Prove by contradiction that there is an $i \in [n]$ such that $x_i \geq 2$ if $x_1,\ldots,x_n \in \mathbb{N} \cup \{0\}$ Prove this statement using a proof by contradiction: 
Let $n$ be a natural number. If $x_1,\ldots,x_n \in \mathbb{N} \cup \{0\}$ and $\sum_{i=1}^{n}{x_i} = n+1$ then there is an $i \in [n]$ such that $x_i \geq 2$
I'm not sure how to approach this problem with the proof by contradiction. So far, I assume that there is no $i \in [n] : x_i \geq 2$. Then, $\sum_{i=1}^{1}{x_i} < 2$. Finally, because $\sum_{i=1}^{1}{x_i} = n + 1 = 2$, I can conclude that by contradiction this is false.
I'm not entirely sure if this logic is a valid proof by contradiction or what a better proof would be. Any help is appreciated
 A: The mistake is that you are focusing only on $\sum_{i=1}^1 x_i = x_1$ to get your contradiction, but we don't know anything about this sum. The sum that we do know a lot about is $\sum_{i=1}^n x_i = x_1 + x_2 + \cdots + x_n = n+1$, so we should try and contradict this assumption.
To be clear, the reason why we don't know anything about $\sum_{i=1}^1 x_i$ is because $n$ was given and is fixed. We don't know what the value of $n$ is, nor do we have control over what value it is, just that we have it
So with proof by contradiction, we would want to try to contradict some assumption or some known fact.
As you pointed out, we assume that there is no $i \in [n]$ such that $x_i \geq 2$. What's another way to say this? Well, this is saying for every $i \in [n], \, x_i < 2$. Now we do have control of this $i$, so we know that $x_i = 0,1$, as each is a nonnegative integer.
Now if we consider the sum $\sum_{i=1}^n x_i = x_1+x_2 + \cdots + x_n$, we note that each $x_i$ could equal $0$ or it could equal $1$. Obviously, the maximal possibility is that each $x_i = 1$, so we have an inequality:
$$\sum_{i=1}^n x_i = \underbrace{x_1}_{\leq 1} + \underbrace{x_2}_{\leq 1} + \cdots + \underbrace{x_n}_{\leq 1} \,\, \leq \,\, \underbrace{1 + 1 + \cdots + 1}_{n-\text{times}} = n.$$
So the sum can be at most $n$. But this is a contradiction, as we know that $\sum_{i=1}^n x_i = n+1$.
A: If $x_i\leq 1$ for all $i$, then $x_1+\ldots+x_n\leq 1+\ldots +1 = n\cdot 1 = n$.
Contraposition:
If $x_1+\ldots+x_n> n$, there is an $i$ with $x_i\geq 2$.
$P\Rightarrow Q \Leftrightarrow \neg Q\Rightarrow\neg P$.
