--- ## Definition > [!tldr] Definition > A **direct proof** is a [[Proof|proof]] applied specifically to a [[Conditional statements|conditional statement]] which is structured as follows: > 1. Assume that the [[Conditional statements|hypothesis]] of the conditional statement is true; > 2. Then use definitions, computations, previously-proven results, or basic facts to explain why the [[Conditional statements|conclusion]] of the conditional statement is true. Notes: - A direct proof uses the logic of the [[Conditional statements|conditional statement]] to structure the argument: According to the [[Truth tables|truth table]] for $P \rightarrow Q$, the statement is true when $P$ is false; so we can begin the argument by assuming that $P$ is true. Then it only requires to show that $Q$ must also be true in that condition. ## Examples and Non-Examples > [!NOTE] Proof involving even and odd integers > **Prove: If $n$ is an even integer, then $n^2$ is also even.** > > **Proof:** We use a direct proof. First, assume that $n$ is an [[Even and odd integers|even]] integer. > > Since $n$ is even, we can write it as $n = 2k$ where $k$ is another integer. (This is the definition of "[[even]]".) Now look at $n^2$: Since $n = 2k$, we know that $n^2 = (2k)^2$, which means $n^2 = 4k^2$ (using basic algebra to expand $(2k)^2$). We can factor off a $2$ from this to get $n^2 = 2(2k^2)$. Now since $k$ is an integer, $k^2$ is also an integer because of [[Closure of the integers|closure]]; and therefore $2k^2$ is an integer also because of [[Closure of the integers|closure]]. Therefore we have shown there exists an integer, namely $2k^2$, such that $n^2$ is 2 times that integer. This shows that $n^2$ is even, which is what we wanted to show. ◾ > [!NOTE] Proof involving [[Divisibility|divisibility]] > **Prove: If $a,b,$ and $c$ are integers and $a|b$ and $b|c$, then $a|c$.** > > **Proof:** Assume the hypothesis, which says that $a|b$ and $b|c$. Using the definition of [[Divisibility|divisibility]], we can then say that there exist integers $x$ and $y$ such that $b = ax$ and $c = by$. We want to show that there is an integer $z$ such that $c = az$. > > Since $b = ax$ (which we assumed to be true) we can make a substitution into $c = by$ (which we also assumed to be true) to get $c = (ax)y$. The [[Properties of arithmetic|associative property]] then says that $c = a(xy)$. Since $x$ and $y$ are integers, by [[Closure of the integers|closure]] so is $xy$. Therefore we have an integer $z = xy$ such that $c = az$. Therefore $a|c$. ◾ ## Resources ![](https://www.youtube.com/watch?v=H8LLINU6ebY) ![](https://www.youtube.com/watch?v=1tCOucLfdh0) ![]() There are a lot more videos with proof examples at the [Grand Valley State University Communicating in Mathematics playlist](https://www.youtube.com/playlist?list=PL2419488168AE7001).