+2)!, prove 2 ^(k+1) < (k+3)! proof 2^(k+1) = 2x2^k < 2x(k+2)! < (k+3)x(k+2)!=(k+3)! x means multiply October 4th, 2011, 09:12 PM   #7
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Re: Mathematical Induction Proof

Hello, doleary22!

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[color=beige]. . . . . . [/color] Tags induction, mathematical, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post alloy Algebra 3 November 14th, 2012 03:54 AM arun Algebra 5 November 23rd, 2011 07:23 AM Lolligirl Number Theory 2 April 4th, 2011 10:28 AM supercali Calculus 0 November 2nd, 2007 12:52 AM firstsin Abstract Algebra 0 December 31st, 1969 04:00 PM

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