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niikofi12

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"

  • one year ago
  • one year ago

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  1. KingGeorge
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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?

    • one year ago
  2. niikofi12
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    i was able to come up with that, but i'm still struggling with the finishing

    • one year ago
  3. niikofi12
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    @KingGeorge....must i equate them? if so then m gona gt (x+y)=(x'+y') but duz equality mean "related"

    • one year ago
  4. KingGeorge
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    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?

    • one year ago
  5. niikofi12
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    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
  6. KingGeorge
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    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 \((x+y)R(x'+y')\).

    • one year ago
  7. KingGeorge
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    Was that clearer?

    • one year ago
  8. niikofi12
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    much better. i appreciate ur help. u r the best!

    • one year ago
  9. KingGeorge
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    You're welcome.

    • one year ago
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