## niikofi12 one year ago Let R be the relation on the set of real numbers defined by {xRy: x-y is an integer}. Prove that if xRx' and yRy' then (x+y)R(x'+y'). N/B:"R" means "related"

1. KingGeorge

Well, look at $$(x+y)−(x′+y′)=(x−x′)+(y−y′).$$ From your assumption, you have that xRx' and yRy'. This means that x−x′ is an integer and y−y′ is an integer. There's basically only one step left. Can you finish it from here?

2. niikofi12

i was able to come up with that, but i'm still struggling with the finishing

3. niikofi12

@KingGeorge....must i equate them? if so then m gona gt (x+y)=(x'+y') but duz equality mean "related"

4. KingGeorge

This relation is not an equality. However, note that $$x-x'$$ is an integer and so is $$y-y'$$. The sum/difference of two integers is still an integer, so $$(x-x')-(y-y')$$ is still an integer. That make sense?

5. niikofi12

makes sense but my headache is how to gt a plus sign between d grouped like terms and conclude that they are related

6. KingGeorge

Well, hold on for bit longer. We have that $$(x-x')+(y-y')$$ is an integer since $$x-x'$$ is an integer and $$y-y'$$ is an integer. But $$(x-x')+(y-y')=(x+y)-(x'+y')$$. So this is an integer. Finally, we go back to the fact that $(x+y)R(x'+y') \iff (x+y)-(x'+y')\;\; \text{is an integer}.$But we just showed that! So $$(x+y)R(x'+y')$$.

7. KingGeorge

Was that clearer?

8. niikofi12

much better. i appreciate ur help. u r the best!

9. KingGeorge

You're welcome.