Sum of $N$ natural numbers is less than $N+2$, then each number is less than $3$. Prove by contradiction: If the sum of $N$ natural numbers is less than $N + 2$ then each of these numbers is less than $3$.
Attempt: I have to assume that the sum of $N$ natural numbers is greater than $N+2$ then each of these numbers is greater than $3$. 
What do I do now? Need help.
Cheers, 
Chris
 A: You have an implication $p\to q$, where $p$ is 

the sum of a certain set of $N$ natural numbers is less than $N+2$,

and $q$ is

each of the $N$ numbers is less than $3$.

To prove such an implication by contradiction, you assume that $p$ is true and $q$ is false, and you try to derive a contradiction from that assumption. Assuming that $p$ is true is assuming that you have $N$ natural numbers, say $a_1,\ldots,a_N$, whose sum is less than $N+2$. Assuming that $q$ is false is assuming that it’s not the case that each of these numbers is less than $3$, which is the same as saying that at least one of them (not all of them) is greater than or equal to $3$ (not greater than $3$).
So our assumptions are:


*

*$a_1,\ldots,a_N$ are natural numbers;  

*$a_1+a_2+\ldots+a_N<N+2$; and  

*at least one of the numbers $a_1,\ldots,a_N$ is $\ge 3$.


We can label the numbers in any order we like, so we might as well assume that $a_1\ge 3$. The other $N-1$ numbers, $a_2,\ldots,a_N$, are all at least one: $a_2\ge 1,a_3\ge 1,\ldots,a_N\ge 1$. Thus,
$$a_1+a_2+\ldots+a_N\ge 3+1+\ldots+1\;.$$
Do you see the contradiction?
