Respuesta :
Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even.
m=2k-n, p=2l-n
Let m+n and n+p be even integers, thus m+n=2k and n+p=2l by definition of even
m+p= 2k-n + 2l-n substitution
= 2k+2l-2n
=2 (k+l-n)
=2x, where x=k+l-n ∈Z (integers)
Hence, m+p is even by direct proof.
m=2k-n, p=2l-n
Let m+n and n+p be even integers, thus m+n=2k and n+p=2l by definition of even
m+p= 2k-n + 2l-n substitution
= 2k+2l-2n
=2 (k+l-n)
=2x, where x=k+l-n ∈Z (integers)
Hence, m+p is even by direct proof.
This is about usage of direct proof.
m + p was proved to be even using direct proof.
- We are given;
m, n and p are integers
m + n and n + p are even integers.
- From direct proof in Maths;
If a is an even integer, then there will exist an integer b such that;
a = 2b
- Applying that to our question, we can say that;
m + n = 2b
n + p = 2c
- Adding both equations together gives;
m + 2n + p = 2b + 2c
- We want to find if m + p is even. Thus, let's rearrange to get them on one side;
m + p = 2b + 2c - 2n
Factorizing, we have;
m + p = 2(b + c - n)
From earlier, b, c and n are integers. Therefore, from earlier formula about direct proof of even integers, we can equally say that m + p is also even.
Read more at; brainly.com/question/17255081