Why is this relation transitive? $R = \{(x,y)\in\Bbb N^2\mid x + 4y = 10\}$
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So you have the linear equation $y=-\frac14 x + \frac52$. Maybe you can try visualizing it from there? – Mason Jun 14 '18 at 22:53
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a cursory trial shows it's not reflexive. – ncmathsadist Jun 14 '18 at 22:55
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The question was presented like this in the textbooks.. and then everything else is fine. I didn't understand how it is transitive. – zion Jun 14 '18 at 22:59
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The point is whenever you can find a $y$ such that $xRy$ and $yRz$, then all $x, y,z$ are forced to equal to $2$, so $xRz$ is also true. – achille hui Jun 14 '18 at 23:05
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Assume x + 4y = 10 and y + 4z = 10. Does x + 4z = 10?
Exercise. Prove for x,y in N, x + 4y = 10, the solutions are
(2,2), (6,1).
Thus if aRb and bRc, then b = 2.
Whence a = b = c = 2, so aRc.
William Elliot
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N is collection of Natural Numbers.. 0 does not exist in N.. So just because it is not having a (y,z) pair.. can lead a relation to be accepted as transitive? – zion Jun 15 '18 at 06:23
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