A direct proof of a mathematical statement is a valid argument, in words, that shows a mathematical statement is tautologically true given a set of assumptions.
Claim: An odd plus an odd is an even.
Proof: One must be very precise with definitions when doing a mathematical proof. In this case, by an odd we mean an integer in the set \(\{1, -1, 3, -3, 5, -5, \dots\}\) and by an even we mean an integer in the set \(\{0, 2, -2, 4, -4, \dots\}.\)
By a quick check, you can see that sometimes and odd plus an odd is an odd. For example, \(1\) and \(3\) are both odds, and \(1+3 = 4\) is even. Similarly, \(-5\) and \(5\) are both odd, and \(-5 + 5 = 0\) is an even. However, it is impossible to check all cases by hand. To show the statement is always true, we need a proof.
Another way to write the set of odds is \(\{x : x = 2k+1 \text{ for some integer }k\},\) and another way to write the set of evens is \(\{x : x = 2k \text{ for some integer }k\}.\) For example, \(5\) is odd and \(5 = 2 \cdot 2 + 1.\) Similarly, \(-7\) is odd and \(-7 = -4 \cdot 2 + 1.\) On the other hand, \(6\) is even and \(6 = 3 \cdot 2,\) and \(0\) is even and \(0 = 0 \cdot 2.\)
Let \(a\) and \(b\) be odd numbers. Then there exists integers \(c\) and \(d\) such that \(a = 2c+1\) and \(b = 2d+1.\) So,
\begin{align}
a+b & = (2c+1)+(2d+1) \\
& = 2c + 2d + 2 \\
& = (c + d + 1) \cdot 2
\end{align}
Since \(c+d+1\) is an integer, \((c+d+1) \cdot 2\) is an even. Since \(a\) and \(b\) could be any odd numbers, any odd plus any odd is an even.
In general, a proof will have the structure above. There will first be a claim, which will be called a Claim, Lemma, Corollary or Theorem. Then there will be the start of a argument, which will be either Proof or Counter Example.