User:Leigh.Samphier/Common/Induction Example

Proof

Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$\ds n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$

Basis for the Induction

$\map P 1$ is true, as this just says $1^2 = 1$.

This is our basis for the induction.

Induction Hypothesis

Now we need to show 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 our induction hypothesis:

$\ds k^2 = \sum_{j \mathop = 1}^k \paren {2 j - 1}$

Then we need to show:

$\ds \paren {k + 1}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}$

Induction Step

This is our induction step:

\(\ds \paren {k + 1}^2\)

\(=\)

\(\ds k^2 + 2 k + 1\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 k + 1\)

Induction Hypothesis

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 \paren {k + 1} - 1\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}\)

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\ds \forall n \in \N: n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$

$\blacksquare$