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21 February 2009 @ 12:50 am

Malproposition:  Suppose that ~ is a relation on a set A.  If ~ is symmetric and transitive, then ~ is reflexive.

Proof:  Let x,y belong to A.  If x ~ y then y ~ x since ~ is symmetric.  Now, x ~ y and y ~ x and since ~ is transitive, we can conclude x ~ x.  Therefore ~ is reflexive.

Find the flaw in this reasoning.

(This one I succeeded in doing.)
Current Mood: busybusy
Stevie: Satchelsstrickl on February 21st, 2009 06:36 am (UTC)
It switches from talking about what x ~ y implies to assuming that x ~ y for our randomly chosen pair of elements (x, y) in A. It is true that *if* x ~ y, then x ~ x, but there may be an element z where for all elements y, not (z ~ y).
Wiseacreewin on February 23rd, 2009 10:10 pm (UTC)
This is very true. ;) As the prof said today, "We don't even know if y is in A!"
Wiseacreewin on February 23rd, 2009 10:11 pm (UTC)
Also this icon is brilliant.