
The first case is when $n$ is even, we can write it as $2k$ . So, $n = 2k, \ k \in \mathbb{Z}$ .
Plugging it in, we get $(2k)^2 + 3(2k) + 2$ , which is $4k^2 + 6k + 2$ .
We can factor out a $2$ , and write it as $2(2k^2 + 3k + 1)$ .
Note that what we have now is also a multiple of $2$ , so we can write it as $2r$ where $r = 2k^2 + 3k + 1$ . But, we don't even have to do that, since it is a multiple of $2$ , the whole thing is even, so our proposition is true for the first case.
The second case is when $n$ is $0$ . We can just plug it in again, and get $(0)^2 + 3(0) + 2$ , which is just $2$ .
$2$ is even, so our proposition is true for the second case as well.
The third case is when $n$ is odd. We can write an odd number as $2k + 1$ . Again, let's plug it in to the formula: $(2k + 1)^2 + 3(2k + 1) + 2$ .
We get $4k^2 + 4k + 1 + 6k + 3 + 2$ . Putting it all together, we have $4k^2 + 10k + 6$ . We can factor out $2$ here, which makes it $2(2k^2 + 5k + 3)$ .
The fact that we are able to factor out a $2$ shows that it is a multiple of $2$ , which means that it is an even number. So, our proposition is true for the third case.
Therefore, we have showed that $n^2 + 3n + 2$ is even if $n$ is an integer. $\blacksquare$

Proof by Contradiction

If the proposition we're dealing with is false, a false fact has to be true. Read that sentence once again.
A false fact cannot be true, therefore the proposition has to be true. And, this is the way of proof by contradiction.
This is an indirect proof.

A well-known example is proving that $\sqrt{2}$ is irrational.
In this case, if the proposition " $\sqrt{2}$ is irrational" is false, then " $\sqrt{2}$ is rational" has to be true. But we're going to show that it cannot be true (a false fact), therefore, our proposition must be true.
Let's take a look at it:


So, assume that $\sqrt{2}$ is rational.
By definition, in order to be rational, it has to be a ratio of integers, that is, we have to write it in the form $\frac{a}{b}$ (where $b \neq 0$ ) in the lowest (simplest) terms: $\sqrt{2} = \frac{a}{b}$
We can take the square of both sides: $2 = \frac{a^2}{b^2}$
Multiplying both sides by $b^2$ we get: $2b^2 = a^2$
Since $a^2$ is an even number, $a$ has to be even (the square of an even number is an even number).
That means we can write $a$ as $2c$ where $c$ is an integer.
So, now we can write the whole thing as: $2b^2 = (2c)^2$
And, continue: $2b^2 = 4c^2$
Simplifying the expression, we get: $b^2 = 2c^2$
Now, since $b^2$ is an even number, $b$ has to be even as well.
We see that both $a$ and $b$ are even — which means they have a common factor, and that means $\frac{a}{b}$ cannot be the most simplified form. We have a contradiction, therefore $\sqrt{2}$ is irrational. $\blacksquare$

NOTE:
You might wonder why we assumed that if " $n^2$ is even, then $n$ is even". Remember that with prime factorization, all integers more than $1$ can be uniquely represented as a product of prime numbers. Since $n$ is one of the divisors of $n^2$ , if $2$ is among the primes that divide $n^2$ , then one of the prime divisors of $n$ has to be $2$ as well. Therefore, if $n^2$ is even, $n$ is even.

Bite-Sized Mathematics for Computer Science

Notes on MIT's 6.042J

(Two) Proof Methods

Proof by Cases

Proof by Contradiction