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 one year ago
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
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"

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KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Well, 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?

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

niikofi12
 one year ago
Best 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"

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1This 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?

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

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Well, 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 \((x+y)R(x'+y')\).

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