(b) Consider the following \"proof\" that any relation that is symmetric and tra
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(b) Consider the following "proof" that any relation that is symmetric and transitive must ~is also reflexive. Proof. Take any z E S. Because ~ is transitive, we know that for any z, y, z E S, if r ~y and y~z then r~z must also hold. In particular, if we let az, this tells us that if~y and y~x then r~r. We also know that our relation~ is symmetric; in other words, by the definition of symmetric we know that if ~y then yx Therefore, if we combine these two observations we get that for any z S that r~r. In other words, our relation is reflexive, as claimed! This proof has a flaw in its logic. Find the logical mistake and explain why it is a mistake.Explanation / Answer
If a relation is symmetric and transitive then it must be reflexive.
Let's see :
By definition of reflexive, Symmetric and Transitive we have :
A relation ~ in S is symmetric if for x, y in S, y ~ x whenever x ~ y
A relation ~ in S is transitive if for x, y and z in S, x ~ z whenever x ~ y and y ~ z
A relation ~ in S is reflexive if x ~ x for every x in S
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The flaw in logic is that, transitive relation definition is written incorrectly
for transitive relation, it is written in the question : x, y, z belongs to S , if x ~ y, y ~ z is true then x ~ z must hold
Correction is : x, y, z belongs to S, if x ~ z, whenever x ~ y and y ~ z
then relation ~ is transitive (In layman's language, if there are a, b, c objects in a room then if (a is related to c) and (a is also related to b) then (b must be also related to c))
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