Prove by induction $7^n$ is an odd number for every natural number n

Question

I have proved the base case already of n=1 where $7 = 2p+1$

Then I assumed $n=k$ for $7^k = 2p+1$ for $k \in N$ and $p \in N$

To prove $7^{k+1} = 2p+1$ I have these steps so far:

$7(7^k) = 2p+1$

$7(2p+1) = 2p+1$

$14p+7 = 2p+1$

I am not sure how to prove from here.

Answer 1

Just rewrite your sum as

$$14p+7=2(7p+3)+1$$

Also, it's not a good idea to use $p$ for all of your steps. If $7^k=2p+1$, then you need to show that $7^{k+1}=2q+1$ for some integer $q$. Otherwise the equality is not actually equal since you are using the same expression as your desired form for different valued expressions. Now since $14p+7=2(7p+3)+1$, let $q=7p+3$. Thus $7^{k+1}=2q+1$. And you're done.

Answer 2

If $7^k$ is odd i.e. $7^k=2p+1$ for some integer $p$, then $$ 7^{k+1}=7(2p+1)=14p+7=2(7p+3)+1 $$ as desired.

Answer 3

More simply: take as given that the product of two odd numbers is odd and that 7 is odd.

Suppose $7^k$ is odd. Then $7^{k+1} = 7^k × 7$ is the product of two odd numbers and therefore also odd.

i.e.: If the proposition is true for $n=k$, it is also true for $n=k+1$.

But it is true for $n=1$, since $7^1=7$ is odd. Therefore by induction it is true for all $n>0$.

(I'm assuming this is an exercise in understanding proof by induction, not in understanding odd numbers.)