5

Problem

Proof with induction that $n^2+n$ is even when $n \in \mathbb{N}$

Attempt to solve

$n^2+n$ is even when it satisfies: $$ n^2+n \mod 2 = 0 $$ meaning it's divisible by 2. I can begin to proof that this statement holds when $n \in \mathbb{N}$.

Base case

the statement holds when $n=0$

$$ 0^2+0 \mod 2 = 0 $$

Which is true since $0$ divided any $n\in \mathbb{N}$ is $0$.

Inductive step

Assume statement holds when $n=k$ (induction hypothesis)

$$ k^2+ k \mod 2 = 0 $$

then let $n = k+1$

$$ (k+1)^2 + (k+1) \mod 2 = 0 $$ $$ \iff k^2+2k+1+k+1 \mod 2 = 0 $$ $$ \iff k^2+k + 2k + 2 \mod 2 = 0 $$

Based on induction hypothesis $k^2+k \mod 2 = 0$. Which implies we can deduce our statement to

$$ \implies 2k + 2 \mod 2 = 0 $$ Which is true for $\forall k \in \mathbb{N}$ since any $k\in \mathbb{N}$ multiplied by 2 is divisible by 2. Adding 2 to this does not alter the divisibility since 2 is also divisible by 2.

Q.E.D

Is my proof correct?

Tuki
  • 2,237

2 Answers2

1

Your proof is correct.

For the base case, you pick $n=0$. I think you assume $\Bbb N=\{0,1,2,\cdots\}$. But many are treat $\Bbb N$ as $\{1,2,\cdots\}$. Anyway, your result is true for this base case $n=1$ also, since $1+1=2$ which is even.

Alternatively, note that $n^2+n=n(n+1)$, a product of two consecutive terms. So one factor must be even and hence the expression is even!

Chinnapparaj R
  • 11,589
1

I don't have much knowledge of the mod notation or proofs of this type, but I know this:

notice that $n^2+n=n(n+1)$ which for $n\in \mathbb{N}$ is the same as multiplying two consecutive integers. Since integers alternate between odd and even, we can see that either $n$ or $n+1$ must be even, and the other must be odd. since $\text{odd}\times\text{even}=\text{even}$ this shows that $n(n+1)$ must also be even.

I believe this is the equivalent of saying:

$$n \mod 2=0 \,\,\text{or}\,\,(n+1)\mod 2=0$$ $$\therefore n(n+1) \mod 2=0$$

Henry Lee
  • 12,215