December 9th, 2016, 02:48 PM  #1  
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  verify by induction that $5^{2n}  1$ is divisible by 24.
Hi, I have done this exercise and kindly I would want to receive some possible corrections and suggestions! Also I have looked at the Peano Postulates or Axioms: Quote:
Exercise: Using the principle of mathematical induction, check the truth of $24  5^{2n}  1$ for each $n \in \mathbb{N}$ My own development: $P(n) : 5^{2n}  1$ is divisible by 24, for each $n \in \mathbb{N}$ Consider a set $K = \left \{ k : k \in \mathbb{N}, P(k) \mbox{ is true } \right \}$ i.e. a set $K$ containing all the $k$ numbers that makes $P(n)$ true when $n=k$ Consider $1 \in \mathbb{N}$ for the first Peano Axiom. Let's check if $P(n)$ is true for $n=1$. If so, $1$ will be an element to include in the set $K$ defined above: $P(1) : 5^{2(1)}  1 = \\ = 5^2  1 = \\ = 25  1 = \\ = 24$ is divisible by $24$. And the value $24$ we just obtained, is divisible by $24$ . so $P(1)$ seems to be true! Therefore we have $1 \in K$. And the Postulate V (a) is fulfilled. Consider now, $k$ any other element in $K$, we suppose that is true the following proposition: $P(k) : 5^{2k}  1$ is divisible by $24$ for each $k \in \mathbb{N}$ we want to know if also the successor of $k$, will be in $K$. So we consider $k^{*}$ as the successor of $k$. We want to check the truth of: $P(k^{*}) : 5^{2k^{*}}  1$ is divisible by $24$, for each $k \in \mathbb{N}$ for one definition of addition in $\mathbb{N}$ we have that $k + 1 = k^{*}$ i.e. the successor of a number is that number added with $1$ so going to substituting we obtain: $P(k+1) : 5^{2(k+1)}  1 = \\ = 5^{2k + 2}  1 = \\ = 5^{2k} \cdot 5^{2}  1 = \\ = 5^{2k} \cdot 25  1 = \\ = 5^{2k} \cdot (24 + 1)  1 = \\ = ( 5^{2k} \cdot 24 ) + (5^{2k}  1),$ is divisible by $24$. We can see that $( 5^{2k} \cdot 24 )$ is divisible by $24$, and also $(5^{2k}  1)$ is is divisible by $24$ by inductive hypotesis, hence $( 5^{2k} \cdot 24 ) + (5^{2k}  1)$ is divisible by $24$. So $P(k+1)$ is true. Since also $k^{*} \in K$, for each $k \in \mathbb{N}$, the Postulate V (b) is fulfilled. It follows that $K = \mathbb{N}$, the Postulate V is fullfilled, and the proposition is valid for every $n \in \mathbb{N}$ What do you think about it? Many thanks!  
December 9th, 2016, 03:04 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,498 Thanks: 755 
I think you can make it about a fifth of its length. You can assume your audience understands how induction works. $P_1: 5^2  1 \pmod{24} = (251)\pmod{24}= 24\pmod{24}=0$ $P_1 =True$ $5^{2(n+1)}1 = 5^2(5^{2n})1$ We assume $P_n$ so $ 5^2(5^{2n})1 = 5^2(24k+1)1 = (25)(24)k + 251 = (25k+1)(24)$ $(25k+1)(24) \pmod{24} = 0$ so $P_{n+1} = True$ QED Last edited by skipjack; December 13th, 2016 at 02:07 AM. 
December 9th, 2016, 03:50 PM  #3 
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  
December 9th, 2016, 04:28 PM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,595 Thanks: 620 Math Focus: Wibbly wobbly timeywimey stuff.  
December 9th, 2016, 07:42 PM  #5 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 
My method: The statement "24 divides $\displaystyle \begin{align*} 5^{2\,n}  1 \end{align*}$ for all $\displaystyle \begin{align*} n \in \mathbf{N} \end{align*}$" is equivalent to "$\displaystyle \begin{align*} 5^{2\,n}  1 = 24\,p \end{align*}$ where $\displaystyle \begin{align*} n, p \in \mathbf{N} \end{align*}$". Base Step: $\displaystyle \begin{align*} 5^{2 \cdot 1}  1 &= 5^2  1 \\ &= 25  1 \\ &= 24 \\ &= 24 \cdot 1 \end{align*}$ The base step is true. Inductive Step: Assume that $\displaystyle \begin{align*} 5^{2\,k}  1 = 24\,m \end{align*}$ is a true statement for some $\displaystyle \begin{align*} k, m \in \mathbf{N} \end{align*}$. Then $\displaystyle \begin{align*} 5^{2\,\left( k + 1 \right) }  1 &= 5^{2\,k + 2}  1 \\ &= 5^{2\,k} \cdot 5^2  1 \\ &= 25 \cdot 5^{2\,k}  1 \\ &= 24 \cdot 5^{2\,k} + 5^{2\,k}  1 \\ &= 24 \cdot 5^{2\,k} + 24\,m \\ &= 24 \,\left( 5^{2\,k} + m \right) \\ &= 24\,p \textrm{ where } p = 5^{2\,k} + m \end{align*}$ The inductive step is true. Thus we have proved that 24 divides $\displaystyle \begin{align*} 5^{2\,n}  1 \end{align*}$ for all $\displaystyle \begin{align*} n \in \mathbf{N} \end{align*}$. 
December 10th, 2016, 05:08 PM  #6 
Senior Member Joined: Dec 2015 From: Earth Posts: 156 Thanks: 21 
$\displaystyle 5^{2n} 1=25^n1=25^n1^n=24(25^{n1}+25^{n2}+....+1)$ $\displaystyle 2424(25^{n1}+25^{n2}+....+1)$ 
December 13th, 2016, 01:15 AM  #7 
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  
December 13th, 2016, 02:15 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,053 Thanks: 1395  
December 13th, 2016, 05:20 AM  #9 
Member Joined: Jun 2014 From: Alberta Posts: 55 Thanks: 2 
I was taught a slightly different but easier way. If (base case) 5^(2(1))  1 = 24, 5^(2n)  1 = 24m, m,n E N > 5^(2n+2)  1 = 24p = 25*5^(2n)  1 = 25(24m + 1)  1, p E N Then 5^(2n)  1 = 24m That should be all you need, except maybe add a little more work to show that 24p = 25(24m + 1)  1 = 25*24m + 24, which is just trivial arithmetic. 
December 13th, 2016, 05:37 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,053 Thanks: 1395  

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$52n, divisible, induction, verify 
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