Divisor of Integer/Examples/80 divides 9^2n - 1/Proof 1
Theorem
Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
- $80 \divides 9^{2 n} - 1$
where $\divides$ denotes divisibility.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the equivalent proposition:
- $80 \divides 9^{2 n} - 1$
where $\divides$ denotes divisbility
$\map P 0$ is the case:
\(\ds 9^{2 \times 0} - 1\) | \(=\) | \(\ds 9^0 - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - 1\) | Zeroth Power of Real Number equals One | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 80\) | \(\divides\) | \(\ds 9^{2 \times 0} - 1\) | Integer Divides Zero |
Thus $\map P 0$ is seen to hold.
Basis for the Induction
$\map P 1$ is the case:
\(\ds 9^{2 \times 1} - 1\) | \(=\) | \(\ds 9^2 - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 81 - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 80\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 80\) | \(\divides\) | \(\ds 9^{2 \times 1} - 1\) | Integer Divides Itself |
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $80 \divides 9^{2 k} - 1$
from which it is to be shown that:
- $80 \divides 9^{2 \paren {k + 1} } - 1$
Induction Step
This is the induction step:
From the induction hypothesis we have that:
- $80 \divides 9^{2 k} - 1$
Hence by definition of divisibility, we have:
- $\exists r \in \Z: 9^{2 k} - 1 = 80 r$
and so:
- $(1): \quad \exists r \in \Z: 9^{2 k} = 80 r + 1$
\(\ds 9^{2 \paren {k + 1} } - 1\) | \(=\) | \(\ds 9^2 \times 9^{2 k} - 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists r \in \Z: \, \) | \(\ds 9^{2 \paren {k + 1} } - 1\) | \(=\) | \(\ds 81 \times \paren {80 r + 1} - 1\) | from $(1)$ | |||||||||
\(\ds \) | \(=\) | \(\ds 81 \times 80 r + 80\) | algebra | |||||||||||
\(\ds \) | \(=\) | \(\ds 80 \paren {81 r + 1}\) | algebra |
Hence by definition of divisibility:
- $80 \divides 9^{2 \paren {k + 1} } - 1$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{\ge 0}: 80 \divides 9^{2 n} - 1$
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Method of Induction: Example $2$