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December 14th, 2016, 02:50 AM   #11
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Quote:
Originally Posted by skipjack View Post
It contains various slips. For example, your reference to "any other element" is inappropriate (as you intend to show that such an element exists, you shouldn't assume it does).
I'm interested about you what you said!

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
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Quote:
Originally Posted by Prove It View Post
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 $\displaystyle \begin{align*} n \in \mathbf{N} \end{align*}$.
@Prove It. I am reading again this post. And, sorry, I have something to ask again. I can add something more:
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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