Let’s refresh the concept of direct proof.

“A direct proof of a conditional statement p → q is constructed when the first step is the assumption that p is true; subsequent steps are constructed using rules of inference, with the final step showing that q must also be true. A direct proof shows that a conditional statement p → q is true by showing that if p is true, then q must also be true, so that the combination p true and q false never occurs.” Discrete Mathematics and its Applications by Rosen.

We can rephrase the exercise as follows: if n is odd, then n is the difference of two squares.

As per the definition of direct proof, we will assume that p (n is odd) is true, and we are going to take subsequent steps, using rules of inference, to show that q is also true.

n=2k+1 for a certain integer k (definition of odd number).

Now, we must prove that 2k+1 is the difference of 2 squares.

2k+1 = 1 (2k+1)

= (k+1-k)( 2k+1)

= 2k^{2}+ k + 2k + 1 -2k^{2} -k

= 2k^{2}+ 3k + 1 – 2k^{2} – k, simplifying 3k-k

= 2k^{2 }+ 2k + 1 – 2k^{2}

= k^{2}+ k^{2 }+ 2k + 1 – k^{2} – k^{2}, simplifying one term k^{2} with one term -k^{2},

= k^{2}+ 2k + 1 – k^{2},

= (k+1)^{2} – k^{2}

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**Related exercises:**

- 3. Show that the square of an even number is an even number using a direct proof
- 4. Show that the additive inverse, or negative, of an even number is an even number using a direct proof
- 5. Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even. What kind of proof did you use?
- 6- Use a direct proof to show that the product of two odd numbers is odd