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December 14th, 2016, 02:50 AM | #11 | |
Senior Member Joined: Jan 2013 From: Italy Posts: 154 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! | |
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December 22nd, 2016, 02:01 PM | #12 | |
Senior Member Joined: Jan 2013 From: Italy Posts: 154 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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