Powers of 16 Modulo 20/Proof 1
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Theorem
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Then:
- $16^n \equiv 16 \pmod {20}$
Proof
Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
- $16^n \equiv 16 \pmod {20}$
Basis for the Induction
$\map P 1$ is the case:
- $16^1 \equiv 16 \pmod {20}$
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:
- $16^k \equiv 16 \pmod {20}$
from which it is to be shown that:
- $16^{k + 1} \equiv 16 \pmod {20}$
Induction Step
This is the induction step:
We have:
\(\ds 16^{k + 1}\) | \(=\) | \(\ds 16^k \times 16\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 16 \times 16\) | \(\ds \pmod {20}\) | Induction Hypothesis | ||||||||||
\(\ds \) | \(\equiv\) | \(\ds 256\) | \(\ds \pmod {20}\) | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds 16\) | \(\ds \pmod {20}\) |
So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that:
- $\forall n \in \Z_{> 0}: 16^n \equiv 16 \pmod {20}$
$\blacksquare$