This problem aims to familiarize us with the **method of poof.** The concept required to solve this problem is related to **discrete mathematics**, including **direct proof** or** proof by contradiction,** and** proof by contrapositive.**

There are multiple methods to write a **proof,** but here we are going to see only two methods, **proof by contradiction** and** proof by contrapositive**. Now proof by **contradiction** is a kind of proof that **demonstrates** the truth or the reality of a proposal, by exhibiting that **considering** the proposal to be incorrect **points** to a contradiction. It is also comprehended as **indirect proof.**

For a **proposal** to be **proved,** the event such as $P$ is assumed to be **false,** or $\sim P$ is said to be **true.**

Whereas the method of **proof by contrapositive** is utilized to prove **conditional statements** of the structure “If $P$, then $Q$”.This is a **conditional** statement that shows that $P \implies Q$. Its **contrapositive** form would be $\sim Q \implies \sim P$.

## Expert Answer

Let’s **suppose** $m\times n$ is even, then we can assume an **integer** $k$ such that we get a **relation:**

\[ m\times n= 2k\]

If we get $m$ to be **even** then there is **nothing** to **prove,** so let’s say that $m$ is **odd.** Then we can set the value of $m$ to be $2j + 1$, where $j$ is some **positive integer:**

\[ m = 2j + 1 \]

Substituting this into the **first equation:**

\[ m\times n= 2k\]

\[ (2j + 1)\times n= 2k\]

\[ 2jn + n = 2k\]

And therefore,

\[ n= 2k – 2jn \]

\[ n= 2(k – jn) \]

Since $k – jn$ is an **integer,** this shows that $n$ would be an **even number.**

**Proof by contraposition:**

Suppose that the **statement** “$m$ is even or $n$ is even” is **not true.** Then both $m$ and $n$ are supposed to be **odd.** Let’s see if the product of **two odd numbers** is an **even** or an **odd number:**

Let $n$ and $m$ be equal to $2a + 1$ and $2b + 1$ respectively, then their **product** is:

\[ (2a+1)(2b+1) = 4ab+2a+2b+1 \]

\[ = 2(2ab+a+b)+1 \]

This shows that the **expression** $2(2ab+a+b)+1$ is of the form $2n+1$, thus the **product** is **odd.** If the **product** of odd numbers is **odd,** then $mn$ is not true to be even. Therefore, in order for $mn$ to be **even,** $m$ must be **even** or $n$ must be an **even number.**

## Numerical Result

In order for $mn$ to be **even,** $m$ must be even or $n$ must be an **even number proved** by **contraposition.**

## Example

Let $n$ be an **integer** and the **expression** $n3 + 5$ is odd, then prove that $n$ is **even** by using **p****roof by contraposition.**

The **contrapositive** is “If $n$ is odd, then $n^3 +5$ is **even.”** Suppose that $n$ is odd. Now we can write $n=2k+1$. Then:

\[n^2+5= (2k+1)3+5 =8k^3+12k^2+6k+1+5\]

\[=8k^3+12k^2+6k+6 = 2(4k^3+6k^2+3k+3)\]

Hence, $n^3+5$ is** twice** some** integer**, thus it is said to be **even** by the **definition** of **even integers.**