Question 1140421

Consecutive even numbers can be represented by {{{x}}}(smaller one) for the first and the next number has to be {{{2}}} more than that. So the next even number is {{{x + 2}}} (greater one). 

if the greater of two consecutive even integers is more than one-half times the smaller, we have

{{{x + 2>(1/2)x}}}

{{{x -(1/2)x>-2}}}

{{{(1/2)x>-2}}}

{{{x>-2/(1/2)}}}

{{{x>-4}}}=> first possible solution for {{{x}}} is {{{-2}}}

then {{{x + 2}}} (greater one) will be {{{-2 + 2=0}}}

the smallest possible integers: {{{-2}}}, {{{0}}}