Why Are Odd Numbers Not Even?

There are even numbers and odd numbers. But for some reason, there aren't any numbers that are even and odd at the same time. Why is that? In this video we give a proof that odd numbers are not even!
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Music: C418 - Pr Department

[@suvrotica]

Can we also show the contradiction by saying that m-n =1/2 here and that cannot be, since m,n are integers and m>n ?

[@MuPrimeMath]

Yes, that would also work! Since m>n, we have m-n ≥ 1 > 1/2, which is a contradiction.

[@matejcataric2259]

Nice stuff! Do more Linear Algebra and calc that is also very interesting :) Love your stuff

[@radhamadhavkr]

Nice explaining Love ❤from India🇮🇳

[@drsonaligupta75]

Welcome back!!!

[@jordyclash3386]

Extrañaba tus videos tan geniales y educativos. Saludos desde Costa Rica

[@cycklist]

Welcome back mate.

[@aashsyed1277]

Gr8 videos!

[@guill3978]

Can you prove that every divisor of 2^n-1 where n is prime is congruent to 1mod(2n) and also congruent to 1mod(8) or -1mod(8)?

[@erikestrella7240]

El final fue mi parte favorita!!! Now... Escribire algo para ti. La conjetura de goldbach dice que: para todo 2k existen p,q tal que p+q = 2k Ok??... Yo encontré un forma de encontrar las soluciones. If 2k = p+q then k^2 > pq Let u = Max(pq) pq = k^2 -i where i natural Ejemplo 2k = 30 k^2= 225 pq = 225-4 =(17)(13) Cool!! Jaja solo es por diversión (;

[@hotlatte1222]

No wonder, in this world, odd things are not treated even. You just proved a social bias. And it is also a prime principle. So sad.

[@angelmendez-rivera351]

I think the manner in which the question is phrased is silly. By definition, the odd integers are those integers which are not the even integers. The question should be rephrased instead as, why can odd numbers be written as 2·m + 1? This comes down to the Euclidean division algorithm. Every integer can be written in the form m = 2·q + r. Notice that r = 0, or r = 1, without loss of generality, because if r = 2, then 2·q + r = 2·q + 2 = 2·(q + 1) = 2·(q + 1) + 0, and so, q + 1 |-> q means that r = 2 |-> r = 0. So every integer m can be written as 2·q + r, where r = 0 or r = 1. This can be proven using some extended induction. From this, it follows that if r = 0, then m is even. Therefore, m is odd if and only if r = 1.

[@MuPrimeMath]

In this video, I define odd numbers as integers of the form 2k+1, not as the numbers that aren't even. The two definitions are equivalent.