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

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KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

niikofi12 Group TitleBest ResponseYou've already chosen the best response.0
i was able to come up with that, but i'm still struggling with the finishing
 one year ago

niikofi12 Group TitleBest ResponseYou've already chosen the best response.0
@KingGeorge....must i equate them? if so then m gona gt (x+y)=(x'+y') but duz equality mean "related"
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
This relation is not an equality. However, note that \(xx'\) is an integer and so is \(yy'\). The sum/difference of two integers is still an integer, so \((xx')(yy')\) is still an integer. That make sense?
 one year ago

niikofi12 Group TitleBest ResponseYou've already chosen the best response.0
makes sense but my headache is how to gt a plus sign between d grouped like terms and conclude that they are related
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Well, hold on for bit longer. We have that \((xx')+(yy')\) is an integer since \(xx'\) is an integer and \(yy'\) is an integer. But \((xx')+(yy')=(x+y)(x'+y')\). 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 \((x+y)R(x'+y')\).
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Was that clearer?
 one year ago

niikofi12 Group TitleBest ResponseYou've already chosen the best response.0
much better. i appreciate ur help. u r the best!
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
You're welcome.
 one year ago
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