Direct Proof

Prove that the product of two consecutive odd numbers is always odd

Let $a$ and $b$ be the two consecutive odd numbers. 1.1. Without loss of generality, assume that $a < b$ , hence $b = a + 2$ 1.2. Now, $a = 2 k + 1$ for some integer $k$ (by definition of odd numbers) 1.3. Similarly, $b = a + 2 = 2 k + 3$ 1.4. Therefore, $ab = (2 k + 1) (2 k + 3) = 4 k^{2} + 8 k + 3 = 2 (2 k^{2} + 4 k + 1) + 1$ (by basic algebra) 1.5. let $m = (2 k^{2} + 4 k + 1)$ which is an integer (by closure of integers under addition and multiplication) 1.6. Then, $ab = 2 m + 1$ , which is odd (by definition of odd numbers)
Therefore, the product of two consecutive odd numbers is always odd

WLOG

“Without loss of generality” may be abbreviated to WLOG

This is used before an assumption in a proof which narrows the premise to some special case
It implies that the proof of this special case can be easily applied to all other cases

Prove that the following statement is not true: The product of two irrational numbers is always irrational.

Let the two irrational numbers be $2$ and $8$ 1.1. Then, $2 \times 8 = 2 \times 8$ (by basic algebra), which $= 16 = 4$ , which is a rational number
Therefore, the statement “the product of two irrational numbers is always irrational” is not true

Prove the following: $\exists x \in Z$ s.t. (such that) $x > 2$ and $x^{2} - 5 x + 6 > 0$

Proof by construction is where you explicitly find the value with the correct properties
It is a form of Direct Proof
No need to explain how you found the result (e.g. 17), you just need to show that the result has the required properties

Assume that ~S (not S) is true
Based on this, use known facts and theorems to arrive at a logical contradiction.
Since every step of the argument thus far is logically correct, the problem must lie in the assumption (that ~S is true)
Thus, it can be concluded that ~S is false, that is, S is true

Prove that $2$ is irrational

Suppose not, that $2$ is rational 1.1. Then, $\exists a, b \in Z, b \neq = 0$ s.t. $2 = \frac{a}{b}$ (by definition of rational numbers) 1.2. Convert $\frac{a}{b}$ to its lowest term, $\frac{m}{n}$ 1.3. $m^{2} = 2 n^{2}$ (by basic algebra) (note: this is done by squaring both sides of $2 = \frac{m}{n}$ ) 1.4. Hence, $m^{2}$ is even (by definition of even numbers, as $n^{2}$ is an integer by closure) 1.5. Hence, $m$ is even (by Proposition 4.6.4 (has been proved somewhere else)) 1.6. Let $m = 2 k$ ; Substituting into 1.3: $4 k^{2} = 2 n^{2}$ , or $n^{2} = 2 k^{2}$ 1.7. Hence, $n^{2}$ is even (by definition of even numbers) 1.8. Hence, $n$ is even (by Proposition 4.6.4) 1.9. So both $m$ and $n$ are even, but this contradicts that $\frac{m}{n}$ is in its lowers term
Therefore, the assumption that $2$ is rational is false
Hence $2$ is irrational

Prove that the difference of two consecutive squares between 30 and 100 is odd

The squares between 30 and 100 are 36, 49, 64 and 81 1.1. Case 1: $49 - 36 = 13$ which is odd 1.2. Case 2: $64 - 49 = 15$ which is odd 1.3. Case 3: $81 - 64 = 17$ which is odd
Therefore, the difference of two consecutive squares between 30 and 100 is odd