I'm stuck trying to prove a basic proposition about the arithmetic mean from an Algebra book. The book asks to prove first a theorem that goes something like this:
Let $A = \frac{a_1 + \cdots + a_n}{n}$ be the arithmetic mean of the real numbers $a_1, \cdots, a_n$. By way of contradiction, show there exists $i$ such that $A\leq a_i$.
My proof:
Proof. BWOC. Let $i$ be an arbitrary natural number such that $1 \leq i\leq n$ and suppose $a_i < A$. Since $i$ is arbitrary, in particular, $a_1< A$, $\cdots$ , $a_n<A$. Then, $$ a_1 + \cdots + a_n < An$$ $$ An < An$$ Which is clearly false, then, there exists $i$ such that $A\leq a_i$.
$\blacksquare$
Then the book asks to prove or disprove there exists $i$ such that $A < a_i$. Which is the part where I'm stuck
So, let $a_1=a_2=1$.Then, $\frac{a_1+a_2}{2}= \frac{1+1}{2}=1=A$ hence $1=a_1=a_2=a_i<A=1$ but $1<1$ is false. Hence no $i$'s satisfy $A < a_i$
Similarly, let $a_1=a_2=0$.Then, $\frac{a_1+a_2}{2}= \frac{0}{2}=0=A$ hence $0=a_1=a_2=a_i<A=0$ but $0<0$ is false.
On the other hand, suppose there does not exists $i$ such that $A < a_i$. Hence, for all $i$ st. $1\leq i \leq n$, it is the case that $a_i \leq A$, but by the theorem above, there exists $i$ such that $A\leq a_i$. So, for all $i$, $a_i \leq A$ is false too.
Where is the flaw on my logic. It seems to me that I'm not negating statements correctly, or Am I?
If there is no $i$ such that $A <a_i$, then in fact all the $a_i$ have to be equal to $A$: in this case, $a_i \le A$ and $A \le a_i$ are true for all $i$.