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 March 21st, 2019, 11:03 PM #1 Newbie   Joined: Oct 2017 From: sweden Posts: 24 Thanks: 0 Doing mathematical induction There is a mathematical induction problem, where I am not sure if the steps I am doing this makes it a valid method. Here is the problem: Prove the following via mathematical induction: $2^n>n+4, n\ge 3$ Proof: We skip the case for $P(3)$. Assume $P(k)$ is true, then we prove $P(k+1)$ also holds true. We state our induction hypothesis: $$2^k>k+4, k\ge 3$$ We start with the expression $2^{k+1}-k-5,$ and show this is greater than $0$: $$2^{k+1}-k-5=2\times 2^k-k-5>2\times(k+4)-k-5$$ This holds from the induction hypothesis. $$2\times 2^k-k-5>2\times(k+4)-k-5=k+3>0$$ Note $k+3>0$ for $k\ge 3$ hence we have: $$2\times 2^k-k-5>0$$ $$2\times 2^k>k+5=(k+1)+4$$ $$2^{k+1}>(k+1)+4$$ Hence by mathematical induction, $P(k+1)$ holds true for when $P(k)$ holds true. Does this method work as how I have approached it? March 22nd, 2019, 04:12 PM #2 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 $2^{k+1}=2(2^k)\gt 2(k+4)=2k+8\gt (k+1)+4$. Is a direct approach. Thanks from topsquark March 23rd, 2019, 09:16 AM #3 Newbie   Joined: Oct 2017 From: sweden Posts: 24 Thanks: 0 Thank you for this, but I know this method works as it is the most standard approach. However the answer you posted is of no use, as it does not answer my question asking about the validity of the method I outlined in my original post. Last edited by heinsbergrelatz; March 23rd, 2019 at 09:23 AM. March 23rd, 2019, 01:37 PM   #4
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 Originally Posted by heinsbergrelatz Thank you for this, but I know this method works as it is the most standard approach. However the answer you posted is of no use, as it does not answer my question asking about the validity of the method I outlined in my original post.
Your method is valid, but what was the point, since the usual method is much simpler. March 23rd, 2019, 07:39 PM #5 Newbie   Joined: Oct 2017 From: sweden Posts: 24 Thanks: 0 Thank you very much, finally there is an answer which tells me it is valid, you have no idea how many people on mathstackexchange made this question so hard by not answering my simple yes or no it is valid question. March 23rd, 2019, 10:48 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2273 Your algebra was correct, but your wording needed improvement. The proposition $P(n)$ is that $2^n > n + 4$, where $n$ is an integer. You have to prove that $P(n)$ is true for $n \geqslant 3$, so your first step (after defining the proposition) should be to show by substitution that the base case, $P(3)$, is true. You should now state the induction hypothesis, that $P(n)$ is true for $3 \leqslant n \leqslant k$. Next, you should prove the induction step, that the induction hypothesis implies that $P(k+1)$ is true. The induction hypothesis includes that $P(k)$ is true, so it often suffices (as in your example) to show that $P(k)$ implies $P(k+1)$. Finally, state that, by induction, $P(n)$ is true for $n \geqslant 3$ (or some equivalent wording). Thanks from heinsbergrelatz Tags induction, mathematical 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 hy2000 Pre-Calculus 3 October 14th, 2018 12:53 AM erblina zeqiri Algebra 1 January 5th, 2015 04:15 PM spuncerj Pre-Calculus 1 November 28th, 2014 04:43 PM remeday86 Applied Math 1 June 20th, 2010 09:52 AM firstsin Abstract Algebra 0 December 31st, 1969 04:00 PM

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