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December 14th, 2016, 03:50 AM  #11  
Senior Member Joined: Jan 2013 From: Italy Posts: 136 Thanks: 7  Quote:
The phrase used in the book for a similar proof is ... $P(1)$ is true and $1 \in K$. Next let $k$ be any element of $K$ ... Take a look at this: I have followed this textbook, in my opinion the only one that made me think math induction with peano axioms!
how can you resolve that slips you have referred to? Thanks!  
December 22nd, 2016, 03:01 PM  #12  
Senior Member Joined: Jan 2013 From: Italy Posts: 136 Thanks: 7  Quote:
If we set: $5^{2(k+1)}1 = 24p$ from the following passage: $24 \cdot 5^{2k} + 5^{2k}  1 = 24p$ we have, that the following quantity is divisible by 24, for inductive hypotesis: $5^{2k}  1 = 24m$ and also the following quantity is divisible by 24: $24 \cdot 5^{2k} = 24s$ therefore, the inductive step, became a sum of two positions $24p = 24m + 24s$ $24p = 24(m+s)$ $p = m+s$ So the desired value for the integer p is m+s. But, what I want to ask is: How can I be sure about the rightness of the result we have found? Maybe, Do I have to substitute that $p$ in the $P(n)$, and try to find something useful like this? $5^{2n}  1 = 24(m+s)$  

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$52n, divisible, induction, verify 
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