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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"

Algebra
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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?
i was able to come up with that, but i'm still struggling with the finishing
@KingGeorge....must i equate them? if so then m gona gt (x+y)=(x'+y') but duz equality mean "related"

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Other answers:

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?
makes sense but my headache is how to gt a plus sign between d grouped like terms and conclude that they are related
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 \((x+y)R(x'+y')\).
Was that clearer?
much better. i appreciate ur help. u r the best!
You're welcome.

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