Question

(20pts) Prove that the product of any two odd numbers is always odd. Make sure to explain your proof​

Answer 1

Answer:

Prove that the product of any two odd numbers is always odd.

Examples:

$3 \cdot 7 = 21 \text{ } \boxed{\text{odd}}\\5 \cdot 9 = 45 \text{ } \boxed{\text{odd}}\\83 \cdot 3 = 249 \text{ } \boxed{\text{odd}}$

Defining an odd number as:

$u= 2k+1, k \in \mathbb{Z}$

$u \cdot u = (2k+1)(2k+1)= 4k^2+4k+1$

$\text{Is } \boxed{4k^2+4k+1} \text{ odd?}$

Yes, it is, because $4k^2 \text{ and } 4k \text{ will always be even considering } k \in \mathbb{Z}$

An even number plus one is an odd number.

Answer 2

Answer:

See below (I hope this helps!)

Step-by-step explanation:

Because odd numbers are always 1 greater than even numbers, we can call the two odd numbers x + 1 and y + 1 where x and y are even integers. Multiplying the two gives us:

(x + 1) * (y + 1)

= x * y + x * 1 + 1 * y + 1 * 1

= xy + x + y + 1

We know that x * y will be even because x and y are also even and the sum of two even numbers will be even, and we also know that x and y are even and that 1 is odd. Since the sum of even and odd numbers is always odd, the product of any two numbers is always odd.

*NOTE: I put a limitation on x and y in my proof (the limitation was that x and y must be EVEN integers) but you don't have to do that, you could make the odd integers 2x + 1 and 2y + 1 where x and y could be any integer from the set Z like mirai123 did. I simply gave this proof because it was the first thing that came to mind. While mirai123's proof and mine are different, they are still both correct.