Task: Prove that the sum of two even integers is even.
Definition: An integer $n$ is even if there exists an integer $k$ such that $n = 2k$.
n is even if there is an integer such that
$n = 2k$ or $n = k + k$ since $2k = k + k$.
If given 2 even integers $A$ and $B$, so by definition, $A = a +a$ and $B = b + b$ given some integer a and b.
therefore $A + B = a+a+b+b = (a+b) + (a+b)$ which is equivlant to $2*(a+b)$ which is the definition of an even integer.
an integer is even if there is an integer k such that n = 2k.
If a = 2a and b=2b than a+b=2k because a+a+b+b = 2a+2b = 2k
$n=2k$ == $k=n/2$
An even number is any number $a$ for which $a=2c$, provided $c$ is an integer. If $n=2x$ and $m=2y$, $m+n = 2x+2y = 2(x+y)$, therefore, for any even $m$ and $n$, $m+n$ is even.
A number n is even if there is an integer $k$ such that $n=2k$.
Let $2a$ and $2b$ be even integers. Let the sum of two even numbers be $2a+2b=2(a+b)$. Let $2(a+b)=2k$. Since the sum of two integers, $a$ and $b$, results in an integer $k$, this follows the form $n=2k$ where $k=a+b$ and $n=2a+2b$.
Therefore, the sum of two even integers $a$ and $b$ is even.
An integer n is even if there is some integer k such that n=2k.
Let $A$, $B$, $C$ and $D$ be integers such that $A=2C$ and $B=2D$
Let $E=A+B$
Substitute 2C in for A and 2D in for B
$E=(2C)+(2D)=(C+C)+(D+D)=(C+D)+(C+D)=2(C+D)$
Because $E=2(C+D)$ and $(C+D)$ is an integer, E is even.
Let $n$ and $m$ be even integers. By definition of “even”, we can say that $n = 2k$ and $m = 2l$ for some integers $k$ and $l$. Then
$$ n + m = 2k + 2l = 2(k + l), $$
which is even by definition, since $k + l$ is also an integer.
Let A and B be even Integers
$A = 2k$ and $B = 2c$ with $k$ and $c$ being any integer
$A$ and $B$ are inherently even integers because of $n = 2k$ being the definition of an even integer
Thus, $A+B$ = Even integer because $2k + 2c = 2(k + c)$ because sum of the integer is a multiple of two
Any Even number is divisible by 2 therefore any even number can be thought of as a specific amount of 2’s. If it is odd it wont have a perfect amount of 2’s in it. Such as 5 has two sets of 2 plus 1 so that isn’t even, but 6 has 2 in it three times. Thus, n is even if there is an integer k such that $n = 2k$.
$a+b=k$
$2*(a+b)=2a+2b$
$2k=2a+2b$
Due to these equations we can prove that starting K at any number and $*/$ by 2 we will always get a even. If a and b are any real numbers the equation below stand.