Induction: $(2a-1)^{n}-1$ is an even number

Question

Just wanted to ask whether there are any mistakes.

Proof by induction:

$\forall n \in \mathbb{N}: (2a-1)^{n}-1 , a \in \mathbb{N}, $ is an even number.

Base case:

$n=1: (2a-1)^{1}-1 = 2a-1-1 = 2a-2 = 2\cdot(a-1) \rightarrow$ even number

Inductive hypothesis:

Assume that $(2a-1)^{n}-1$ is an even number. So we can say $(2a-1)^{n}-1 = 2k$, for some $k \in \mathbb{N}$.

Inductive step:

$n \rightarrow n+1$, so we have to show $(2a-1)^{n+1}-1 = 2k$

$(2a-1)^{n+1}-1\\ = (2a-1)^{n}\cdot(2a-1)-1 \\= (2a-1)^{n}\cdot(2a-1)-(2a-1)+(2a-1)-1 \\= (2a-1)\cdot[(2a-1)^{n} -1]+(2a-1)-1 \\=(2a-1)\cdot 2k+2a-2 \\ =2\cdot[(2a-1)\cdot k+a-1]$

$\Longrightarrow$ for $n+1$ is an even number

Answer 1

It is correct, as alternative in the induction step we can also use the induction hypotesis as follows

$$(2a-1)^{n+1}-1=(2a-1)\color{red}{(2a-1)^{n}}-1=(2a-1)\color{red}{(2k+1)}-1=2k(2a-1)+2a-2$$

Answer 2

not sure why going this way, but it easier to show that $(2a-1)^n$ is always an odd number, also in induction (it's elementary, doesn't sure if it can be counted as induction), going something like that:

base: $n=1$ : for every $a\in \Bbb Z$, $2a-1$ is an odd number

induction hypothesis: for every $n'\le n \in \Bbb N$, $(2a-1)^{n'}$ is odd.

inductive step: we will show that $(2a-1)^{n+1}$ is an odd number: $$(2a-1)^{n+1}=(2a-1)(2a-1)^n$$ by hypothesis, both expressions are odd, and multiplication of odd number are odd. then for every $n\in \Bbb N$, $(2a-1)^n$ is odd and therfore $(2a-1)^n-1$ are even

Answer 3

The statement is equivalent to $$(2a-1)^n$$ is odd for all $n\in N$

Note that product of two odd integers is odd.

Now we start the induction.

It is true for $n=1$ because $2a-1$ is odd.

If true for $n$ then $$ (2a-1)^{n+1} = (2a-1)^n(2a-1)$$ is the product of two odd integers so it is odd.