Prove that for every even integer n, there is an even integer k, such that: n <

Question

Prove that for every even integer n, there is an even integer k, such that: n < k < n+3.
Remember that assuming existence of k and deriving a value for it does not prove existence.
You may assume the addition rules for even and odd numbers (even + even is even, even + odd is odd, etc.) without proof. Prove that for every even integer n, there is an even integer k, such that: n < k < n+3.
Remember that assuming existence of k and deriving a value for it does not prove existence.
You may assume the addition rules for even and odd numbers (even + even is even, even + odd is odd, etc.) without proof.
Remember that assuming existence of k and deriving a value for it does not prove existence.
You may assume the addition rules for even and odd numbers (even + even is even, even + odd is odd, etc.) without proof.

Explanation / Answer

ANSWER 1:

n < k < n +3

add 3

n+3 < k+3 < n+6

where n+3 is odd

also k+3 is odd

also n+6 is even

it holds true that after two odd number if there is even number also it is true beween even and odd number one even number exists.

example btween 4 and 7

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