Today we will discuss how to prove statements with “for all/every” or “there exists.”
Warm-up:
Prove that for every even integer $n$, the integer $n + 2$ is even.
Start with “Suppose that $n$ is an even integer.” Then prove that $n + 2$ is even.
Prove that there exists an integer $n$ which is even.
Proof. The integer 2 is even by definition, since is $2 = 2(1)$.
Recipe for proving a “for all” statement:
<aside>
Proposition. For all $x$, $P(x)$.
Proof. Suppose $x$. … Therefore $P(x)$.
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Recipe for proving a “there exists” statement.
<aside>
Proposition. There exists an $x$ such that $P(x)$.
Proof. Let $x$ be… [describe a specific example]. … Therefore $P(x)$.
</aside>
Today we’ll work with the definition of a derivative:
Definition. Let $f$ be a function $\mathbb{R} \to \mathbb{R}$, and let $c, L \in \mathbb{R}$. We say that $f'(c) = L$ (i.e. the derivative of $f$ at $c$ is $L$ if for all $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x \in \mathbb{R}$, if $|x - c| < \delta$, then
$$ \left| \frac{f(x) - f(c)}{x - c} - L\right| < \varepsilon \,. $$
Let’s prove that $f'(1) = 10$, where $f(x) = 2x^2 + 6x + 9$.
Scratch work: Let’s compute
$$ \begin{aligned} \frac{f(x) - f(c)}{x - c} &= \frac{(2x^2 + 6x + 9) - f(1)}{x - 1} \\ &= \frac{2x^2 + 6x + 9 - 17}{x-1} \\ &= \frac{2x^2 + 6x - 8}{x-1} \\ &= \frac{(x-1)(2x+8)}{x-1} \\ &= 2x + 8 \,. \end{aligned} $$
Then