It's 2 Easy 2 Be Even

$Odd \times Odd \neq Even$

Proof:- Let $\displaystyle 2m+1$ and $\displaystyle 2n+1$ be two odd numbers.

Their product is:

$(2m+1)(2n+1) = 4mn + 2m + 2n +1$

$\displaystyle \Rightarrow (2m+1)(2n+1) = 2(2mn + m + n) + 1$

$\displaystyle \Rightarrow (2m+1)(2n+1) = 2k + 1 = Odd$

All of the given identities are correct except for "odd x odd = even."

Counterexample: $3 \times 5 = 15$

The product of two odd numbers is actually always and odd number: "odd x odd = odd"

Proof:

Consider the prime factorization of any two odd numbers. The prime factorization of the product of the two odd numbers will be the product of the two prime factor lists. Therefore, unless there's a 2 in one of the prime factorizations of the original numbers, there won't be a 2 in the prime factorization of the product. Therefore, if the original two numbers are odd, the product will be odd as well.

And I'll leave proving that all of the other given identities are correct to other solution writers! :)

even+even=even 2+2=4 odd+odd=even 3+3=6 even x odd= even 6x5=30 even x even= even 4x4=16 odd x odd= even X <-

Let $2k$ means even and $2k+1$ means odd.Let's check

1. $2k \times 2k=4k^2=2(2k^2)$ which is even

2. $(2k+1)+(2k+1)=4k+2=2(k+1)$ which is even

3. $2k \times (2k+1)=4k^2+2k=2(2k^2+k)=2(k(2k+1))$ which is even

4. $2k+2k=4k=2(2k)$ which is even

5. $(2k+1) \times (2k+1)=4k^2+4k+1=2(2k^2+2k)+1=2(2(k^2+k))=2(2k(k+1))+1$ which is odd,not even